Codeforces Round #313 B. Gerald is into Art（简单题）

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本博客介绍了一个解决艺术布局的问题：如何将两幅非常罕见的抽象艺术画作合理地放置在一个特殊设计的矩形板上，而不让它们重叠或超出边界。通过详细的输入输出规范，读者可以了解如何判断是否能够成功地将画作放置在板上，并通过示例测试验证解决方案的有效性。

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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:

$\text{[math]}$

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

$\text{[math]}$

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\n");
        }
        else
        {
            printf("NO\n");
        }

    }
    return 0;
}