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

124815

 

Sample Output

1.000001.250001.423611.527421.58044

 

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;}