HDU-5879 Cure（精度）（极限）

Cure

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7400    Accepted Submission(s): 1099

Problem Description

Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.

 

Input

There are multiple cases.
For each test case, there is a single line, containing a single positive integer n. 
The input file is at most 1M.

 

Output

The required sum, rounded to the fifth digits after the decimal point.

 

Sample Input

124815

 

Sample Output

1.000001.250001.423611.527421.58044

首先要知道∑1/(i*i)有极限，所以计算到一定数值以后结果会不再变化（小数点后后5位），然后找到边界1e6。

注意1M的输入。and atoi()

//下面盗用队友代码

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>using namespace std;char str[10000005];int main(){    while(~scanf("%s",str))    {        int len = strlen(str);        double tosum = 0;        if(len <= 6)        {            if(len < 6 || str[0] == '1')            {                int n = atoi(str);                for(int i = 1;i <= n;i++)                {                    double tmp = i;                    tosum += 1/(tmp*tmp);                }                printf("%.5f\n",tosum);            }            else                printf("1.64493\n");         }         else if(len > 6)                printf("1.64493\n");    }    return 0;}