2016 ACM/ICPC Asia Regional Qingdao Online 1002 Cure

Posted 2020-08-09 Annetree的博客

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

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstdlib>
 5 #include<cstring>
 6 using namespace std;
 7 double a[1000005];
 8 char ch[1000005];
 9 int n;
10 int main()
11 {
12 
13     for(int i=1000000;i>=1;i--)
14      a[i]=a[i+1]+1.0/pow((double)i,2.0);
15 
16     while(~scanf("%s",&ch))
17     {
18         int len=strlen(ch);
19         int n=0;
20         for(int i=0;i<len;i++) {n=n*10+ch[i]-‘0‘; if (n>1000000) break;}
21 
22         if(n>=1000000) printf("%.5lf\n",a[1]);
23           else printf("%.5lf\n",a[1]-a[n+1]);
24     }
25     return 0;
26 }