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HDU5879(打表)

2024-08-13 16:48:56   218人阅读

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 }

 

HDU5879(打表)


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