【45.65%】【codeforces 560B】Gerald is into Art
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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?
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 and b3 — 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).
Examples
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.

【题目链接】:http://codeforces.com/contest/560/problem/B
【题解】
先固定第一个矩形在左下角(靠边);这样肯定是最好的方法;
然后枚举第二个矩形要放在哪里;
两个矩形都有两种形态.即竖着放还是横着放;
处理一下就好;
时间复杂度是O(N^2);
->第二个矩形只有判断有没有越界就可以了;
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int MAXN = 1e3+100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
int n,x;
int a1,b1,a2,b2,a3,b3;
bool bo[MAXN][MAXN];
bool get_ans()
{
rep1(i,1,a1)
rep1(j,1,b1)
if (bo[i][j])
{
if (i+a3-1<=a1 && j+b3-1<=b1) return true;
if (i+b3-1<=a1 && j+a3-1<=b1) return true;
// if (get_ans1(i,j,a3,b3)) return true;
// if (get_ans1(i,j,b3,a3)) return true;
}
return false;
}
bool chushi(int ta,int tb)
{
memset(bo,false,sizeof bo);
rep1(i,1,a1)
rep1(j,1,b1)
bo[i][j] = true;
rep1(i,1,ta)
rep1(j,1,tb)
if (!bo[i][j])
return false;
else
bo[i][j] = false;
return true;
}
int main()
{
//freopen("F:\\rush.txt","r",stdin);
rei(a1);rei(b1);
rei(a2);rei(b2);
rei(a3);rei(b3);
if (chushi(a2,b2))
if (get_ans())
{
puts("YES");
return 0;
}
if (chushi(b2,a2))
if (get_ans())
{
puts("YES");
return 0;
}
puts("NO");
return 0;
}
【45.65%】【codeforces 560B】Gerald is into Art的更多相关文章
- 【 BowWow and the Timetable CodeForces - 1204A 】【思维】
题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制 ...
- 【codeforces 734F】Anton and School
[题目链接]:http://codeforces.com/problemset/problem/734/F [题意] 给你两个数组b和c; 然后让你找出一个非负数组a满足题中所给关系; [题解] 有个 ...
- 【codeforces 65A】Harry Potter and Three Spells
[题目链接]:http://codeforces.com/problemset/problem/65/A [题意] 你有3种魔法; 1.可以将a单位的石头变成b单位的铅 2.可以将c单位的铅变成d单位 ...
- 【codeforces 750D】New Year and Fireworks
time limit per test2.5 seconds memory limit per test256 megabytes inputstandard input outputstandard ...
- 【codeforces 415D】Mashmokh and ACM(普通dp)
[codeforces 415D]Mashmokh and ACM 题意:美丽数列定义:对于数列中的每一个i都满足:arr[i+1]%arr[i]==0 输入n,k(1<=n,k<=200 ...
- 【搜索】【并查集】Codeforces 691D Swaps in Permutation
题目链接: http://codeforces.com/problemset/problem/691/D 题目大意: 给一个1到N的排列,M个操作(1<=N,M<=106),每个操作可以交 ...
- 【中途相遇法】【STL】BAPC2014 K Key to Knowledge (Codeforces GYM 100526)
题目链接: http://codeforces.com/gym/100526 http://acm.hunnu.edu.cn/online/?action=problem&type=show& ...
- 【链表】【模拟】Codeforces 706E Working routine
题目链接: http://codeforces.com/problemset/problem/706/E 题目大意: 给一个N*M的矩阵,Q个操作,每次把两个同样大小的子矩阵交换,子矩阵左上角坐标分别 ...
- 【数论】【扩展欧几里得】Codeforces 710D Two Arithmetic Progressions
题目链接: http://codeforces.com/problemset/problem/710/D 题目大意: 两个等差数列a1x+b1和a2x+b2,求L到R区间内重叠的点有几个. 0 < ...
随机推荐
- MYSQL存储过程中 使用变量 做表名--转
原文地址:http://blog.csdn.net/business122/article/details/7528859 今天写一个对数据库做快照的存储过程,用到了动态表名,突然发现MYSQL不支持 ...
- LoadRunner IP欺骗使用
- [Javascript AST] 3. Continue: Write ESLint rule
The rule we want to write is show warning if user using console method: // valid foo.console() conso ...
- 本地 oracle 安装文件夹满触发 ORA-7445 [_memmove()+64] 导致Instance Crashed 的事故
近期处理了一个问题,原因是因为命中ORA-600 [kole_t2u], [34] - description, bugs 导致 在udump 文件夹下大量转储 出cdmp 文件, 然后这些 cdmp ...
- Linux下常用的中文输入法平台有IBus、fcitx和scim
Linux下常用的中文输入法平台有IBus.fcitx和scim.scim现在维护滞后,不推荐使用. IBus ("Intelligent Input Bus") 是一个 输入法框 ...
- lighttpd + php for android 安卓上的WEB服务器
lighttpd + php for android 安卓上的WEBSER 这个项目在 http://hex.ro/wp/blog/php-and-lighttpd-for-android 目前不支持 ...
- DataTable +chart控件
//这是仿你的DataTable //-----开始--------- DataTable dataTable1 = new System.Data.DataTable(); dataTable1.C ...
- crm2013 查看下拉框的选项
在CRM2011中,我们非常easy查看下拉框的选择.打开页面,按F12.把光标对准目标,就会显示出详细的选项,如图:' watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi ...
- git的安装及其使用
在Windows上安装Git的快捷方式: 工具:1.Windows的console工具:ConEmu(https://conemu.github.io/)多窗口.记录log.多theme选择,操作Gi ...
- UVA 11090 Going in Cycle!!(Bellman-Ford推断负圈)
题意:给定一个n个点m条边的加权有向图,求平均权值最小的回路. 思路:使用二分法求解.对于每个枚举值mid,推断每条边权值减去mid后有无负圈就可以. #include<cstdio> # ...