【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
- Android开发经验一判断当前屏幕是全屏还是非全屏
public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView ...
- android图片特效处理之光晕效果
这篇将讲到图片特效处理的图片光晕效果.跟前面一样是对像素点进行处理,本篇实现的思路可参见android图像处理系列之九--图片特效处理之二-模糊效果和android图像处理系列之十三--图片特效处理之 ...
- IIS特殊字符设置
简介:[iis7]请求筛选模块被配置为拒绝包含双重转义序列的请求.HTTP 错误 404.11 - Not Found 特殊字符最好替换成其他的字符,主要的特殊字符有”*”.”&”.”%”.” ...
- 计科1111-1114班第一次实验作业(NPC问题——回溯算法、聚类分析)
实验课安排 地点: 科技楼423 时间: 计科3-4班---15周周一上午.周二下午 计科1-2班---15周周一下午.周二晚上(晚上时间从18:30-21:10) 请各班学委在实验课前飞信通知大家 ...
- h.264硬件解码
// H264HWDecoder.m // H264EncoderDecoder // // Created by lujunjie on 2016/11/28. // Copyright © 201 ...
- Maven搭建Spring Security3.2项目详解
本来是打算在上一篇SpringMVC+Hibernate上写的,结果发现上面那篇 一起整合的,结果发现上一篇内容实在是太长了,就另起一篇,这篇主要是采用 Maven搭建Spring+SpringMVC ...
- 【河南省多校脸萌第六场 A】分班级
[链接]点击打开链接 [题意] 在这里写题意 [题解] 最大的给了最小的,实际上就对应了,最大值减1,最小值加1. 那么二分最后班级人数最小的最大可能是几->temp1; 二分最后班级人数最大的 ...
- 右键菜单添加带图标的Notepad++
给Notepad++ 加带图标右键菜单 方式一: 拷贝以下代码建立一个reg文件,替换相关路径,保存,双击运行加入注册表 Windows Registry Editor Version 5.00 [H ...
- 【Codeforces Round #442 (Div. 2) B】Nikita and string
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 枚举中间那一段从哪里开始.哪里结束就好 注意为空的话,就全是a. 用前缀和优化一下. [代码] #include <bits/ ...
- 【CS Round #48 (Div. 2 only)】Game of Chance
[链接]h在这里写链接 [题意] 在这里写题意 [题解] 在这里写题解 [错的次数] 0 [反思] 在这了写反思 [代码] #include <bits/stdc++.h> using n ...