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-);

int main()
{

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

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

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

   return ;
}