Codeforces 492E Vanya and Field

E. Vanya and Field

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya decided to walk in the field of size n × n cells. The field contains m apple trees, the i-th apple tree is at the cell with coordinates(xi, yi). Vanya moves towards vector (dx, dy). That means that if Vanya is now at the cell (x, y), then in a second he will be at cell . The following condition is satisfied for the vector: , where  is the largest integer that divides both a and b. Vanya ends his path when he reaches the square he has already visited.

Vanya wonders, from what square of the field he should start his path to see as many apple trees as possible.

Input

The first line contains integers n, m, dx, dy(1 ≤ n ≤ 106, 1 ≤ m ≤ 105, 1 ≤ dx, dy ≤ n) — the size of the field, the number of apple trees and the vector of Vanya's movement. Next m lines contain integers xi, yi (0 ≤ xi, yi ≤ n - 1) — the coordinates of apples. One cell may contain multiple apple trees.

Output

Print two space-separated numbers — the coordinates of the cell from which you should start your path. If there are several answers you are allowed to print any of them.

Sample test(s)

input

5 5 2 3
0 0
1 2
1 3
2 4
3 1

output

1 3

input

2 3 1 1
0 0
0 1
1 1

output

0 0

Note

In the first sample Vanya's path will look like: (1, 3) - (3, 1) - (0, 4) - (2, 2) - (4, 0) - (1, 3)

In the second sample: (0, 0) - (1, 1) - (0, 0)

题目给出  gcd( n , dx ) = gcd( n , dy ) = 1 , 意味着（0~n-1）每个数都可以遍历到。

然后从x,y轴方向0坐标开始分别预处理出的n个数。

对于两颗苹果树，如果他们 y坐标到0坐标的距离 与 x坐标到0坐标的距离 两者的差值相同。

那么这两个坐标是来自同一个环的。

扫一遍m个点以后取出最大的就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <algorithm>
 
using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int N =  ;
const int M =  ;
const int inf = 1e9+;
#define X first
#define Y second
 
int n , m , dx , dy , x[N] , y[N] ,  cnt[N] , id ;
vector<pii>e;
void test() {
    for( int i =  ; i < n ; ++i ) cout << x[i] << ' ' ; cout <<endl ;
    for( int i =  ; i < n ; ++i ) cout << y[i] << ' ' ; cout <<endl ;
}
void Run() {
 
    memset( cnt ,  , sizeof cnt ) ;
    x[] =  ; id =  ; for( int i = dx ; i !=  ; i = (i+dx)%n ) x[i] = id++;
    y[] =  ; id =  ; for( int i = dy ; i !=  ; i = (i+dy)%n ) y[i] = id++;
//    test();
    e.resize(m);
    for( int i =  ; i < m ; ++i ) {
        cin >> e[i].X >> e[i].Y ;
        int disx = ( x[e[i].X] + n ) % n ;
        int disy = ( y[e[i].Y] + n ) % n ;
        cnt[ (disx + n - disy)%n  ] ++ ;
    }
    LL ans =  ;
    for( int i = ; i < n ; ++i ) {
        if( cnt[ans] < cnt[i] ) ans = i;
    }
    cout << ( ans*dx ) % n << ""<< endl ;
}
int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    ios::sync_with_stdio(false);
    while( cin >> n >> m >> dx >> dy ) Run();
}