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