Gym - 100971G Repair

原题地址：http://codeforces.com/gym/100971/problem/G

G. Repair
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Alex is repairing his country house. He has a rectangular metal sheet of size a × b. He has to cut two rectangular sheets of sizes a1 × b1 and a2 × b2 from it. All cuts must be parallel to the sides of the initial sheet. Determine if he can do it.

Input
The first line contains two space-separated integers a and b (1 ≤ a, b ≤ 109) — the sizes of the initial sheet.

The second line contains two space-separated integers a1 and b1 (1 ≤ a1, b1 ≤ 109) — the sizes of the first sheet to cut out.

The third line contains two space-separated integers a2 and b2 (1 ≤ a2, b2 ≤ 109) — the sizes of the second sheet to cut out.

Output
Output «YES» (without quotes), if it’s possible to cut two described sheets from the initial sheet, or «NO» (without quotes), if it’s not possible.

Examples
input
12 20
14 7
5 6
output
YES
input
12 20
11 9
20 7
output
NO

题目分析

判断一个矩形版能不能分割成两个矩形板，矩形板相邻的多种组合，比较组合后的长和宽和矩形的长和宽判断能否分割

程序代码

#include
#include
#include
#include
#include
using namespace std;

int main()
{
    long long a, b;
    long long a1, b1;
    long long a2, b2;
    long long Max, Min, max1, min1, max2, min2;
    cin>>a>>b>>a1>>b1>>a2>>b2;
    if(a1+a2<=a && max(b1, b2)<=b)
    {
        cout<<"YES";
    }
    else if(a1+b2<=a && max(b1, a2)<=b)
    {
        cout<<"YES";
    }
    else if(b1+b2<=b && max(a1, a2)<=a)
    {
        cout<<"YES";
    }
    else if(b1+a2<=b && max(a1, b2)<=a)
    {
        cout<<"YES";
    }
    /********************************/
    else if(a1+a2<=b && max(b1, b2)<=a)
    {
        cout<<"YES";
    }
    else if(a1+b2<=b && max(b1, a2)<=a)
    {
        cout<<"YES";
    }
    else if(b1+b2<=a && max(a1, a2)<=b)
    {
        cout<<"YES";
    }
    else if(b1+a2<=a && max(a1, b2)<=b)
    {
        cout<<"YES";
    }
    else
    {
        cout<<"NO";
    }
    return 0;
}