【威爾遜定理】HDOJ Zball in Tina Town 5391

Zball in Tina Town

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1356    Accepted Submission(s): 705

Problem Description

Tina Town is a friendly place. People there care about each other.

Tina has a ball called zball. Zball is magic. It grows larger every day. On the first day, it becomes 1 time as large as its original size. On the second day,it will become2 times as large as the size on the first day. On the n-th day,it will become n times as large as the size on the (n-1)-th day. Tina want to know its size on the (n-1)-th day modulo n.

 

Input

The first line of input contains an integer T, representing the number of cases.

The following T lines, each line contains an integer n, according to the description.
T≤105,2≤n≤109

 

Output

For each test case, output an integer representing the answer.

 

Sample Input

2 3 10

 

Sample Output

2 0

 

Source

BestCoder Round #51 (div.2)

 

題意：

計算(n-1)! mod  n  的值

解題思路：

威爾遜定理：p是素數 則( p -1 )! ≡ -1 ( mod p )

p不是素數  結果是0. 注意4的情況！

AC代碼：

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

using namespace std;

bool is_prime(int n)
{
    if(n==1||n==2) return true;
    if(!(n&1)) return false;
    int m=sqrt(n);
    for(int i=2;i<=m;i++){
        if(n%i==0) return false;
    }
    return true;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        if(n==4){
            printf("2\n");
            continue;
        }
        if(is_prime(n)) printf("%d\n",n-1);
        else printf("0\n");
    }
    return 0;
}