Cure(1+12+1419+116+....+1n*n)

Cure

Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 8 Accepted Submission(s) : 2 Font: Times New Roman | Verdana | Georgia Font Size: ← → 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

笔记：错了好几次，坑点1：输入一定要用字符串； 坑点2：将每个答案先存储起来； 坑点3：大概当n=150000时，后面的数约等于0。

正确代码：

#include<iostream> #include<stdio.h> #include<cmath> using namespace std; long long maxn=1500008; double num[1500208]; void init(){ num[1]=1; for(long long i=2;i<=maxn;i++){ num[i]=num[i-1]+((1.0/i)*(1.0/i)); } } int main(){ init(); string s; while(cin>>s){ int len=s.length(); int f=0; long long n=0; for(int i=0;i<len;i++){ n=n*10+(s[i]-'0'); if(n>150000){ f=1; break; } } if(f){ printf("%.5f\n",num[150000]); } else{ printf("%.5f\n",num[n]); } } return 0; }