题解 | #牛牛学数列3#
牛牛学数列3
https://www.nowcoder.com/practice/f65c726d081c4160a9356eabf0dc21d9
#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;
}
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