题解 | 牛牛学数列3

#include <stdio.h>
int main() {
    int n = 0; 
    int i, j = 0;
    int sum = 0;
    double result = 0;
    int flag = 1;
    scanf("%d",&n);
    for (i = 1; i <= n; i++) 
    {
        sum = 0;
        flag = 1;
        for (j = 1; j <= i; j++) {
            sum += flag * (2 * j - 1);
            flag = flag * -1;
        }
        result = result + 1 * 1.0 / sum;
    }
    printf("%.3lf",result);
    return 0;
}