【威爾遜定理】HDOJ YAPTCHA 2973

YAPTCHA

                                           Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                   Total Submission(s): 756    Accepted Submission(s): 396

Problem Description

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

13 1 2 3 4 5 6 7 8 9 10 100 1000 10000

 

Sample Output

0 1 1 2 2 2 2 3 3 4 28 207 1609

 

Source

Central European Programming Contest 2008

題意：

計算題目中給的式子。

解題思路：

當3k+7不是素數時，可以得到((3k+6)!+1)/(3k+7)=[(3k+6)!/(3k+7)]，此時括號內的值爲0.

當3k+7是素數時，由威爾遜定理知(3k+6)! = -1 (mod 3k+7) ，可以得到((3k+6)!+1)/(3k+7)=[(3k+6)!/(3k+7)]+1，此時括號內的值爲1.

AC代碼：

#include <stdio.h>
#include <math.h>
#include <algorithm>

using namespace std;

const int MAXN = 3100000+10;

int prime[MAXN];
int E[1000010];

bool is_prime()
{
    prime[0]=prime[1]=false;
    for(int i=2;i<MAXN;i++){
        if(!prime[i]){
            for(int j=i*2;j<MAXN;j+=i){
                prime[j]=true;
            }
        }
    }
}

int init()
{
    is_prime();
    E[1]=0;
    int res=0;
    for(int i=2;i<=1000000;i++){
        if(!prime[3*i+7]) E[i]=E[i-1]+1;
        else E[i]=E[i-1];
    }
}


int main()
{
    int t;
    init();
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        printf("%d\n",E[n]);
    }
    return 0;
}