zoukankan      html  css  js  c++  java
  • Codeforces Round #313 B. Gerald is into Art(简单题)

    B. Gerald is into Art
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Gerald bought two very rare paintings at the Sotheby's auction and he now wants to hang them on the wall. For that he bought a special board to attach it to the wall and place the paintings on the board. The board has shape of an a1 × b1 rectangle, the paintings have shape of a a2 × b2 and a3 × b3 rectangles.

    Since the paintings are painted in the style of abstract art, it does not matter exactly how they will be rotated, but still, one side of both the board, and each of the paintings must be parallel to the floor. The paintings can touch each other and the edges of the board, but can not overlap or go beyond the edge of the board. Gerald asks whether it is possible to place the paintings on the board, or is the board he bought not large enough?

    Input

    The first line contains two space-separated numbers a1 and b1 — the sides of the board. Next two lines contain numbers a2, b2, a3 andb3 — the sides of the paintings. All numbers ai, bi in the input are integers and fit into the range from 1 to 1000.

    Output

    If the paintings can be placed on the wall, print "YES" (without the quotes), and if they cannot, print "NO" (without the quotes).

    Sample test(s)
    input
    3 2
    1 3
    2 1
    
    output
    YES
    
    input
    5 5
    3 3
    3 3
    
    output
    NO
    
    input
    4 2
    2 3
    1 2
    
    output
    YES
    
    Note

    That's how we can place the pictures in the first test:

    And that's how we can do it in the third one.




      题意:给出一个大的矩形,然后再给出两个小的矩形,问能不能把这两个小的矩

    形在不重叠的情况下放进大的矩形中,能的话输出YES。否则输出NO



    #include<iostream>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    
    using namespace std;
    
    int main()
    {
        int n,m;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            int a,b;
            int c,d;
            scanf("%d%d",&a,&b);
            scanf("%d%d",&c,&d);
            if(((a+c)<=n &&max(b,d)<=m)||((a+c)<=m &&max(b,d)<=n) || ((a+d)<=n &&max(b,c)<=m) || ((a+d)<=m &&max(b,c)<=n) || ((b+d)<=n &&max(a,c)<=m)||((b+d)<=m &&max(a,c)<=n) || ((b+c)<=n &&max(a,d)<=m) || ((b+c)<=m &&max(a,d)<=n))
            {
                printf("YES
    ");
            }
            else
            {
                printf("NO
    ");
            }
    
        }
        return 0;
    }


  • 相关阅读:
    模块之间的消息传递优势与问题
    cp命令
    gimp的python控制台生成UI代码
    three.js(六) 地形法向量生成
    mysql 主从复制配置
    Three.js(九)perlin 2d噪音的生成
    three.js(五) 地形纹理混合
    mongodb 复制
    pycparse python的c语法分析器
    ajax出现 所需的数据还不可用
  • 原文地址:https://www.cnblogs.com/liguangsunls/p/6852644.html
Copyright © 2011-2022 走看看