zzuli OJ 1040: 数列求和1

Description

Input

Output

Sample Input

Sample Output

HINT

Source

#include<stdio.h>

int main()
{
int i, n;
double sum, deno;

scanf("%d", &n);
sum = 0.0;
deno = 1.0; //初始值为第一项的分母

for(i = 1; i <= n; i++) //循环n次
{
sum += 1.0 / deno;  //计算并累加当前项
deno += 2;        //计算下一项分母
}

printf("%.2f\n", sum);
return 0;
}