题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5879
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.
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
Hint:
题意：
给 n，求 ∑k=1n1k2sum_{k=1}^{n}frac{1}{k^2}∑k=1nk21。
题解：
n很大的时候答案趋近于π26frac{pi^2}{6}6π2，nnn 小的时候预处理即可。
其实可以直接暴力直接打表过，但是自己当时比较2，没想到，直接用了大数的相加。还有就是前导0自己忘记考虑了，哎，自己要做的事情还有很多啊。
代码：
#include <cmath>
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e6;
#define met(a,b) memset(a,b,sizeof(a))
#define pi acos(-1.0)
char s[maxn+1000];
double num[maxn];
int main()
{
met(num,0);
num[1]=1;
for(int i=2;i<=maxn;i++)
num[i]=num[i-1]+1.0/i/i;
while(scanf("%s",s)!=EOF)
{
int cnt=0;
int l=strlen(s);
int l1=l;
while(s[cnt]=='0'||cnt==l)
cnt++;
l-=cnt;
if(l>6)
printf("%.5lf
",pi*pi/6.0);
else
{
int sum=0;
for(int i=0;i<l1;i++)
sum=sum*10+s[i]-'0';
if(sum>maxn)
printf("%.5lf
",pi*pi/6.0);
else
printf("%.5lf
",num[sum]);
}
}
}