2015 Multi-University Training Contest 2 1002

Buildings

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

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.

 

题意：有一个nxm的矩阵，除去一个1x1的小正方形，问能将矩阵填满且有一边在矩阵上的矩形的最大面积的最小值是多少？

分析：（以下都是n<=m便于处理）

1、除去小正方形：

矩阵的的最优解是ans=(n + 1) / 2;

 

2、不除去小正方形

先从特殊情况考虑开始：n % 2 == 1 && x == y && n == m && x == n / 2 + 1

这种情况下 ans = ans - 1;

 

其次就是n!=m的情况

如果不放小正方形的话，就相当于用矩形填充整个矩阵，由于把小正方形加进去了，原来放在这里的矩形就不能放了，除了小正方形的地方就要用其他矩形去填补，那么如何去填补？

假设正方形处于任意位置(x,y)

那么就要在小正方形的上下先判断max(x - 1,n - x)再和m-y+1和y进行比较（可以想象成此处矩形是竖着放的）

#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
int n,m,x,y;
int main()
{
    while(scanf("%d %d %d %d",&n,&m,&x,&y) != EOF)
    {
        if(n>m)
        {
            swap(n,m);
            swap(x,y);
        }
        int ans=(n+1)/2;
        if(n%2==1 && x==y && n==m && x==n/2+1)
            ans=ans-1;
        int len=max(x-1,n-x);
        int ans1=min(m-y+1,min(len,y));
        ans=max(ans1,ans);
        printf("%d\n",ans);
    }
    return 0;
}