题解 | #牛牛学数列3#

#include <stdio.h>
int main() {
    int n;
    scanf("%d", &n);
    int k = -1;//-1的n次方作循环变量
    double sum = 0.0;
    int s = 0;
    double t = 0;
    for (int j = 1; j <= 2 * n - 1 ; j = j + 2)  {
        k = (-1) * k;
        s = s + j * k;
        t = 1.0 / s;
        sum = sum + t;
    }
    printf("%.3lf", sum);
    return 0;
}