# C 程序：檢查一個數字是否可以表示為兩個質數之和 > 原文： [https://www.programiz.com/c-programming/examples/sum-prime-numbers](https://www.programiz.com/c-programming/examples/sum-prime-numbers) #### 在此示例中，您將學習檢查用戶輸入的整數是否可以表示為所有可能組合的兩個質數之和。 要理解此示例，您應該了解以下 [C 編程](/c-programming "C tutorial")主題： * [C `if...else`語句](/c-programming/c-if-else-statement) * [C `for`循環](/c-programming/c-for-loop) * [C 函數](/c-programming/c-functions) * [C 用戶定義的函數](/c-programming/c-user-defined-functions) * * * 為了完成此任務，我們將創建一個名為`checkPrime()`的函數。 如果傳遞給函數的數字是[質數](https://www.programiz.com/c-programming/examples/prime-number)，則`checkPrime()`返回 1。 * * * ## 作為兩個質數之和的整數 ```c #include <stdio.h> int checkPrime(int n); int main() { int n, i, flag = 0; printf("Enter a positive integer: "); scanf("%d", &n); for (i = 2; i <= n / 2; ++i) { // condition for i to be a prime number if (checkPrime(i) == 1) { // condition for n-i to be a prime number if (checkPrime(n - i) == 1) { printf("%d = %d + %d\n", n, i, n - i); flag = 1; } } } if (flag == 0) printf("%d cannot be expressed as the sum of two prime numbers.", n); return 0; } // function to check prime number int checkPrime(int n) { int i, isPrime = 1; for (i = 2; i <= n / 2; ++i) { if (n % i == 0) { isPrime = 0; break; } } return isPrime; } ``` **輸出** ```c Enter a positive integer: 34 34 = 3 + 31 34 = 5 + 29 34 = 11 + 23 34 = 17 + 17 ```