幸运数是最小的整数m>1，对于给定的正整数n，p N +m是质数。这里是p N 是前n个素数的乘积，即素数阶乘（或 原始人 )第n级。 例如：
null
p3 = 2 × 3 × 5 = 30p4 = 2 × 3 × 5 × 7 = 210p5 = 2 × 3 × 5 × 7 × 11 = 2310
现在，素阶乘p之间的最小差m N 第一个素数大于p N 其中（m>1）是一个素数。 例如：
Input : n = 3Output : 7Explanation : 7 must be added to the productof first n prime numbers to make the product prime. 2 x 3 x 5 = 30, need to add 7 to make it 37, which is a primeInput : n = 5Output : 23
方法 ：要找到第n个幸运数，请计算前n个素数的乘积（primorial）。让这个乘积为p，然后我们发现素数大于p，并返回找到的素数和p之间的差。
p4 + 13 = 223, where m = 13, a fortunate numberp5 + 23 = 2333, where m = 23, a fortunate numberp6 + 17 = 30047, where m = 17, a fortunate number
C++
// C++ program to find n-th Fortunate number #include <bits/stdc++.h> using namespace std; bool isPrime( int n) { // Corner cases if (n <= 1) return false ; if (n <= 3) return true ; // This is checked so that we can skip // middle five numbers in below loop if (n%2 == 0 || n%3 == 0) return false ; for ( int i=5; i*i<=n; i=i+6) if (n%i == 0 || n%(i+2) == 0) return false ; return true ; } // Function to Find primorial of order n // (product of first n prime numbers). long long int primorial( long long int n) { long long int p = 2; n--; for ( int i = 3; n != 0; i++) { if (isPrime(i)) { p = p * i; n--; } i++; } return p; } // Function to find next prime number greater // than n long long int findNextPrime( long long int n) { // Note that difference (or m) should be // greater than 1. long long int nextPrime = n + 2; // loop continuously until isPrime // returns true for a number above n while ( true ) { // Ignoring the prime number that // is 1 greater than n if (isPrime(nextPrime)) break ; nextPrime++; } return nextPrime; } // Returns n-th Fortunate number long long int fortunateNumber( int n) { long long int p = primorial(n); return findNextPrime(p) - p; } // Driver function int main() { long long int n = 5; cout << fortunateNumber(n) << "" ; return 0; } |
JAVA
// Java program to find n-th Fortunate number import java.lang.*; import java.util.*; class GFG { public static boolean isPrime( int n) { // Corner cases if (n <= 1 ) return false ; if (n <= 3 ) return true ; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0 ) return false ; for ( int i = 5 ; i * i <= n; i = i + 6 ) if (n % i == 0 || n % (i + 2 ) == 0 ) return false ; return true ; } // Function to Find primorial of order n // (product of first n prime numbers). public static int primorial( int n) { int p = 2 ; n--; for ( int i = 3 ; n != 0 ; i++) { if (isPrime(i) == true ) { p = p * i; n--; } i++; } return p; } // Function to find next prime number greater // than n public static int findNextPrime( int n) { // Note that difference (or m) should be // greater than 1. int nextPrime = n + 2 ; // loop continuously until isPrime // returns true for a number above n while ( true ) { // Ignoring the prime number that // is 1 greater than n if (isPrime(nextPrime) == true ) break ; nextPrime++; } return nextPrime; } // Returns n-th Fortunate number public static int fortunateNumber( int n) { int p = primorial(n); return findNextPrime(p)-p; } //Driver function public static void main (String[] args) { int n = 5 ; System.out.println(fortunateNumber(n)); } } /*This code is contributed by Akash Singh*/ |
Python3
# Python3 program to find # n-th Fortunate number def isPrime(n): # Corner cases if (n < = 1 ): return False if (n < = 3 ): return True # This is checked so that we can skip # middle five numbers in below loop if (n % 2 = = 0 or n % 3 = = 0 ): return False i = 5 while (i * i < = n): if (n % i = = 0 or n % (i + 2 ) = = 0 ): return False i + = 6 return True # Function to Find primorial of order n # (product of first n prime numbers). def primorial(n): p = 2 ; n - = 1 ; i = 3 while (n ! = 0 ): if (isPrime(i)): p = p * i n - = 1 i + = 1 return p # Function to find next prime # number greater than n def findNextPrime(n): # Note that difference (or m) # should be greater than 1. nextPrime = n + 2 # loop continuously until isPrime # returns true for a number above n while ( True ): # Ignoring the prime number that # is 1 greater than n if (isPrime(nextPrime)): break nextPrime + = 1 return nextPrime # Returns n-th Fortunate number def fortunateNumber(n): p = primorial(n) return findNextPrime(p) - p # Driver Code n = 5 print (fortunateNumber(n)) # This code is contributed by Anant Agarwal. |
C#
// C# program to find // n-th Fortunate number using System; class GFG { public static bool isPrime( int n) { // Corner cases if (n <= 1) return false ; if (n <= 3) return true ; // This is checked so that // we can skip middle five // numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false ; for ( int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false ; return true ; } // Function to Find primorial // of order n (product of first // n prime numbers). public static int primorial( int n) { int p = 2; n--; for ( int i = 3; n != 0; i++) { if (isPrime(i) == true ) { p = p * i; n--; } i++; } return p; } // Function to find next // prime number greater than n public static int findNextPrime( int n) { // Note that difference (or m) // should be greater than 1. int nextPrime = n + 2; // loop continuously until // isPrime returns true // for a number above n while ( true ) { // Ignoring the prime number // that is 1 greater than n if (isPrime(nextPrime) == true ) break ; nextPrime++; } return nextPrime; } // Returns n-th // Fortunate number public static int fortunateNumber( int n) { int p = primorial(n); return findNextPrime(p) - p; } // Driver Code public static void Main () { int n = 5; Console.WriteLine(fortunateNumber(n)); } } // This code is contributed // by anuj_67. |
PHP
<?php // PHP program to find n-th // Fortunate number function isPrime( $n ) { // Corner cases if ( $n <= 1) return false; if ( $n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if ( $n % 2 == 0 || $n % 3 == 0) return false; for ( $i = 5; $i * $i <= $n ; $i = $i + 6) if ( $n % $i == 0 || $n % ( $i + 2) == 0) return false; return true; } // Function to Find primorial of order n // (product of first n prime numbers). function primorial( $n ) { $p = 2; $n --; for ( $i = 3; $n != 0; $i ++) { if (isPrime( $i )) { $p = $p * $i ; $n --; } $i ++; } return $p ; } // Function to find next prime // number greater than n function findNextPrime( $n ) { // Note that difference (or m) // should be greater than 1. $nextPrime = $n + 2; // loop continuously until isPrime // returns true for a number above n while (true) { // Ignoring the prime number that // is 1 greater than n if (isPrime( $nextPrime )) break ; $nextPrime ++; } return $nextPrime ; } // Returns n-th Fortunate number function fortunateNumber( $n ) { $p = primorial( $n ); return findNextPrime( $p ) - $p ; } // Driver Code $n = 5; echo fortunateNumber( $n ) , "" ; // This code is contributed by ajit ?> |
Javascript
<script> // JavaScript program to find n-th Fortunate number function isPrime(n) { // Corner cases if (n <= 1) return false ; if (n <= 3) return true ; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false ; for (let i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false ; return true ; } // Function to Find primorial of order n // (product of first n prime numbers). function primorial(n) { let p = 2; n--; for (let i = 3; n != 0; i++) { if (isPrime(i) == true ) { p = p * i; n--; } i++; } return p; } // Function to find next prime number greater // than n function findNextPrime(n) { // Note that difference (or m) should be // greater than 1. let nextPrime = n + 2; // loop continuously until isPrime // returns true for a number above n while ( true ) { // Ignoring the prime number that // is 1 greater than n if (isPrime(nextPrime) == true ) break ; nextPrime++; } return nextPrime; } // Returns n-th Fortunate number function fortunateNumber(n) { let p = primorial(n); return findNextPrime(p)-p; } // Driver Code let n = 5; document.write(fortunateNumber(n)); </script> |
输出：
23
优化 ：上述解决方案可以使用 埃拉托斯坦筛 .
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