hdu5879 Cure（水）

思路：打表可知n>=1e6的时候会收敛到一个值，那么直接打表就好了

坑点：没说n多大，只给了输入文件大小，所以n可能很大很大，直接用字符串就好了

#include<bits/stdc++.h>using namespace std;const int maxn = 1e6+7;double sum[maxn];string n;void init(){sum[0]=0;    for(int i = 1;i<=1e6;i++)      sum[i]=sum[i-1]+double(1.0/i/i);}int main(){init();while(cin >> n){        if(n.size()>=7)printf("%.5lf\n",sum[1000000]);else{int len = n.size();int a = 0;for(int i = 0;i<len;i++)a = a*10+n[i]-'0';printf("%.5lf\n",sum[a]);}}}

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