ural 1204. Idempotents

1204. Idempotents

Time limit: 1.0 second
Memory limit: 64 MB

 

The number x is called an idempotent modulo n if

x*x = x (mod n)

Write the program to find all idempotents modulo n, where n is a product of two distinct primes p and q.

Input

First line contains the number k of test cases to consider (1 ≤ k ≤ 1000). Each of the following k lines contains one number n < 10⁹.

Output

Write on the i-th line all idempotents of i-th test case in increasing order. Only nonnegative solutions bounded by n should be printed.

Sample

input	output
3 6 15 910186311	0 1 3 4 0 1 6 10 0 1 303395437 606790875

Problem Author: Pavel Atnashev
Problem Source: USU Internal Contest, March 2002

大意：给出n, 求x, 使得x*x = x(mod n)

思路：x1 = 0, x2 = 1时公式成立

x*x = x (mod n)

n = q*p

x*(x-1) = (a*b)*(q*p) , a, b为整数

(1)x = a*q, x-1 = b*p,

a*q +b*p = x + (x-1) = 1;

用拓展欧几里得算法求出a, b， x3 = a*q;

(2) x-1 = a*q, x = b*p

x4 = b*p

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <cstring>
#define FOR(i, a, b)  for(int i = (a); i <= (b); i++)
#define RE(i, n) FOR(i, 1, n)
#define FORP(i, a, b) for(int i = (a); i >= (b); i--)
#define REP(i, n) for(int i = 0; i <(n); ++i)
#define SZ(x) ((int)(x).size )
#define ALL(x) (x).begin(), (x.end())
#define MSET(a, x) memset(a, x, sizeof(a))
using namespace std;


typedef long long int ll;
typedef pair<int, int> P;
int read() {
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
const double pi=3.14159265358979323846264338327950288L;
const double eps=1e-6;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 100005;
const int xi[] = {0, 0, 1, -1};
const int yi[] = {1, -1, 0, 0};

int N, T;
int a[MAXN], prime[MAXN], cnt;
void init() {
    cnt = 0;
    for(int i = 2; i < MAXN; i++) a[i] = true;
    for(int i = 2; i < MAXN; i++) {
        if(a[i]) {
            prime[++cnt] = i;
        }
        for(int j = 1; j <= cnt; j++) {
            if(i*prime[j] >= MAXN) break;
            a[i*prime[j]] = false;
            if(i%prime[j] == 0) break;
        }
    }
}
int exgcd(int a, int b, int &x, int &y) {
    int d = a;
    if(b != 0) {
        d = exgcd(b, a%b, y, x);
        y -= (a/b)*x;
    } else {
        x = 1, y = 0;
    }
    return d;
}
int main() {
    //freopen("in.txt", "r", stdin);
    init();
    int t, n;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        int p = - 1, q = -1;
        for(int i = 1; i <= cnt; i++) {
            if(n%prime[i] == 0) {
                int k = n/prime[i];
                int flag = 1;
                for(int j = 2; j*j <= k; j++){
                    if(k%j == 0){
                        flag = 0;
                        break;
                    }
                }
                if(flag){
                    p = prime[i], q = n/prime[i];
                    break;
                }
            }
        }
      //  printf("%d %d
", p, q);
        //  if(p > q) swap(p, q);
        int x, y, res1, res2, d;
        d = exgcd(p, q, x, y);
       // printf("%d %d
", x, y);
        res1 = p*x;
        if(res1 <= 0) res1 += n;
      //  printf("%d
",res1);
        d = exgcd(q, p, x, y);
       // printf("%d %d
", x, y);
        res2 = q*x;
        if(res2 <= 0) res2 += n;
        //printf("%d
", res2);
        if(res2 < res1) swap(res1, res2);
        printf("0 1 %d %d
", res1, res2);
    }
    return 0;
}