求100-200之间的所有素数之和

＃include “stdio.h”
main()
{
int n, i, s＝0;
}

#include <stdio.h>

int main() {
    int sum = 0;
    for (int i = 100; i <= 200; i++) {
        int is_prime = 1;
        for (int j = 2; j*j <= i; j++) {
            if (i % j == 0) {
                is_prime = 0;
                break;
            }
        }
        if (is_prime) {
            sum += i;
        }
    }
    printf("%d", sum);
    return 0;
}


#include <stdio.h>

int is_prime(int num) {
    if (num < 2) {
        return 0;
    }
    for (int i = 2; i * i <= num; i++) {
        if (num % i == 0) {
            return 0;
        }
    }
    return 1;
}

int main() {
    int sum = 0;
    for (int i = 100; i <= 200; i++) {
        if (is_prime(i)) {
            sum += i;
        }
    }
    printf("100到200之间素数之和为：%d\n", sum);
    return 0;
}

计算100到200之间所有素数的和：

#include <stdio.h>

int main() {
    int n, i, s = 0;

    for (n = 100; n <= 200; n++) {
        int isPrime = 1; // Assume n is prime

        // Check if n is divisible by any number from 2 to n-1
        for (i = 2; i < n; i++) {
            if (n % i == 0) {
                isPrime = 0; // n is not prime
                break;
            }
        }

        if (isPrime) {
            s += n; // Add prime number to the sum
        }
    }

    printf("和=:%d\n", s);

    return 0;
}