Buildings 分类： ACM 多校 2015-07-23 22:09 8人阅读 评论(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

Case 1 :
http://acm.hdu.edu.cn/data/images/C590-1002-2.jpg

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

Case 2:
http://acm.hdu.edu.cn/data/images/C590-1002-3.jpg

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.

题意是尽量把n*m的区域划分为多个，然后求最大的区域的面积。
要分成多个那么每个区域都可以表示成1*k。先看若是没有那个黑格子的话，可以得知答案就是(min(n,m)+1)/2;
但是由黑格子，现在就看黑格子在那个位置是否会对答案产生影响，详情看代码。然后还有一点容易忽略的是当n和m相等且同为奇数的话，要看看黑格子是否在正中，在正中位置答案减一。。

#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
",ans);
        int left=b,right=y-b+1;//黑格子距离左边，右边的位置 
        int up=a-1,down=x-a; //黑格子距离上面，下面的距离 
        int ans1=min(left,right);
        int ANS=0;
        if(ans1>ans&&a-1!=x-a)//若是大于则区域的划分有 变化 
        {
            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
",ANS);
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。