查找数组中的所有对（a，b），使a%b=k-yiteyi-C++库

给定一个具有不同元素的数组，任务是在数组中查找对，使a%b=k，其中k是给定的整数。

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例如：

Input  :  arr[] = {2, 3, 5, 4, 7}                 k = 3Output :  (7, 4), (3, 4), (3, 5), (3, 7)7 % 4 = 33 % 4 = 33 % 5 = 33 % 7 = 3

建议：请在“上解决” 实践 “首先，在讨论解决方案之前。

A. 天真的解决方案 就是一对一对地做，检查它们的模是否等于k。如果等于k，则打印该对。

C++

                         // C++ implementation to find such pairs                       
                         #include <bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         // Function to find pair such that (a % b = k)                       
                         bool                                     printPairs(                                     int                                     arr[],                                     int                                     n,                                     int                                     k)                       
                         {                       
                                                 bool                                     isPairFound =                                     true                                     ;                       
           
                                                 // Consider each and every pair                       
                                                 for                                     (                                     int                                     i = 0; i < n; i++) {                       
                                                 for                                     (                                     int                                     j = 0; j < n; j++) {                       
                                                 // Print if their modulo equals to k                       
                                                 if                                     (i != j && arr[i] % arr[j] == k) {                       
                                                 cout <<                                     "("                                     << arr[i] <<                                     ", "                       
                                                 << arr[j] <<                                     ")"                       
                                                 <<                                     " "                                     ;                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
                                                 }                       
                                                 }                       
           
                                                 return                                     isPairFound;                       
                         }                       
           
                         // Driver program                       
                         int                                     main()                       
                         {                       
                                                 int                                     arr[] = { 2, 3, 5, 4, 7 };                       
                                                 int                                     n =                                     sizeof                                     (arr) /                                     sizeof                                     (arr[0]);                       
                                                 int                                     k = 3;                       
           
                                                 if                                     (printPairs(arr, n, k) ==                                     false                                     )                       
                                                 cout <<                                     "No such pair exists"                                     ;                       
           
                                                 return                                     0;                       
                         }                       

JAVA

                         // Java implementation to find such pairs                       
           
                         class                                     Test {                       
                                                 // method to find pair such that (a % b = k)                       
                                                 static                                     boolean                                     printPairs(                                     int                                     arr[],                                     int                                     n,                                     int                                     k)                       
                                                 {                       
                                                 boolean                                     isPairFound =                                     true                                     ;                       
           
                                                 // Consider each and every pair                       
                                                 for                                     (                                     int                                     i =                                     0                                     ; i < n; i++) {                       
                                                 for                                     (                                     int                                     j =                                     0                                     ; j < n; j++) {                       
                                                 // Print if their modulo equals to k                       
                                                 if                                     (i != j && arr[i] % arr[j] == k) {                       
                                                 System.out.print(                                     "("                                     + arr[i] +                                     ", "                                     + arr[j] +                                     ")"                       
                                                 +                                     " "                                     );                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
                                                 }                       
                                                 }                       
           
                                                 return                                     isPairFound;                       
                                                 }                       
           
                                                 // Driver method                       
                                                 public                                     static                                     void                                     main(String args[])                       
                                                 {                       
                                                 int                                     arr[] = {                                     2                                     ,                                     3                                     ,                                     5                                     ,                                     4                                     ,                                     7                                     };                       
                                                 int                                     k =                                     3                                     ;                       
           
                                                 if                                     (printPairs(arr, arr.length, k) ==                                     false                                     )                       
                                                 System.out.println(                                     "No such pair exists"                                     );                       
                                                 }                       
                         }                       

Python3

                         # Python3 implementation to find such pairs                       
           
                         # Function to find pair such that (a % b = k)                       
                         def                                     printPairs(arr, n, k):                       
           
                                                 isPairFound                                     =                                     True                       
           
                                                 # Consider each and every pair                       
                                                 for                                     i                                     in                                     range                                     (                                     0                                     , n):                       
                                   
                                                 for                                     j                                     in                                     range                                     (                                     0                                     , n):                       
                                   
                                                 # Print if their modulo equals to k                       
                                                 if                                     (i !                                     =                                     j                                     and                                     arr[i]                                     %                                     arr[j]                                     =                                     =                                     k):                       
                                   
                                                 print                                     (                                     "("                                     , arr[i],                                     ", "                                     , arr[j],                                     ")"                                     ,                       
                                                 sep                                     =                                     "                                     ", end = "                                     ")                       
                                                 isPairFound                                     =                                     True                       
                                   
                                                 return                                     isPairFound                       
           
                         # Driver Code                       
                         arr                                     =                                     [                                     2                                     ,                                     3                                     ,                                     5                                     ,                                     4                                     ,                                     7                                     ]                       
                         n                                     =                                     len                                     (arr)                       
                         k                                     =                                     3                       
                         if                                     (printPairs(arr, n, k)                                     =                                     =                                     False                                     ):                       
                                                 print                                     (                                     "No such pair exists"                                     )                       
           
                         # This article is contributed by Smitha Dinesh Semwal.                       

C#

                         // C# implementation to find such pair                       
                         using                                     System;                       
           
                         public                                     class                                     GFG {                       
                                   
                                                 // method to find pair such that (a % b = k)                       
                                                 static                                     bool                                     printPairs(                                     int                                     [] arr,                                     int                                     n,                                     int                                     k)                       
                                                 {                       
                                                 bool                                     isPairFound =                                     true                                     ;                       
           
                                                 // Consider each and every pair                       
                                                 for                                     (                                     int                                     i = 0; i < n; i++) {                       
                                                 for                                     (                                     int                                     j = 0; j < n; j++)                       
                                                 {                       
                                                 // Print if their modulo equals to k                       
                                                 if                                     (i != j && arr[i] % arr[j] == k)                       
                                                 {                       
                                                 Console.Write(                                     "("                                     + arr[i] +                                     ", "                       
                                                 + arr[j] +                                     ")"                                     +                                     " "                                     );                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
                                                 }                       
                                                 }                       
           
                                                 return                                     isPairFound;                       
                                                 }                       
           
                                                 // Driver method                       
                                                 public                                     static                                     void                                     Main()                       
                                                 {                       
                                                 int                                     [] arr = { 2, 3, 5, 4, 7 };                       
                                                 int                                     k = 3;                       
           
                                                 if                                     (printPairs(arr, arr.Length, k) ==                                     false                                     )                       
                                                 Console.WriteLine(                                     "No such pair exists"                                     );                       
                                                 }                       
                         }                       
           
                         // This code is contributed by Sam007                       

PHP

                         <?php                       
                         // PHP implementation to                       
                         // find such pairs                       
           
                         // Function to find pair                       
                         // such that (a % b = k)                       
                         function                                     printPairs(                                     $arr                                     ,                                     $n                                     ,                                     $k                                     )                       
                         {                       
                                                 $isPairFound                                     = true;                       
           
                                                 // Consider each and every pair                       
                                                 for                                     (                                     $i                                     = 0;                                     $i                                     <                                     $n                                     ;                                     $i                                     ++)                       
                                                 {                       
                                                 for                                     (                                     $j                                     = 0;                                     $j                                     <                                     $n                                     ;                                     $j                                     ++)                       
                                                 {                       
                                                 // Print if their modulo                       
                                                 // equals to k                       
                                                 if                                     (                                     $i                                     !=                                     $j                                     &&                                     $arr                                     [                                     $i                                     ] %                       
                                                 $arr                                     [                                     $j                                     ] ==                                     $k                                     )                       
                                                 {                       
                                                 echo                                     "("                                     ,                                     $arr                                     [                                     $i                                     ] ,                                     ", "                                     ,                       
                                                 $arr                                     [                                     $j                                     ] ,                                     ")"                                     ,                                     " "                                     ;                       
                                                 $isPairFound                                     = true;                       
                                                 }                       
                                                 }                       
                                                 }                       
           
                                                 return                                     $isPairFound                                     ;                       
                         }                       
           
                         // Driver Code                       
                         $arr                                     =                                     array                                     (2, 3, 5, 4, 7);                       
                         $n                                     = sizeof(                                     $arr                                     );                       
                         $k                                     = 3;                       
           
                         if                                     (printPairs(                                     $arr                                     ,                                     $n                                     ,                                     $k                                     ) == false)                       
                                                 echo                                     "No such pair exists"                                     ;                       
           
                         // This code is contributed by ajit                       
                         ?>                       

Javascript

                         <script>                       
                                                 // Javascript implementation to find such pair                       
                                   
                                                 // method to find pair such that (a % b = k)                       
                                                 function                                     printPairs(arr, n, k)                       
                                                 {                       
                                                 let isPairFound =                                     true                                     ;                       
                                   
                                                 // Consider each and every pair                       
                                                 for                                     (let i = 0; i < n; i++) {                       
                                                 for                                     (let j = 0; j < n; j++)                       
                                                 {                       
                                                 // Print if their modulo equals to k                       
                                                 if                                     (i != j && arr[i] % arr[j] == k)                       
                                                 {                       
                                                 document.write(                                     "("                                     + arr[i] +                                     ", "                                     + arr[j] +                                     ")"                                     +                                     " "                                     );                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
                                                 }                       
                                                 }                       
                                   
                                                 return                                     isPairFound;                       
                                                 }                       
                                   
                                                 let arr = [ 2, 3, 5, 4, 7 ];                       
                                                 let k = 3;                       
           
                                                 if                                     (printPairs(arr, arr.length, k) ==                                     false                                     )                       
                                                 document.write(                                     "No such pair exists"                                     );                       
           
                         </script>                       

输出：

(3, 5) (3, 4) (3, 7) (7, 4)

时间复杂性： O（n） ^2. ) 一 有效解决方案 基于以下观察结果：

如果k本身存在于arr[]中，则k与所有元素arr[i]形成一对，其中k
对于大于或等于k的所有元素，我们使用以下事实。

   If arr[i] % arr[j] = k,    ==> arr[i] = x * arr[j] + k   ==> (arr[i] - k) = x * arr[j]  We find all divisors of (arr[i] - k)  and see if they are present in arr[].

为了快速检查数组中是否存在元素，我们使用哈希。

C++

                         // C++ program to find all pairs such that                       
                         // a % b = k.                       
                         #include <bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         // Utility function to find the divisors of                       
                         // n and store in vector v[]                       
                         vector<                                     int                                     > findDivisors(                                     int                                     n)                       
                         {                       
                                                 vector<                                     int                                     > v;                       
           
                                                 // Vector is used to store  the divisors                       
                                                 for                                     (                                     int                                     i = 1; i <=                                     sqrt                                     (n); i++) {                       
                                                 if                                     (n % i == 0) {                       
                                                 // If n is a square number, push                       
                                                 // only one occurrence                       
                                                 if                                     (n / i == i)                       
                                                 v.push_back(i);                       
                                                 else                                     {                       
                                                 v.push_back(i);                       
                                                 v.push_back(n / i);                       
                                                 }                       
                                                 }                       
                                                 }                       
                                                 return                                     v;                       
                         }                       
           
                         // Function to find pairs such that (a%b = k)                       
                         bool                                     printPairs(                                     int                                     arr[],                                     int                                     n,                                     int                                     k)                       
                         {                       
                                                 // Store all the elements in the map                       
                                                 // to use map as hash for finding elements                       
                                                 // in O(1) time.                       
                                                 unordered_map<                                     int                                     ,                                     bool                                     > occ;                       
                                                 for                                     (                                     int                                     i = 0; i < n; i++)                       
                                                 occ[arr[i]] =                                     true                                     ;                       
           
                                                 bool                                     isPairFound =                                     false                                     ;                       
                                                 for                                     (                                     int                                     i = 0; i < n; i++) {                       
                                                 // Print all the pairs with (a, b) as                       
                                                 // (k, numbers greater than k) as                       
                                                 // k % (num (> k)) = k i.e. 2%4 = 2                       
                                                 if                                     (occ[k] && k < arr[i]) {                       
                                                 cout <<                                     "("                                     << k <<                                     ", "                                     << arr[i] <<                                     ") "                                     ;                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
           
                                                 // Now check for the current element as 'a'                       
                                                 // how many b exists such that a%b = k                       
                                                 if                                     (arr[i] >= k) {                       
                                                 // find all the divisors of (arr[i]-k)                       
                                                 vector<                                     int                                     > v = findDivisors(arr[i] - k);                       
           
                                                 // Check for each divisor i.e. arr[i] % b = k                       
                                                 // or not, if yes then print that pair.                       
                                                 for                                     (                                     int                                     j = 0; j < v.size(); j++) {                       
                                                 if                                     (arr[i] % v[j] == k && arr[i] != v[j] && occ[v[j]]) {                       
                                                 cout <<                                     "("                                     << arr[i] <<                                     ", "                       
                                                 << v[j] <<                                     ") "                                     ;                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
                                                 }                       
           
                                                 // Clear vector                       
                                                 v.clear();                       
                                                 }                       
                                                 }                       
           
                                                 return                                     isPairFound;                       
                         }                       
           
                         // Driver program                       
                         int                                     main()                       
                         {                       
                                                 int                                     arr[] = { 3, 1, 2, 5, 4 };                       
                                                 int                                     n =                                     sizeof                                     (arr) /                                     sizeof                                     (arr[0]);                       
                                                 int                                     k = 2;                       
           
                                                 if                                     (printPairs(arr, n, k) ==                                     false                                     )                       
                                                 cout <<                                     "No such pair exists"                                     ;                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // Java program to find all pairs such that                       
                         // a % b = k.                       
           
                         import                                     java.util.HashMap;                       
                         import                                     java.util.Vector;                       
           
                         class                                     Test {                       
                                                 // Utility method to find the divisors of                       
                                                 // n and store in vector v[]                       
                                                 static                                     Vector<Integer> findDivisors(                                     int                                     n)                       
                                                 {                       
                                                 Vector<Integer> v =                                     new                                     Vector<>();                       
           
                                                 // Vector is used to store  the divisors                       
                                                 for                                     (                                     int                                     i =                                     1                                     ; i <= Math.sqrt(n); i++) {                       
                                                 if                                     (n % i ==                                     0                                     ) {                       
                                                 // If n is a square number, push                       
                                                 // only one occurrence                       
                                                 if                                     (n / i == i)                       
                                                 v.add(i);                       
                                                 else                                     {                       
                                                 v.add(i);                       
                                                 v.add(n / i);                       
                                                 }                       
                                                 }                       
                                                 }                       
                                                 return                                     v;                       
                                                 }                       
           
                                                 // method to find pairs such that (a%b = k)                       
                                                 static                                     boolean                                     printPairs(                                     int                                     arr[],                                     int                                     n,                                     int                                     k)                       
                                                 {                       
                                                 // Store all the elements in the map                       
                                                 // to use map as hash for finding elements                       
                                                 // in O(1) time.                       
                                                 HashMap<Integer, Boolean> occ =                                     new                                     HashMap<>();                       
                                                 for                                     (                                     int                                     i =                                     0                                     ; i < n; i++)                       
                                                 occ.put(arr[i],                                     true                                     );                       
           
                                                 boolean                                     isPairFound =                                     false                                     ;                       
                                                 for                                     (                                     int                                     i =                                     0                                     ; i < n; i++) {                       
                                                 // Print all the pairs with (a, b) as                       
                                                 // (k, numbers greater than k) as                       
                                                 // k % (num (> k)) = k i.e. 2%4 = 2                       
                                                 if                                     (occ.get(k) && k < arr[i]) {                       
                                                 System.out.print(                                     "("                                     + k +                                     ", "                                     + arr[i] +                                     ") "                                     );                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
           
                                                 // Now check for the current element as 'a'                       
                                                 // how many b exists such that a%b = k                       
                                                 if                                     (arr[i] >= k) {                       
                                                 // find all the divisors of (arr[i]-k)                       
                                                 Vector<Integer> v = findDivisors(arr[i] - k);                       
           
                                                 // Check for each divisor i.e. arr[i] % b = k                       
                                                 // or not, if yes then print that pair.                       
                                                 for                                     (                                     int                                     j =                                     0                                     ; j < v.size(); j++) {                       
                                                 if                                     (arr[i] % v.get(j) == k && arr[i] != v.get(j) && occ.get(v.get(j))) {                       
                                                 System.out.print(                                     "("                                     + arr[i] +                                     ", "                       
                                                 + v.get(j) +                                     ") "                                     );                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
                                                 }                       
           
                                                 // Clear vector                       
                                                 v.clear();                       
                                                 }                       
                                                 }                       
           
                                                 return                                     isPairFound;                       
                                                 }                       
           
                                                 // Driver method                       
                                                 public                                     static                                     void                                     main(String args[])                       
                                                 {                       
                                                 int                                     arr[] = {                                     3                                     ,                                     1                                     ,                                     2                                     ,                                     5                                     ,                                     4                                     };                       
                                                 int                                     k =                                     2                                     ;                       
           
                                                 if                                     (printPairs(arr, arr.length, k) ==                                     false                                     )                       
                                                 System.out.println(                                     "No such pair exists"                                     );                       
                                                 }                       
                         }                       

Python3

                         # Python3 program to find all pairs                       
                         # such that a % b = k.                       
                                   
                         # Utility function to find the divisors                       
                         # of n and store in vector v[]                       
                         import                                     math as mt                       
           
                         def                                     findDivisors(n):                       
           
                                                 v                                     =                                     []                       
           
                                                 # Vector is used to store the divisors                       
                                                 for                                     i                                     in                                     range                                     (                                     1                                     , mt.floor(n                                     *                                     *                                     (.                                     5                                     ))                                     +                                     1                                     ):                       
                                                 if                                     (n                                     %                                     i                                     =                                     =                                     0                                     ):                       
                                   
                                                 # If n is a square number, push                       
                                                 # only one occurrence                       
                                                 if                                     (n                                     /                                     i                                     =                                     =                                     i):                       
                                                 v.append(i)                       
                                                 else                                     :                       
                                                 v.append(i)                       
                                                 v.append(n                                     /                                     /                                     i)                       
                                   
                                                 return                                     v                       
           
                         # Function to find pairs such that (a%b = k)                       
                         def                                     printPairs(arr, n, k):                       
           
                                                 # Store all the elements in the map                       
                                                 # to use map as hash for finding elements                       
                                                 # in O(1) time.                       
                                                 occ                                     =                                     dict                                     ()                       
                                                 for                                     i                                     in                                     range                                     (n):                       
                                                 occ[arr[i]]                                     =                                     True                       
           
                                                 isPairFound                                     =                                     False                       
           
                                                 for                                     i                                     in                                     range                                     (n):                       
                                   
                                                 # Print all the pairs with (a, b) as                       
                                                 # (k, numbers greater than k) as                       
                                                 # k % (num (> k)) = k i.e. 2%4 = 2                       
                                                 if                                     (occ[k]                                     and                                     k < arr[i]):                       
                                                 print                                     (                                     "("                                     , k,                                     ","                                     , arr[i],                                     ")"                                     , end                                     =                                     " "                                     )                       
                                                 isPairFound                                     =                                     True                       
           
                                                 # Now check for the current element as 'a'                       
                                                 # how many b exists such that a%b = k                       
                                                 if                                     (arr[i] >                                     =                                     k):                       
                                   
                                                 # find all the divisors of (arr[i]-k)                       
                                                 v                                     =                                     findDivisors(arr[i]                                     -                                     k)                       
           
                                                 # Check for each divisor i.e. arr[i] % b = k                       
                                                 # or not, if yes then print that pair.                       
                                                 for                                     j                                     in                                     range                                     (                                     len                                     (v)):                       
                                                 if                                     (arr[i]                                     %                                     v[j]                                     =                                     =                                     k                                     and                       
                                                 arr[i] !                                     =                                     v[j]                                     and                       
                                                 occ[v[j]]):                       
                                                 print                                     (                                     "("                                     , arr[i],                                     ","                                     , v[j],                       
                                                 ")"                                     , end                                     =                                     " "                                     )                       
                                                 isPairFound                                     =                                     True                       
           
                                                 return                                     isPairFound                       
           
                         # Driver Code                       
                         arr                                     =                                     [                                     3                                     ,                                     1                                     ,                                     2                                     ,                                     5                                     ,                                     4                                     ]                       
                         n                                     =                                     len                                     (arr)                       
                         k                                     =                                     2                       
           
                         if                                     (printPairs(arr, n, k)                                     =                                     =                                     False                                     ):                       
                                                 print                                     (                                     "No such pair exists"                                     )                       
           
                         # This code is contributed by mohit kumar                       

C#

                         // C# program to find all pairs                       
                         // such that a % b = k.                       
                         using                                     System;                       
                         using                                     System.Collections.Generic;                       
           
                         class                                     GFG                       
                         {                       
                         // Utility method to find the divisors                       
                         // of n and store in vector v[]                       
                         public                                     static                                     List<                                     int                                     > findDivisors(                                     int                                     n)                       
                         {                       
                                                 List<                                     int                                     > v =                                     new                                     List<                                     int                                     >();                       
           
                                                 // Vector is used to store                       
                                                 // the divisors                       
                                                 for                                     (                                     int                                     i = 1;                       
                                                 i <= Math.Sqrt(n); i++)                       
                                                 {                       
                                                 if                                     (n % i == 0)                       
                                                 {                       
                                                 // If n is a square number,                       
                                                 // push only one occurrence                       
                                                 if                                     (n / i == i)                       
                                                 {                       
                                                 v.Add(i);                       
                                                 }                       
                                                 else                       
                                                 {                       
                                                 v.Add(i);                       
                                                 v.Add(n / i);                       
                                                 }                       
                                                 }                       
                                                 }                       
                                                 return                                     v;                       
                         }                       
           
                         // method to find pairs such                       
                         // that (a%b = k)                       
                         public                                     static                                     bool                                     printPairs(                                     int                                     [] arr,                       
                                                 int                                     n,                                     int                                     k)                       
                         {                       
                                                 // Store all the elements in the                       
                                                 // map to use map as hash for                       
                                                 // finding elements in O(1) time.                       
                                                 Dictionary<                                     int                                     ,                       
                                                 bool                                     > occ =                                     new                                     Dictionary<                                     int                                     ,                       
                                                 bool                                     >();                       
                                                 for                                     (                                     int                                     i = 0; i < n; i++)                       
                                                 {                       
                                                 occ[arr[i]] =                                     true                                     ;                       
                                                 }                       
           
                                                 bool                                     isPairFound =                                     false                                     ;                       
                                                 for                                     (                                     int                                     i = 0; i < n; i++)                       
                                                 {                       
                                                 // Print all the pairs with (a, b) as                       
                                                 // (k, numbers greater than k) as                       
                                                 // k % (num (> k)) = k i.e. 2%4 = 2                       
                                                 if                                     (occ[k] && k < arr[i])                       
                                                 {                       
                                                 Console.Write(                                     "("                                     + k +                                     ", "                                     +                       
                                                 arr[i] +                                     ") "                                     );                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
           
                                                 // Now check for the current element                       
                                                 // as 'a' how many b exists such that                       
                                                 // a%b = k                       
                                                 if                                     (arr[i] >= k)                       
                                                 {                       
                                                 // find all the divisors of (arr[i]-k)                       
                                                 List<                                     int                                     > v = findDivisors(arr[i] - k);                       
           
                                                 // Check for each divisor i.e.                       
                                                 // arr[i] % b = k or not, if                       
                                                 // yes then print that pair.                       
                                                 for                                     (                                     int                                     j = 0; j < v.Count; j++)                       
                                                 {                       
                                                 if                                     (arr[i] % v[j] == k &&                       
                                                 arr[i] != v[j] && occ[v[j]])                       
                                                 {                       
                                                 Console.Write(                                     "("                                     + arr[i] +                       
                                                 ", "                                     + v[j] +                                     ") "                                     );                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
                                                 }                       
           
                                                 // Clear vector                       
                                                 v.Clear();                       
                                                 }                       
                                                 }                       
           
                                                 return                                     isPairFound;                       
                         }                       
           
                         // Driver Code                       
                         public                                     static                                     void                                     Main(                                     string                                     [] args)                       
                         {                       
                                                 int                                     [] arr =                                     new                                     int                                     [] {3, 1, 2, 5, 4};                       
                                                 int                                     k = 2;                       
           
                                                 if                                     (printPairs(arr, arr.Length, k) ==                                     false                                     )                       
                                                 {                       
                                                 Console.WriteLine(                                     "No such pair exists"                                     );                       
                                                 }                       
                         }                       
                         }                       
           
                         // This code is contributed by Shrikant13                       

Javascript

                         <script>                       
           
                         // JavaScript program to find all pairs such that                       
                         // a % b = k.                       
                                   
                                                 // Utility method to find the divisors of                       
                                                 // n and store in vector v[]                       
                                                 function                                     findDivisors(n)                       
                                                 {                       
                                                 let v = [];                       
                                   
                                                 // Vector is used to store  the divisors                       
                                                 for                                     (let i = 1; i <= Math.sqrt(n); i++)                       
                                                 {                       
                                                 if                                     (n % i == 0) {                       
                                                 // If n is a square number, push                       
                                                 // only one occurrence                       
                                                 if                                     (n / i == i)                       
                                                 v.push(i);                       
                                                 else                                     {                       
                                                 v.push(i);                       
                                                 v.push(Math.floor(n / i));                       
                                                 }                       
                                                 }                       
                                                 }                       
                                                 return                                     v;                       
                                                 }                       
                                   
                                                 // method to find pairs such that (a%b = k)                       
                                                 function                                     printPairs(arr,n,k)                       
                                                 {                       
                                                 // Store all the elements in the map                       
                                                 // to use map as hash for finding elements                       
                                                 // in O(1) time.                       
                                                 let occ =                                     new                                     Map();                       
                                                 for                                     (let i = 0; i < n; i++)                       
                                                 occ.set(arr[i],                                     true                                     );                       
                                   
                                                 let isPairFound =                                     false                                     ;                       
                                                 for                                     (let i = 0; i < n; i++) {                       
                                                 // Print all the pairs with (a, b) as                       
                                                 // (k, numbers greater than k) as                       
                                                 // k % (num (> k)) = k i.e. 2%4 = 2                       
                                                 if                                     (occ.get(k) && k < arr[i]) {                       
                                                 document.write(                                     "("                                     + k +                                     ", "                                     +                       
                                                 arr[i] +                                     ") "                                     );                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
                                   
                                                 // Now check for the current element as 'a'                       
                                                 // how many b exists such that a%b = k                       
                                                 if                                     (arr[i] >= k) {                       
                                                 // find all the divisors of (arr[i]-k)                       
                                                 let v = findDivisors(arr[i] - k);                       
                                   
                                                 // Check for each divisor                       
                                                 // i.e. arr[i] % b = k                       
                                                 // or not, if yes then                       
                                                 // print that pair.                       
                                                 for                                     (let j = 0; j < v.length; j++)                       
                                                 {                       
                                                 if                                     (arr[i] % v[j] == k &&                       
                                                 arr[i] != v[j] &&                       
                                                 occ.get(v[j]))                       
                                                 {                       
                                                 document.write(                                     "("                                     + arr[i] +                                     ", "                       
                                                 + v[j] +                                     ") "                                     );                       
                                                 isPairFound =                                     true                                     ;                       
                                                 }                       
                                                 }                       
                                   
                                                 // Clear vector                       
                                                 v=[];                       
                                                 }                       
                                                 }                       
                                   
                                                 return                                     isPairFound;                       
                                                 }                       
                                   
                                                 // Driver method                       
                                                 let arr=[3, 1, 2, 5, 4 ];                       
                                                 let k = 2;                       
                                                 if                                     (printPairs(arr, arr.length, k) ==                                     false                                     )                       
                                                 document.write(                                     "No such pair exists"                                     );                       
                                   
           
                         // This code is contributed by unknown2108                       
           
                         </script>                       

输出：

(2, 3) (2, 5) (5, 3) (2, 4)

时间复杂性： O（n*sqrt（max））其中max是数组中的最大元素。 参考： https://stackoverflow.com/questions/12732939/find-pairs-in-an-array-such-that-ab-k-where-k-is-a-given-integer

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