最小的x，使得1*n，2*n，…x*n有1到9的所有数字-yiteyi-C++库

给定一个正数n，我们需要找到x，使得1*n，2*n，3*n…。。x*n至少给出10位数字一次。如果没有这样的x是可能的打印-1。 例如：

null

Input : n = 1692Output : 3Explanation:n = 1692, we got the digits- 1, 2, 6, 92*n = 3384, we got the digits- 1, 2, 3, 4, 6, 8, 9.3*n = 5076, we got the digits- 1, 2, 3, 4, 5, 6, 7, 8, 9.At this step we got all the digits at leastonce. Therefore our answer is 3.Input  : 1Output : 10Input  : 0Output :-1

推荐：请尝试你的方法 {IDE} 首先，在进入解决方案之前。

这里使用的想法很简单。我们从1开始，一直与n相乘，直到我们至少一次没有得到所有的10位数。为了跟踪每次迭代中出现的所有数字，我们使用了一个临时数组，大小为10，初始值为0。每当我们第一次得到一个数字时，我们将用1初始化它在数组中的索引。当所有数字访问一次时，我们就完成了。下面是它的实现。

C++

                         // CPP program to find x such that 1*n, 2*n, 3*n                       
                         // ...x * n have all digits from 1 to 9 at least                       
                         // once                       
                         #include <bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         // Returns smallest value x such that 1*n, 2*n,                       
                         // 3*n ...x * n have all digits from 1 to 9 at                       
                         // least once                       
                         int                                     smallestX(                                     int                                     n)                       
                         {                       
                                                 // taking temporary array and variable.                       
                                                 int                                     temp[10] = { 0 };                       
           
                                                 if                                     (n == 0)                       
                                                 return                                     -1;                       
           
                                                 // iterate till we get all the 10 digits                       
                                                 // at least once                       
                                                 int                                     count = 0, x = 0;                       
                                                 for                                     (x = 1; count < 10; x++) {                       
                                                 int                                     y = x * n;                       
                                   
                                                 // checking all the digits                       
                                                 while                                     (y) {                       
                                                 if                                     (temp[y % 10] ==                                     false                                     ) {                       
                                                 count++;                       
                                                 temp[y % 10] =                                     true                                     ;                       
                                                 }                       
                                                 y /= 10;                       
                                                 }                       
                                                 }                       
                                                 return                                     x - 1;                       
                         }                       
           
                         // driver function                       
                         int                                     main()                       
                         {                       
                                                 int                                     n = 5;                       
                                                 cout <<smallestX(n);                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // Java program to find x such                       
                         // that 1*n, 2*n, 3*n...x * n                       
                         // have all digits from 1 to 9                       
                         // at least once                       
                         import                                     java.io.*;                       
                         import                                     java.util.*;                       
           
                         class                                     GFG                       
                         {                       
                                   
                         // Returns smallest value x                       
                         // such that 1*n, 2*n, 3*n                       
                         // ...x * n have all digits                       
                         // from 1 to 9 at least once                       
                         public                                     static                                     int                                     smallestX(                                     int                                     n)                       
                         {                       
                                                 // taking temporary                       
                                                 // array and variable.                       
                                                 int                                     [] temp =                                     new                                     int                                     [                                     10                                     ];                       
                                                 for                                     (                                     int                                     i =                                     0                                     ; i <                                     10                                     ; i++)                       
                                                 temp[i] =                                     0                                     ;                       
           
                                                 if                                     (n ==                                     0                                     )                       
                                                 return                                     -                                     1                                     ;                       
           
                                                 // iterate till we get                       
                                                 // all the 10 digits                       
                                                 // at least once                       
                                                 int                                     count =                                     0                                     , x =                                     0                                     ;                       
                                                 for                                     (x =                                     1                                     ; count <                                     10                                     ; x++)                       
                                                 {                       
                                                 int                                     y = x * n;                       
                                   
                                                 // checking all                       
                                                 // the digits                       
                                                 while                                     (y >                                     0                                     )                       
                                                 {                       
                                                 if                                     (temp[y %                                     10                                     ] ==                                     0                                     )                       
                                                 {                       
                                                 count++;                       
                                                 temp[y %                                     10                                     ] =                                     1                                     ;                       
                                                 }                       
                                                 y /=                                     10                                     ;                       
                                                 }                       
                                                 }                       
                                                 return                                     x -                                     1                                     ;                       
                         }                       
           
                         // Driver Code                       
                         public                                     static                                     void                                     main(String args[])                       
                         {                       
                                                 int                                     n =                                     5                                     ;                       
                                                 System.out.print(smallestX(n));                       
                         }                       
                         }                       
           
                         // This code is contributed                       
                         // by Akanksha Rai(Abby_akku)                       

Python3

                         # Python3 program to find x such                       
                         # that 1*n, 2*n, 3*n ...x * n                       
                         # have all digits from 1 to 9                       
                         # at least once                       
           
                         # Returns smallest value x such                       
                         # that 1*n, 2*n, 3*n ...x * n                       
                         # have all digits from 1 to 9                       
                         # at least once                       
                         def                                     smallestX(n):                       
                                                 # taking temporary                       
                                                 # array and variable.                       
                                                 temp                                     =                                     [                                     0                                     ]                                     *                                     10                       
           
                                                 if                                     (n                                     =                                     =                                     0                                     ):                       
                                                 return                                     -                                     1                       
           
                                                 # iterate till we get                       
                                                 # all the 10 digits                       
                                                 # at least once                       
                                                 count                                     =                                     0                       
                                                 x                                     =                                     1                       
                                                 while                                     (count <                                     10                                     ):                       
                                                 y                                     =                                     x                                     *                                     n                       
                                   
                                                 # checking all                       
                                                 # the digits                       
                                                 while                                     (y>                                     0                                     ):                       
                                                 if                                     (temp[y                                     %                                     10                                     ]                                     =                                     =                                     0                                     ):                       
                                                 count                                     +                                     =                                     1                       
                                                 temp[y                                     %                                     10                                     ]                                     =                                     1                       
                                                 y                                     =                                     int                                     (y                                     /                                     10                                     )                       
                                                 x                                     +                                     =                                     1                       
           
                                                 return                                     x                                     -                                     1                       
           
           
                         # Driver code                       
                         if                                     __name__                                     =                                     =                                     '__main__'                                     :                       
                                                 n                                     =                                     5                       
                                                 print                                     (smallestX(n))                       
           
                         # This code is contributed                       
                         # by mits                       

C#

                         // C# program to find x such                       
                         // that 1*n, 2*n, 3*n...x * n                       
                         // have all digits from 1 to 9                       
                         // at least once                       
                         using                                     System;                       
           
                         class                                     GFG                       
                         {                       
                                   
                         // Returns smallest value x                       
                         // such that 1*n, 2*n, 3*n                       
                         // ...x * n have all digits                       
                         // from 1 to 9 at least once                       
                         public                                     static                                     int                                     smallestX(                                     int                                     n)                       
                         {                       
                                                 // taking temporary                       
                                                 // array and variable.                       
                                                 int                                     [] temp =                                     new                                     int                                     [10];                       
                                                 for                                     (                                     int                                     i = 0; i < 10; i++)                       
                                                 temp[i] = 0;                       
           
                                                 if                                     (n == 0)                       
                                                 return                                     -1;                       
           
                                                 // iterate till we get                       
                                                 // all the 10 digits                       
                                                 // at least once                       
                                                 int                                     count = 0, x = 0;                       
                                                 for                                     (x = 1; count < 10; x++)                       
                                                 {                       
                                                 int                                     y = x * n;                       
                                   
                                                 // checking all the digits                       
                                                 while                                     (y > 0)                       
                                                 {                       
                                                 if                                     (temp[y % 10] == 0)                       
                                                 {                       
                                                 count++;                       
                                                 temp[y % 10] = 1;                       
                                                 }                       
                                                 y /= 10;                       
                                                 }                       
                                                 }                       
                                                 return                                     x - 1;                       
                         }                       
           
                         // dDriver Code                       
                         static                                     void                                     Main()                       
                         {                       
                                                 int                                     n = 5;                       
                                                 Console.Write(smallestX(n));                       
                         }                       
                         }                       
           
                         // This code is contributed by mits                       

PHP

                         <?php                       
                         // PHP program to find x such                       
                         // that 1*n, 2*n, 3*n ...x * n                       
                         // have all digits from 1 to 9                       
                         // at least once                       
           
                         // Returns smallest value x such                       
                         // that 1*n, 2*n, 3*n ...x * n                       
                         // have all digits from 1 to 9                       
                         // at least once                       
                         function                                     smallestX(                                     $n                                     )                       
                         {                       
                                                 // taking temporary                       
                                                 // array and variable.                       
                                                 $temp                                     =                                     array_fill                                     (0, 10, false);                       
           
                                                 if                                     (                                     $n                                     == 0)                       
                                                 return                                     -1;                       
           
                                                 // iterate till we get                       
                                                 // all the 10 digits                       
                                                 // at least once                       
                                                 $count                                     = 0;                       
                                                 $x                                     = 0;                       
                                                 for                                     (                                     $x                                     = 1;                                     $count                                     < 10;                                     $x                                     ++)                       
                                                 {                       
                                                 $y                                     =                                     $x                                     *                                     $n                                     ;                       
                                   
                                                 // checking all                       
                                                 // the digits                       
                                                 while                                     (                                     $y                                     )                       
                                                 {                       
                                                 if                                     (                                     $temp                                     [                                     $y                                     % 10] == false)                       
                                                 {                       
                                                 $count                                     ++;                       
                                                 $temp                                     [                                     $y                                     % 10] = true;                       
                                                 }                       
                                                 $y                                     = (int)(                                     $y                                     / 10);                       
                                                 }                       
                                                 }                       
                                                 return                                     $x                                     - 1;                       
                         }                       
           
                         // Driver code                       
                         $n                                     = 5;                       
                         echo                                     smallestX(                                     $n                                     );                       
           
                         // This code is contributed                       
                         // by mits                       
                         ?>                       

Javascript

                         <script>                       
           
                         // Javascript program to find x such                       
                         // that 1*n, 2*n, 3*n...x * n                       
                         // have all digits from 1 to 9                       
                         // at least once                       
                                   
                         // Returns smallest value x                       
                         // such that 1*n, 2*n, 3*n                       
                         // ...x * n have all digits                       
                         // from 1 to 9 at least once                       
                         function                                     smallestX(n)                       
                         {                       
                                                 // taking temporary                       
                                                 // array and variable.                       
                                                 let temp = Array.from({length: 10},                       
                                                 (_, i) => 0);                       
                                                 for                                     (let i = 0; i < 10; i++)                       
                                                 temp[i] = 0;                       
                                   
                                                 if                                     (n == 0)                       
                                                 return                                     -1;                       
                                   
                                                 // iterate till we get                       
                                                 // all the 10 digits                       
                                                 // at least once                       
                                                 let count = 0, x = 0;                       
                                                 for                                     (x = 1; count < 10; x++)                       
                                                 {                       
                                                 let y = x * n;                       
                                   
                                                 // checking all                       
                                                 // the digits                       
                                                 while                                     (y > 0)                       
                                                 {                       
                                                 if                                     (temp[y % 10] == 0)                       
                                                 {                       
                                                 count++;                       
                                                 temp[y % 10] = 1;                       
                                                 }                       
                                                 y /= 10;                       
                                                 }                       
                                                 }                       
                                                 return                                     x - 1;                       
                         }                       
           
                         // driver program                       
           
                                                 let n = 5;                       
                                                 document.write(smallestX(n));                       
                                   
                         </script>                       

输出：

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