标签：each blank ota problem section other aced ges auc
提交链接：http://codeforces.com/contest/560/problem/B
题面：
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 ana1?×?b1 rectangle, the paintings have shape of aa2?×?b2 anda3?×?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 andb1 — the sides of the board. Next two lines contain numbersa2,?b2,?a3 andb3 — the sides of the paintings. All numbersai,?bi in the input are integers and fit into the range from1 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.
就是看能不能把两块画。放进板内，要求画必须平行于板。
代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int main()
{
int l1,w1,l2,w2,l3,w3;
scanf("%d%d",&l1,&w1);
scanf("%d%d",&l2,&w2);
scanf("%d%d",&l3,&w3);
bool flag=false;
if((l2+l3)<=l1&&(w2<=w1&&w3<=w1))
flag=true;
else if((l2+w3)<=l1&&(w2<=w1)&&(l3<=w1))
flag=true;
else if((w2+l3)<=l1&&(l2<=w1)&&(w3<=w1))
flag=true;
else if((w2+w3)<=l1&&(l2<=w1&&l3<=w1))
flag=true;
else if((w2+w3)<=w1&&(l2<=l1)&&(l3<=l1))
flag=true;
else if((w2+l3)<=w1&&(l2<=l1)&&(w3<=l1))
flag=true;
else if((l2+w3)<=w1&&(w2<=l1)&&(l3<=l1))
flag=true;
else if((l2+l3)<=w1&&(w2<=l1)&&(w3<=l1))
flag=true;
if(flag)printf("YES\n");
else printf("NO\n");
return 0;
} 【打CF，学算法——二星级】Codeforces Round #313 (Div. 2) B. Gerald is into Art（水题）
标签：each blank ota problem section other aced ges auc
原文地址：http://www.cnblogs.com/llguanli/p/6761259.html
