【CF簡單介紹】

提交鏈接：http://codeforces.com/contest/560/problem/B

題面：

B. Gerald is into Art time limit per test 2 seconds memory limit per test 256 megabytes input standard input output 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 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?

Input

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 numbersa_i,?b_i in the input are integers and fit into the range from1 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:

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

解題：

??? 就是看能不能把兩塊畫。放進板內(nèi)，要求畫必須平行于板。

代碼：

#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; }

總結

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