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

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.

題意是盡量把n*m的區域划分為多個,然后求最大的區域的面積。
要分成多個那么每個區域都可以表示成1*k。先看若是沒有那個黑格子的話,可以得知答案就是(min(n,m)+1)/2;
但是由黑格子,現在就看黑格子在那個位置是否會對答案產生影響,詳情看代碼。然后還有一點容易忽略的是當n和m相等且同為奇數的話,要看看黑格子是否在正中,在正中位置答案減一。。

#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\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;
}

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