数论 - 119. Magic Pairs

 Magic Pairs

Problem's Link

 

Mean: 

已知N、A0、B0，对于给定X、Y，若A0X+B0Y能被N整除，则AX+BY也能被N整除，求所有的A、B.（0<=A、B<N)

analyse:

这道题目就是先把A0和B0和N都除以他们的最大公约数，然后就是枚举A0和B0的0到N-1倍.

最后快排，注意一下有0的情况.

Time complexity: O(N)

 

view code

/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
*       Author: crazyacking
*       Date  : 2016-01-08-10.51
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);

int main()
{

    pair < int , int > ans [ 10010 ];
    int n , a0 , b0;
    scanf( "%d%d%d" , &n , & a0 , & b0);

    int i = a0 %= n;
    int j = b0 %= n;
    int num = 0;

    do
    {
        ans [ num ++ ] = make_pair( i , j);
        i = ( i + a0) %n;
        j = ( j + b0) %n;
    }
    while( i != a0 || j != b0);

    sort( ans , ans + num);
    printf( "%d \n " , num);
    for( i = 0; i < num; i ++)
        printf( "%d %d \n " , ans [ i ]. first , ans [ i ]. second);

    return 0;
}