hdu5391 Zball in Tina Town(威尔逊定理)

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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): 219    Accepted Submission(s): 144

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

题意：要求(n-1)! mod n
根据威尔逊定理，n为素数的时候(n-1)! mod n = -1 = n - 1，这是个充要条件

在n不是素数的时候，可以发现，只有4不为0，其余的时候都为0

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author xyiyy @https://github.com/xyiyy
 */

#include <iostream>
#include <fstream>

//#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>

using namespace std;

//
// Created by xyiyy on 2015/8/7.
//

#ifndef ICPC_SCANNER_HPP
#define ICPC_SCANNER_HPP

#endif //ICPC_SCANNER_HPP

class hdu5391 {
public:
    void solve(std::istream &in, std::ostream &out) {
        int t;
        in >> t;
        while (t--) {
            int n;
            in >> n;
            if (check(n))out << n - 1 << endl;
            else if (n == 4)out << 2 << endl;
            else out << 0 << endl;
        }
    }

    bool check(int x) {
        for (int i = 2; i * i <= x; i++) {
            if (x % i == 0) {
                return 0;
            }
        }
        return 1;
    }
};

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    hdu5391 solver;
    std::istream &in(std::cin);
    std::ostream &out(std::cout);
    solver.solve(in, out);
    return 0;
}