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;
}