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[54000];

int main()
{
    int n;
    double ans;
    ans = 0;
    sum[0] = 0;
    for(int i = 1; i <= 53000; i++)
    {
        ans += (1.0/i)*(1.0/i);    
        sum[i] = ans;
    }
    string s;
    while(cin>>s)
    {
        int len = s.length();
        n = 0;
        for(int i = 0; i < len; i++)
        {
            n = n*10+s[i]-'0';
            if(n > 120000)break;
        }
        if(n >= 110291)ans = 1.64493;
        else if(n >= 52447)ans = 1.64492;
        else ans = sum[n];
        printf("%.5lf
", ans);
    }

    return 0;
}