思路一:所以奇数的和 - 所有偶数的和

#include <stdio.h>

int main() {
    int odd = 0, even = 0, num = 0;
    scanf("%d", &num);
    for(int i = 1; i <= num; i++) {
        i % 2 == 0 ? even += i : odd += i;
    }
    printf("%d\n", odd - even);
    
    return 0;
}

思路二:找公式的规律 发现输入奇数和偶数有各自的规律后,其实不用循环也可以算出结果:

#include <stdio.h>

int main() {
    int num = 0;
    scanf("%d", &num);
    if (num % 2 == 0) {
        printf("%d\n", - (num / 2));
    }else {
        printf("%d\n", (1 + num) / 2);
    }
    
    return 0;
}