题解 | #牛牛学数列3#

#include <stdio.h>
#include <math.h>
int main() {
    int n;
    scanf("%d",&n);
    long double sum = 0.0;
    for(int i = 1;i <= n;i++){
        sum = sum + (1.0 / (pow(-1,i-1) * i)); 
        // (1-3+5-...((-1)^(n-1))*(2n-1)) 可以转变为 -1的n-1次方乘以n;
    }
    printf("%.3Lf",sum);
}