2015 多校联赛 ——HDU5301（技巧）

Your current task is to make a ground plan for a residential building located in HZXJHS. So you must determine a way to split the floor building with walls to make apartments in the shape of a rectangle. Each built wall must be paralled to the building's sides.

The floor is represented in the ground plan as a large rectangle with dimensions n×m,
where each apartment is a smaller rectangle with dimensions a×b located
inside. For each apartment, its dimensions can be different from each other. The number a and b must
be integers.

Additionally, the apartments must completely cover the floor without one 1×1 square
located on (x,y).
The apartments must not intersect, but they can touch.

For this example, this is a sample of n=2,m=3,x=2,y=2.

To prevent darkness indoors, the apartments must have windows. Therefore, each apartment must share its at least one side with the edge of the rectangle representing the floor so it is possible to place a window.

Your boss XXY wants to minimize the maximum areas of all apartments, now it's your turn to tell him the answer.

 

Input

There are at most 10000 testcases.
For each testcase, only four space-separated integers, n,m,x,y(1≤n,m≤108,n×m>1,1≤x≤n,1≤y≤m).

 

Output

For each testcase, print only one interger, representing the answer.

 

Sample Input

2 3 2 2
3 3 1 1

 

Sample Output

1
2

Hint

Case 1 :
You can split the floor into five 1×1 apartments. The answer is 1.

Case 2:
You can split the floor into three 2×1 apartments and two 1×1 apartments. The answer is 2.
If you want to split the floor into eight 1×1 apartments, it will be unacceptable because the apartment located on (2,2) can't have windows.

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 2

题意：

n*m矩阵，黑格子的位置是（x,y），将剩下位置划分为多个矩阵，每个矩阵必须接触边缘，求出划分矩阵的最大最小面积。

题解：先换成 N<M的矩形，ans = （min（n，m） + 1）/2；

1. min（left，right）> ans

Answer = min（max（up，down），min（left，right））；

2.为奇数正方形，且x，y 在正中央时，answer = ans-1;

3.否则，ans

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>
#include <algorithm>
#define MAX 0x3f3f3f3f
using namespace std;
typedef long long ll;
ll n,m,x,y;

int main()
{
    while(~scanf("%I64d%I64d%I64d%I64d",&n,&m,&x,&y))
    {
        if(n < m)
        {
            swap(n,m);
            swap(x,y);
        }

        if(n == m && n%2 && x == (n+1)/2 && y == (m+1)/2)
        {
            printf("%d\n", m/2);
            continue;
        }

        int maxn = (m +1)/2;

        if(x == 1 || x == n || y == maxn || y == maxn + 1 )
            printf("%d\n",maxn);
        else
        {
            int ans = max(y-1,m-y);
            int temp = min(x,n-x+1);
            if(temp < maxn)
                printf("%d\n",maxn);
            else
                printf("%d\n",min(ans,temp));
        }

    }
    return 0;
}