HDU5879(打表)

Posted 2020-08-09 Penn000

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。

 1 //2016.9.17
 2 #include <iostream>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 
 7 double sum[54000];
 8 
 9 int main()
10 {
11     int n;
12     double ans;
13     ans = 0;
14     sum[0] = 0;
15     for(int i = 1; i <= 53000; i++)
16     {
17         ans += (1.0/i)*(1.0/i);    
18         sum[i] = ans;
19     }
20     string s;
21     while(cin>>s)
22     {
23         int len = s.length();
24         n = 0;
25         for(int i = 0; i < len; i++)
26         {
27             n = n*10+s[i]-‘0‘;
28             if(n > 120000)break;
29         }
30         if(n >= 110291)ans = 1.64493;
31         else if(n >= 52447)ans = 1.64492;
32         else ans = sum[n];
33         printf("%.5lf\n", ans);
34     }
35 
36     return 0;
37 }