### Buildings 分類： ACM 多校 2015-07-23 22:09 8人閱讀 評論(0) 收藏

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

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

You can split the floor into three 2×1 apartments and two 1×1 apartments. The answer is 2.

http://acm.hdu.edu.cn/data/images/C590-1002-4.jpg

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.

``````#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int x,y,a,b;
while(scanf("%d %d %d %d",&x,&y,&a,&b)!=EOF)
{
if(x>y)
{
swap(x,y);
swap(a,b);
}
//cout<<a<<" "<<b<<endl;
int ans=(min(x,y)+1)/2;
//printf("%d\n",ans);
int left=b,right=y-b+1;//黑格子距離左邊，右邊的位置
int up=a-1,down=x-a; //黑格子距離上面，下面的距離
int ans1=min(left,right);
int ANS=0;
if(ans1>ans&&a-1!=x-a)//若是大於則區域的划分有 變化
{
int d=min(left,right);
int c=max(up,down);
//ANS++;
//cout<<c<< " "<<d<<endl;
ANS=min(d,c) ;
cout<<ANS<<endl;
continue;
}
if(x==y&&x%2==1&&a==b&&a==(x+1)/2)
{
ANS=x/2;
cout<<ANS<<endl;
continue;
}
else
{
ANS=ans;
cout<<ANS<<endl;
continue;
}
//printf("%d\n",ANS);
}
return 0;
}``````