HDU5879(打表)

Cure

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 293    Accepted Submission(s): 96

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

1

2

4

8

15

 

Sample Output

1.00000

1.25000

1.42361

1.52742

1.58044

 

Source

2016 ACM/ICPC Asia Regional Qingdao Online

 

n没有给出范围，意思就是默认无限大。。。。。比赛时被坑了，不停RE。

 //2016.9.17
 #include <iostream>
 #include <cstdio>

 using namespace std;

 double sum[];

 int main()
 {
     int n;
     double ans;
     ans = ;
     sum[] = ;
     for(int i = ; i <= ; i++)
     {
         ans += (1.0/i)*(1.0/i);
         sum[i] = ans;
     }
     string s;
     while(cin>>s)
     {
         int len = s.length();
         n = ;
         for(int i = ; i < len; i++)
         {
             n = n*+s[i]-'';
             if(n > )break;
         }
         if(n >= )ans = 1.64493;
         else if(n >= )ans = 1.64492;
         else ans = sum[n];
         printf("%.5lf\n", ans);
     }

     return ;
 }