【威尔逊定理】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

1312345678910100100010000

 

Sample Output

0112222334282071609

 

Source

Central European Programming Contest 2008

题意：

计算题目中给的式子。

解题思路：

当3k+7不是素数时，可以得到((3k+6)!+1)/(3k+7)=[(3k+6)!/(3k+7)]，此时括号内的值为0.

当3k+7是素数时，由威尔逊定理知(3k+6)! = -1 (mod 3k+7) ，可以得到((3k+6)!+1)/(3k+7)=[(3k+6)!/(3k+7)]+1，此时括号内的值为1.

AC代码：

#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;}