为了完成此任务,我们将创建一个名为checkPrime()
的函数 。
如果传递给函数的数字是质数,则checkPrime()
返回1。
整数作为两个质数之和
#include
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;
}
输出
Enter a positive integer: 34
34 = 3 + 31
34 = 5 + 29
34 = 11 + 23
34 = 17 + 17