hdu 5391 （数论）_hdu数论-优快云博客

本文深入探讨了TinaTown中Zball的魔法成长过程，揭示了其每天增长倍数的数学原理，并通过实例展示了如何计算特定天数后的Zball大小。特别关注了质数与合数在计算中的不同表现，引入了威尔逊定理来解释某些特殊条件下的结果。

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Zball in Tina Town

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1401 Accepted Submission(s): 727

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 become $2$ 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 \leq {10}^{5},2 \leq n \leq {10}^{9} $

Output

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

Sample Input

2
3
10

Sample Output

2
0

题意：求（n-1）%n，

合数为0，因为1 ~ n-1中必定有积为n（除了4）

质数为n-1，威尔逊定理( p -1 )! ≡ -1 ( mod p )

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define N 100005
#define mod 258280327
#define MIN 0
#define MAX 1000001

bool prim(int u)
{
    for(int i = 2; i*i <= u; i++)
        if(u % i == 0)
            return false;
    return true;
}

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