zoukankan      html  css  js  c++  java
  • 2015杭电多校第二场

    Buildings

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 1622    Accepted Submission(s): 460


    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(1n,m108,n×m>1,1xn,1ym).
     
    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.
     
    Source
     
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int  main()
    {
        //fropen("in","r",stdin);
       // fropen("out","w",stdin);
        int n,m,a,b;
    
        while(scanf("%d%d%d%d",&n,&m,&a,&b)!=EOF)
        {
            int d,f,h;
            int v1,v2,v3,v4,k,l;
            v1=n/2+n%2;
            f=m/2+m%2;
            h=min(v1,f);
            d=max(v1,f);
            if(n==m&&n%2==1&&m%2==1&&h==min(a,b)&&d==max(a,b))
            {
                printf("%d
    ",h-1);
                continue;
            }
            if(h==v1)
            {
                k=max(a-1,n-a);
    
                v2=min(k,min(m-b+1,b));
            }
            else
            {
                k=max(b-1,m-b);
    
                v2=min(k,min(n-a+1,a));
            }
            v2=max(v2,h);
            printf("%d
    ",v2);
        }
        return 0;
    }
    View Code
  • 相关阅读:
    JavaScript基础知识-forEach循环
    JavaScript基础知识-数组的练习
    JavaScript基础知识-数组的遍历
    JavaScript基础知识-数组的常用方法
    JavaScript基础知识-数组基于索引访问
    JavaScript基础知识-数组的定义方式
    JavaScript基础知识-垃圾回收
    JavaScript基础知识-toString()
    JavaScript基础知识-原型(prototype)
    JavaScript基础知识-构造函数(也称为"类")定义
  • 原文地址:https://www.cnblogs.com/ya-cpp/p/4678006.html
Copyright © 2011-2022 走看看