Codeforces Round #313 (Div. 2) B. Gerald is into Art 水题
B. Gerald is into Art
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/560/problem/B
Description
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 Input
3 2
1 3
2 1
Sample Output
YES
HINT
题意
给你n*m的矩形,问你能否同事摆下n1*m1和n2*m2的矩形
题解:
就判断判断……
就搞啊搞
看代码吧
代码
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
const int maxn=;
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//**************************************************************************************
int a1,b1,a2,b2,a3,b3; int main(){
cin>>a1>>b1>>a2>>b2>>a3>>b3;
if (max(a2,a3)<=a1&&b2+b3<=b1||
max(a2,a3)<=b1&&b2+b3<=a1||
max(a2,b3)<=a1&&b2+a3<=b1||
max(a2,b3)<=b1&&b2+a3<=a1||
max(b2,a3)<=a1&&a2+b3<=b1||
max(b2,a3)<=b1&&a2+b3<=a1||
max(b2,b3)<=a1&&a2+a3<=b1||
max(b2,b3)<=b1&&a2+a3<=a1) cout << "YES\n";
else cout << "NO\n";
}
Codeforces Round #313 (Div. 2) B. Gerald is into Art 水题的更多相关文章
- 【打CF,学算法——二星级】Codeforces Round #313 (Div. 2) B. Gerald is into Art(水题)
[CF简单介绍] 提交链接:http://codeforces.com/contest/560/problem/B 题面: B. Gerald is into Art time limit per t ...
- Codeforces Round #313 (Div. 2) A. Currency System in Geraldion 水题
A. Currency System in Geraldion Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/c ...
- Codeforces Round #368 (Div. 2) A. Brain's Photos (水题)
Brain's Photos 题目链接: http://codeforces.com/contest/707/problem/A Description Small, but very brave, ...
- Codeforces Round #373 (Div. 2) C. Efim and Strange Grade 水题
C. Efim and Strange Grade 题目连接: http://codeforces.com/contest/719/problem/C Description Efim just re ...
- Codeforces Round #185 (Div. 2) A. Whose sentence is it? 水题
A. Whose sentence is it? Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/ ...
- Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 水题
A. Vitya in the Countryside 题目连接: http://codeforces.com/contest/719/problem/A Description Every summ ...
- Codeforces Round #371 (Div. 2) A. Meeting of Old Friends 水题
A. Meeting of Old Friends 题目连接: http://codeforces.com/contest/714/problem/A Description Today an out ...
- Codeforces Round #355 (Div. 2) B. Vanya and Food Processor 水题
B. Vanya and Food Processor 题目连接: http://www.codeforces.com/contest/677/problem/B Description Vanya ...
- dp - Codeforces Round #313 (Div. 1) C. Gerald and Giant Chess
Gerald and Giant Chess Problem's Link: http://codeforces.com/contest/559/problem/C Mean: 一个n*m的网格,让你 ...
随机推荐
- liux下ftp链接服务器的常用命令
FTP命令是Internet用户使用最频繁的命令之一,不论是在DOS还是UNIX操作系统下使用 FTP,都会遇到大量的FTP内部命令.熟悉并灵活应用FTP的内部命令,可以大大方便使用者,并收到事半功倍 ...
- IOS文章地址暂时记录
动画 http://www.jianshu.com/p/1c6a2de68753 iOS App性能优化 http://www.hrchen.com/2013/05/performance-wit ...
- CABasicAnimation
几个可以用来实现热门APP应用PATH中menu效果的几个方法 +(CABasicAnimation *)opacityForever_Animation:(float)time //永久闪烁的动画 ...
- OpenGL超级宝典第5版&&缓冲区
缓冲区有很多用途:可以保存顶点数据,像素数据,纹理数据,着色器处理的输入,不同着色器阶段的输出. 缓冲区保存在GPU内存中,提供高速有效的访问. 像素缓冲区对象: GLuint pixBuffer ...
- bzoj 1009 [HNOI2008]GT考试(DP+KMP+矩阵乘法)
[题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=1009 [题意] 给定一个字符串T,问长度为n且不包含串T的字符串有多少种. [思路] ...
- hive优化之------控制hive任务中的map数和reduce数
一. 控制hive任务中的map数: 1. 通常情况下,作业会通过input的目录产生一个或者多个map任务. 主要的决定因素有: input的文件总个数,input的文件大小,集群设置的 ...
- leetcode@ [354] Russian Doll Envelopes (Dynamic Programming)
https://leetcode.com/problems/russian-doll-envelopes/ You have a number of envelopes with widths and ...
- 【WPF】【火车站点信息查询】
全文涉及到的是C#和XAML 如果这两门语言并非你喜欢的语言,那可以关闭本网页了 本文介绍的是什么? 一个火车站点信息查询软件 本文涉及到的WPF基本知识 Task async await WebCl ...
- Homework-10 BASIC
对于本次作业: 我的整体思路如下: 1.首先修改二维数组求最大子数组和的C语言代码,加入分步骤的当前最优解边界值,局部最优解的记录,使之支持分步执行,连续执行,回滚等功能. 2.将程序改写为Javas ...
- 对"一维最大子数组和"问题的思考(homework-01)
一维最大子数组和问题,即给定一个数组,在它所有的连续子数组的和中,求最大的那个和.“最大子数组和”是一个很好的IT面试考题,在<编程之美>一书中同时阐述了一维数组和二维数组的讨论.本篇博客 ...