2016 ACM/ICPC Asia Regional Qingdao Online 1002 Cure

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

 

n范围很大，需要用字符串读入

 #include <iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstdlib>
 #include<cstring>
 using namespace std;
 double a[];
 char ch[];
 int n;
 int main()
 {

     for(int i=;i>=;i--)
      a[i]=a[i+]+1.0/pow((double)i,2.0);

     while(~scanf("%s",&ch))
     {
         int len=strlen(ch);
         int n=;
         for(int i=;i<len;i++) {n=n*+ch[i]-''; if (n>) break;}

         if(n>=) printf("%.5lf\n",a[]);
           else printf("%.5lf\n",a[]-a[n+]);
     }
     return ;
 }