hdoj5879Cure

Cure

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 229    Accepted Submission(s): 81


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

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
using namespace std;
typedef long long LL;
typedef pair<int,int>pii;
const int maxn=1000010;
double ans[maxn];
char s[1000010];
int main()
{
for(LL i=1;i<=1000000;++i){
ans[i]=ans[i-1]+1.0/(i*i);
}
while(scanf("%s",s)!=EOF){
int len=strlen(s);
if(len>=7){
printf("%.5lf\n",ans[1000000]);
}
else {
int c=0;
for(int i=0;i<len;++i){
c=c*10+(s[i]-'0');
}
printf("%.5lf\n",ans[c]);
}
}
return 0;
}