暴力枚举大水题

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=43311#problem/B

Description

You've got two rectangular tables with sizes n_a × m_a and n_b × m_b cells. The tables consist of zeroes and ones. We will consider the rows and columns of both tables indexed starting from 1. Then we will define the element of the first table, located at the intersection of the i-th row and the j-th column, as a_i, j; we will define the element of the second table, located at the intersection of the i-th row and thej-th column, as b_i, j.

We will call the pair of integers (x, y) a shift of the second table relative to the first one. We'll call the overlap factor of the shift (x, y)value:

where the variables i, j take only such values, in which the expression a_i, j·b_{i + x, j + y} makes sense. More formally, inequalities 1 ≤ i ≤ n_a, 1 ≤ j ≤ m_a, 1 ≤ i + x ≤ n_b, 1 ≤ j + y ≤ m_b must hold. If there are no values of variables i, j, that satisfy the given inequalities, the value of the sum is considered equal to 0.

Your task is to find the shift with the maximum overlap factor among all possible shifts.

Input

The first line contains two space-separated integers n_a, m_a(1 ≤ n_a, m_a ≤ 50) — the number of rows and columns in the first table. Then n_a lines contain m_a characters each — the elements of the first table. Each character is either a "0", or a "1".

The next line contains two space-separated integers n_b, m_b(1 ≤ n_b, m_b ≤ 50) — the number of rows and columns in the second table. Then follow the elements of the second table in the format, similar to the first table.

It is guaranteed that the first table has at least one number "1". It is guaranteed that the second table has at least one number "1".

Output

Print two space-separated integers x, y(|x|, |y| ≤ 10⁹) — a shift with maximum overlap factor. If there are multiple solutions, print any of them.

Sample Input

Input

3 2
01
10
00
2 3
001
111

Output

0 1

Input

3 3
000
010
000
1 1
1

Output

-1 -1

题目要求就是让b平移多少能够是ab中重合的“1”最多.数据很小，我直接枚举了，哈哈

#include <stdio.h>
#include <iostream>
using namespace std;
const int N=52;
char a[N][N],b[N][N];
int n,m,x,y;
int abcd(int xx,int yy)
{
    int sum=0;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
              if(a[i][j]=='1'&&b[i+xx][j+yy]=='1'&&i+xx<=x&&xx+i>=1&&yy+j<=y&&yy+j>0)//一定要判断边界，我因为这里错了好几遍
                  sum++;
    return sum;
}
int main()
{
    int xx,yy;
    while(~scanf("%d%d",&n,&m))
    {
        getchar();
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                cin >>a[i][j];
        scanf("%d%d",&x,&y);
        getchar();
        for(int i=1;i<=x;i++)
           for(int j=1;j<=y;j++)
               cin >> b[i][j];
        int maxx=-1;
        for(int i=-51;i<=51;i++)
           for(int j=-51;j<=51;j++)
               if(abcd(i,j)>maxx)
               {
                  maxx=abcd(i,j);
                  xx=i;
                  yy=j;
               }
        printf("%d %d\n",xx,yy);
    }
    return 0;
}