hdu

The math department has been having problems lately. Due to immense amount of unsolicited automated programs which were crawling across their pages, they decided to put Yet-Another-Public-Turing-Test-to-Tell-Computers-and-Humans-Apart on their webpages. In short, to get access to their scientific papers, one have to prove yourself eligible and worthy, i.e. solve a mathematic riddle. 


However, the test turned out difficult for some math PhD students and even for some professors. Therefore, the math department wants to write a helper program which solves this task (it is not irrational, as they are going to make money on selling the program). 

The task that is presented to anyone visiting the start page of the math department is as follows: given a natural n, compute 

where [x] denotes the largest integer not greater than x.

Input

The first line contains the number of queries t (t <= 10^6). Each query consist of one natural number n (1 <= n <= 10^6).

Output

For each n given in the input output the value of Sn.

Sample Input

1312345678910100100010000

Sample Output

0112222334282071609

题意就是给出n，按公式求sn，观察公式中的一个和项，当被减数为整数时，此和项的值为1，当被减数不是整数时，此和项的值为0；那么怎么判断一个和项是否为整数呢，

这里用威尔逊定理：当且仅当p为素数时：( p -1 )! ≡ -1 ( mod p )  而令被减数中（3*k+6）=p,则被减数化为（(p-1)!+1）/p,从而得到当p为素数时（(p-1)!+1）mod p等于0,即被减数（(p-1)!+1）/p为整数，对应和项为1，

即当3*k+1为素数时，Si=1，否则Si=0；

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define M 3000007int n;bool prime[M+10];int ans[M+10];void init(){    int m=sqrt(M+0.5);    int i,j;    for(i=2;i<=m;i++)    {        if(!prime[i])        {            for(j=i*i;j<=M;j+=i)            {                prime[j]=true;            }        }    }    ans[0]=ans[1]=0;    for(i=1;i<=1000000;i++)    {        if(!prime[i*3+7])            ans[i]=ans[i-1]+1;        else            ans[i]=ans[i-1];    }}int main(){    init();int T,i;scanf("%d",&T);while(T--)    {        scanf("%d",&n);        printf("%d\n",ans[n]);    }return 0;}