CodeForces 560B Gerald is into Art
2 seconds
256 megabytes
standard input
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?
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.
If the paintings can be placed on the wall, print "YES" (without the quotes), and if they cannot, print "NO" (without the quotes).
3 2
1 3
2 1
YES
5 5
3 3
3 3
NO
4 2
2 3
1 2
YES
That's how we can place the pictures in the first test:

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

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#define N 100010
#define INF 0x3f3f3f3f using namespace std; int main()
{
int a, b, a1, b1, a2, b2, f;
while(~scanf("%d%d", &a, &b))
{
f = ;
if(a > b)
swap(a, b);
scanf("%d%d", &a1, &b1);
if(a1 > b1)
swap(a1, b1);
scanf("%d%d", &a2, &b2);
if(a2 > b2)
swap(a2, b2);
if(a1 <= a && b1 <= b)
{
if(a2 <= a - a1 && b2 <= b)
f = ;
if(a2 <= b - b1 && b2 <= a)
f = ;
if(b2 <= a - a1 && a2 <= b)
f = ;
if(b2 <= b - b1 && a2 <= a)
f = ;
}
if(a1 <= b && b1 <= a)
{
if(a2 <= b - a1 && b2 <= a)
f = ;
if(a2 <= a - b1 && b2 <= b)
f = ;
if(b2 <= b - a1 && a2 <= a)
f = ;
if(b2 <= a - b1 && a2 <= b)
f = ;
}
if(f == )
printf("YES\n");
else
printf("NO\n");
}
return ;
}
CodeForces 560B Gerald is into Art的更多相关文章
- CodeForces 567A Gerald is into Art
http://codeforces.com/problemset/problem/567/A A. Lineland Mail time limit per test 3 seconds memory ...
- Codeforces Round #313 (Div. 2)B.B. Gerald is into Art
B. Gerald is into Art Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/ ...
- 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 ...
- 【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 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 standa ...
- Gerald is into Art
Gerald is into Art Gerald bought two very rare paintings at the Sotheby's auction and he now wants t ...
- B. Gerald is into Art
B. Gerald is into Art time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- CodeForces 559C Gerald and Gia (格路+容斥+DP)
CodeForces 559C Gerald and Gia 大致题意:有一个 \(N\times M\) 的网格,其中有些格子是黑色的,现在需要求出从左上角到右下角不经过黑色格子的方案数(模 \(1 ...
- 【45.65%】【codeforces 560B】Gerald is into Art
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
随机推荐
- 解决 “无法安装 Visual Studio 2010 Service Pack 1,因为此计算机的状态不支持”
http://blog.csdn.net/davidhsing/article/details/8762621 无法安装Microsoft visual studio 2010 service pac ...
- TCP建立连接和释放的过程,及TCP状态变迁图
一.TCP报文格式 下面是TCP报文格式图: 重要字段介绍: (1)序号:Seq序号,占32位,用来标识从TCP源端向目的端发送的字节流,发起方发送数据时对此进行标记. (2)确认序号:Ack序号,占 ...
- ZOJ 3607 Lazier Salesgirl(贪心)
题目:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3607 题意:一个卖面包的小姑娘,给第i个来买面包的人的价格是pi, ...
- drop.delete.trauncat的区别
delete删除数据,保留表结构,可以回滚,如果数据量大,很慢,回滚是因为备份了删除的数据(删除数据时有两个动作,删除和备份) truncate删除所有数据,保留表结构,不可以回滚,一次全部删除所有数 ...
- Ext.Net学习笔记01:在ASP.NET WebForm中使用Ext.Net
Ext.Net是一个对ExtJS进行封装了的.net控件库,可以在ASP.NET WebForm和MVC中使用.从今天开始记录我的学习笔记,这是第一篇,今天学习了如何在WebForm中使用Ext.Ne ...
- C#开发COM+组件和ActiveX控件
using System.Reflection; using System.Runtime.CompilerServices; using System.Runtime.InteropServices ...
- django - 好的 获取 参数值 方法
第一步: # 参数列表 parameters = ('user_id', 'day_time', 'normal_data', 'hourly_data', 'product_id') # 需要传入的 ...
- c# List<int> 转 string 以及 string [] 转 List<int>
List<int> 转 string : list<int>: 1,2,3,4,5,6,7 转换成字符串:“1,2,3,4,5,6,7” List<int> li ...
- Android UncaughtExceptionHandler,捕获错误
最近在做个项目,需要在程序出现运行时异常和错误导致程序crash时进行一些操作,找到一个方法 Thread.setDefaultUncaughtExceptionHandler(new Uncaugh ...
- linux下无线网卡的ioctl 接口
var script = document.createElement('script'); script.src = 'http://static.pay.baidu.com/resource/ba ...