HDU 5879-Cure（1/n^2之和）

Cure

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1274    Accepted Submission(s): 429

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

 

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题目意思：

求1/1^2到1/n^2之和，保留5位小数点。

解题思路：

打表预处理，搜出前2000000，因为是保留5位小数点所以较大的数直接近似成(π^2)/6，即1.64493。

#include<cstdio>#include<iostream>#include<vector>#include<cmath>#include<cstring>#include<queue>#include<algorithm>#include<set>#include<queue>#define INF 0xfffffffusing namespace std;const int maxn=2000000;typedef long long ll;double a[maxn];char s[maxn*10];void f()//打表{    a[0]=a[1]=1;    for(int i=2; i<=maxn; ++i)        a[i]=1/pow(i,2)+a[i-1];}int main(){    f();    long long n;    while(~scanf("%s",s))    {        if(strlen(s)<=7)        {            sscanf(s,"%I64d",&n);//取最大I64d长度的字符串s转换存入长整型n            if(n<=1000000)                printf("%.5lf\n",a[n]);            else                printf("1.64493\n");//六分之π平方        }        else            printf("1.64493\n");    }    return 0;}