C - C HDU 5391

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本文介绍了一个计算Z球在特定天数时大小的数学问题，并给出了一种利用威尔逊定理来简化计算的方法。对于素数和合数的情况进行了区分，通过判断输入值是否为素数来直接得出答案。

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C - C

Time Limit:1500MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 5391

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

time as large as its original size. On the second day,it will become

times as large as the size on the first day. On the n-th day,it will become

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

, representing the number of cases.

The following

lines, each line contains an integer

, according to the description.

Output

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

Sample Input

2
3
10

Sample Output

2
0

题解：对，你没看错，这就是求（N-1）的阶乘%N，但令人蛋疼的是你不能用阶乘，因为数据太大会超时。这时就是考验你的脑子的时候了，这里可以用到数论四大定理之一的威尔逊定理：如果N为素数，则（N-1）！+1可以被N正除。那当N为合数时，该怎么办呢？很简单，当N为合数时它的因子一定能在（n-1）！里找到，即（N-1）！
%N=0；注意：当N为4时应该返回2.

具体代码如下：

#include<stdio.h>     
#include<algorithm>
#include<math.h>
using namespace std;
int judge(int n)
{
    int flag=1;
    for(int i=2;i<(int)sqrt(n);i++)
    {
        if(n%i==0)
           {flag=0;break;

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