题解 | #牛牛学数列3#

#include <stdio.h>

int main() {
    int n;
    double s = 0.0;   //储存总数
    int denominator = 0;  //表示分母
    int sign = 1;         //表示变化的符号
    scanf("%d",&n);       
    for (int i = 1; i <=n; i++) {
       denominator = denominator+sign * (2*i-1);//0+1*（2*1-1）=1
       s = s+1.0/denominator;    //s = 0 + 1/1 = 1;
       sign = sign*(-1);
    }
    printf("%.3f\n",s);
    return 0;
}