HDU 5391 Zball in Tina Town

Zball in Tina Town

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1371    Accepted Submission(s): 708

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≤105,2≤n≤109

 

 

Output

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

 

 

Sample Input

2

3

10

 

Sample Output

2

0

找规律 然后大整数判断素数这里有个大整数判断素数模板

 

/* ***********************************************
Author        :guanjun
Created Time  :2015/10/13 20:30:38
File Name     :1.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#include <ctime>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
    return a>b;
}
ll gcd(ll a,ll b){
    return a==0?b:gcd(b%a,a);
}
ll mod_mul(ll a,ll b,ll n){
    ll res=0;
    while(b){
        if(b&1)res=(res+a)%n;
        a=(a+a)%n;
        b>>=1;
    }
    return res;
}
ll mod_exp(ll a,ll b,ll n){
    ll res=1;
    while(b){
        if(b&1)res=mod_mul(res,a,n);
        a=mod_mul(a,a,n);
        b>>=1;
    }
    return res;
}
bool miller_rabin(ll n){
    if(n==2||n==3||n==5||n==7||n==11)return true;
    if(n==1||!(n%2)||!(n%3)||!(n%5)||!(n%7)||!(n%11))return false;
    ll x,pre,u;
    int i,j,k=0;
    u=n-1;
    while(!(u&1)){
        k++;u>>=1;
    }
    srand((ll)time(0));
    for(i=0;i<5;i++){//5次足够AC了
        x=rand()%(n-2)+2;
        if((x%n)==0)continue;
        x=mod_exp(x,u,n);
        pre=x;
        for(j=0;j<k;j++){
            x=mod_mul(x,x,n);
            if(x==1&&pre!=1&&pre!=n-1)return false;
            pre=x;
        }
        if(x!=1)return false;
    }
    return true;
}
ll w[21]={0,1, 2, 6, 24, 120, 720, 5040,40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000,355687428096000, 6402373705728000, 121645100408832000, 2432902008176640000};
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int t;
    cin>>t;
    ll n;
    while(t--){
        scanf("%I64d",&n);
        if(n<=21)printf("%I64d\n",w[n-1]%n);
        else {
            if(!miller_rabin(n))printf("%d\n",0);
            else printf("%I64d\n",n-1);
        }
    }
    return 0;
}