题解 | #牛牛学数列3#

#include <stdio.h>
//输入一个整数n,计算 1+1/（1-3）+1/（1-3+5）+...+1/(1-3+5-...((-1)^(n-1))*(2n-1))的值
//1  1-1/3  1-1/3+1/5...1-1/3+1/5-...
int main() {
    int n = 0;
    scanf("%d", &n);
    int tmp = 0;
    double sum = 0;
    int flag=1;
    for (int i = 1; i <= n; i++) {
        tmp += flag * (2 * i - 1);
        sum += 1.0 / tmp;
        flag = -flag;
    }
    printf("%.3lf\n", sum);
    return 0;
}