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

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

 

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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 0xfffffff
using 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
);
else
printf("1.64493\n");//六分之π平方
}
else
printf("1.64493\n");
}
return 0;
}