CF581D Three Logos 暴力

最新推荐文章于 2019-06-19 14:14:00 发布
转载 最新推荐文章于 2019-06-19 14:14:00 发布 · 63 阅读 · 0 ·
CC 4.0 BY-SA版权
原文链接：http://www.cnblogs.com/zxyqzy/p/10293805.html

探讨了三个公司如何在正方形广告牌上无重叠地放置各自的品牌标志，且广告牌不留空余空间的问题。通过编程解决方案，检查是否可以满足条件，并展示可能的布局方案。

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Three companies decided to order a billboard with pictures of their logos. A billboard is a big square board. A logo of each company is a rectangle of a non-zero area.

Advertisers will put up the ad only if it is possible to place all three logos on the billboard so that they do not overlap and the billboard has no empty space left. When you put a logo on the billboard, you should rotate it so that the sides were parallel to the sides of the billboard.

Your task is to determine if it is possible to put the logos of all the three companies on some square billboard without breaking any of the described rules.

Input

The first line of the input contains six positive integers x1, y1, x2, y2, x3, y3 (1 ≤ x1, y1, x2, y2, x3, y3 ≤ 100), where xi and yi determine the length and width of the logo of the i-th company respectively.

Output

If it is impossible to place all the three logos on a square shield, print a single integer "-1" (without the quotes).

If it is possible, print in the first line the length of a side of square n, where you can place all the three logos. Each of the next n lines should contain n uppercase English letters "A", "B" or "C". The sets of the same letters should form solid rectangles, provided that:

  • the sizes of the rectangle composed from letters "A" should be equal to the sizes of the logo of the first company,
  • the sizes of the rectangle composed from letters "B" should be equal to the sizes of the logo of the second company,
  • the sizes of the rectangle composed from letters "C" should be equal to the sizes of the logo of the third company,

Note that the logos of the companies can be rotated for printing on the billboard. The billboard mustn't have any empty space. If a square billboard can be filled with the logos in multiple ways, you are allowed to print any of them.

See the samples to better understand the statement.

Examples
Input
Copy
5 1 2 5 5 2
Output
Copy
5
AAAAA
BBBBB
BBBBB
CCCCC
CCCCC
Input
Copy
4 4 2 6 4 2
Output
Copy
6
BBBBBB
BBBBBB
AAAACC
AAAACC
AAAACC
AAAACC
暴力就完事了；
代码应该是看不下去的；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }


/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int h[4], w[4];
char mp[300][300];
char ch[3] = { 'A','B','C' };
int main() {
	//ios::sync_with_stdio(0);
	for (int i = 1; i <= 3; i++) {
		cin >> h[i] >> w[i]; if (h[i] > w[i])swap(h[i], w[i]);
	}
	if ((w[1] == w[2]&&w[2] == w[3]) && h[1] + h[2] + h[3] == w[1]) {
		cout << w[1] << endl;
		for (int i = 1; i <= w[1]; i++) {
			for (int j = 1; j <= h[1]; j++)mp[i][j] = 'A';
		//	cout << endl;
		}
		for (int i = 1; i <= w[1]; i++) {
			for (int j = h[1] + 1; j <= h[2] + h[1]; j++)mp[i][j] = 'B';
		//	cout << endl;
		}
		for (int i = 1; i <= w[1]; i++) {
			for (int j = 1 + h[1] + h[2]; j <= w[1]; j++)mp[i][j] = 'C';
		//	cout << endl;
		}
		for (int i = 1; i <= w[1]; i++) {
			for (int j = 1; j <= w[1]; j++)cout << mp[i][j];
			cout << endl;
		}
		return 0;
	}
	else {
		int pos = 0;
		int maxx = 0;
		for (int i = 1; i <= 3; i++) {
			if (w[i] > maxx) {
				maxx = w[i]; pos = i;
			}
		}
		if (pos == 1) {
			if (h[2] + h[3] == w[1]) {
				if (w[3] == w[2] && w[2] + h[1] == w[1]) {
					cout << w[1] << endl;
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1; j <= h[1]; j++)mp[i][j] = 'A';
						//	cout << endl;
					}
					for (int i = 1; i <= h[2]; i++) {
						for (int j = 1 + h[1]; j <= w[1]; j++)mp[i][j] = 'B';
						//	cout << endl;
					}
					for (int i = 1 + h[2]; i <= w[1]; i++)
						for (int j = 1 + h[1]; j <= w[1]; j++)mp[i][j] = 'C';
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1; j <= w[1]; j++)cout << mp[i][j];
						cout << endl;
					}
					return 0;
				}
			}
			else if (h[2] + w[3] == w[1]) {
				if (w[2] == h[3] && w[2] + h[1] == w[1]) {
					cout << w[1] << endl;
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1; j <= h[1]; j++)mp[i][j] = 'A';
						//	cout << endl;
					}
					for (int i = 1; i <= h[2]; i++) {
						for (int j = 1 + h[1]; j <= w[1]; j++)mp[i][j] = 'B';
						//	cout << endl;
					}
					for (int i = 1 + h[2]; i <= w[1]; i++)
						for (int j = 1 + h[1]; j <= w[1]; j++)mp[i][j] = 'C';
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1; j <= w[1]; j++)cout << mp[i][j];
						cout << endl;
					}
					return 0;
				}
			}
			else if (w[2] + h[3] == w[1]) {
				if (h[2] == w[3] && h[2] + h[1] == w[1]) {
					cout << w[1] << endl;
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1; j <= h[1]; j++)mp[i][j] = 'A';
						//	cout << endl;
					}
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1 + h[1]; j <= w[1]; j++)mp[i][j] = 'B';
						//	cout << endl;
					}
					for (int i = 1 + w[2]; i <= w[1]; i++)
						for (int j = 1 + h[1]; j <= w[1]; j++)mp[i][j] = 'C';
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1; j <= w[1]; j++)cout << mp[i][j];
						cout << endl;
					}
					return 0;
				}
			}
			else if (w[2] + w[3] == w[1]) {
				if (h[2] == h[3] && h[2] + h[1] == w[1]) {
					cout << w[1] << endl;
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1; j <= h[1]; j++)mp[i][j] = 'A';
						//	cout << endl;
					}
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1 + h[1]; j <= w[1]; j++)mp[i][j] = 'B';
						//	cout << endl;
					}
					for (int i = 1 + w[2]; i <= w[1]; i++)
						for (int j = 1 + h[1]; j <= w[1]; j++)mp[i][j] = 'C';
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1; j <= w[1]; j++)cout << mp[i][j];
						cout << endl;
					}
					return 0;
				}
			}
		}

		else if (pos == 2) {
			if (h[1] + h[3] == w[2]) {
				if (w[3] == w[1] && w[1] + h[2] == w[2]) {
					cout << w[2] << endl;
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1; j <= h[2]; j++)mp[i][j] = 'B';
						//	cout << endl;
					}
					for (int i = 1; i <= h[1]; i++) {
						for (int j = 1 + h[2]; j <= w[2]; j++)mp[i][j] = 'A';
						//	cout << endl;
					}
					for (int i = 1 + h[1]; i <= w[2]; i++)
						for (int j = 1 + h[2]; j <= w[2]; j++)mp[i][j] = 'C';
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1; j <= w[2]; j++)cout << mp[i][j];
						cout << endl;
					}
					return 0;
				}
			}
			if (h[1] + w[3] == w[2]) {
				if (h[3] == w[1] && w[1] + h[2] == w[2]) {
					cout << w[2] << endl;
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1; j <= h[2]; j++)mp[i][j] = 'B';
						//	cout << endl;
					}
					for (int i = 1; i <= h[1]; i++) {
						for (int j = 1 + h[2]; j <= w[2]; j++)mp[i][j] = 'A';
						//	cout << endl;
					}
					for (int i = 1 + h[1]; i <= w[2]; i++)
						for (int j = 1 + h[2]; j <= w[2]; j++)mp[i][j] = 'C';
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1; j <= w[2]; j++)cout << mp[i][j];
						cout << endl;
					}
					return 0;
				}
			}
			if (w[1] + h[3] == w[2]) {
				if (w[3] == h[1] && h[1] + h[2] == w[2]) {
					cout << w[2] << endl;
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1; j <= h[2]; j++)mp[i][j] = 'B';
						//	cout << endl;
					}
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1 + h[2]; j <= w[2]; j++)mp[i][j] = 'A';
						//	cout << endl;
					}
					for (int i = 1 + w[1]; i <= w[2]; i++)
						for (int j = 1 + h[2]; j <= w[2]; j++)mp[i][j] = 'C';
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1; j <= w[2]; j++)cout << mp[i][j];
						cout << endl;
					}
					return 0;
				}
			}
			if (w[1] + w[3] == w[2]) {
				if (h[3] == h[1] && h[1] + h[2] == w[2]) {
					cout << w[2] << endl;
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1; j <= h[2]; j++)mp[i][j] = 'B';
						//	cout << endl;
					}
					for (int i = 1; i <= w[1]; i++) {
						for (int j = 1 + h[2]; j <= w[2]; j++)mp[i][j] = 'A';
						//	cout << endl;
					}
					for (int i = 1 + w[1]; i <= w[2]; i++)
						for (int j = 1 + h[2]; j <= w[2]; j++)mp[i][j] = 'C';
					for (int i = 1; i <= w[2]; i++) {
						for (int j = 1; j <= w[2]; j++)cout << mp[i][j];
						cout << endl;
					}
					return 0;
				}
			}
		}
		else {
		if (h[1] + h[2] == w[3]) {
			if (w[2] == w[1] && w[1] + h[3] == w[3]) {
				cout << w[3] << endl;
				for (int i = 1; i <= w[3]; i++) {
					for (int j = 1; j <= h[3]; j++)mp[i][j] = 'C';
					//	cout << endl;
				}
				for (int i = 1; i <= h[1]; i++) {
					for (int j = 1 + h[3]; j <= w[3]; j++)mp[i][j] = 'A';
					//	cout << endl;
				}
				for (int i = 1 + h[1]; i <= w[3]; i++)
					for (int j = 1 + h[3]; j <= w[3]; j++)mp[i][j] = 'B';
				for (int i = 1; i <= w[3]; i++) {
					for (int j = 1; j <= w[3]; j++)cout << mp[i][j];
					cout << endl;
				}
				return 0;
			}
		}
		if (h[1] + w[2] == w[3]) {
			if (h[2] == w[1] && w[1] + h[3] == w[3]) {
				cout << w[3] << endl;
				for (int i = 1; i <= w[3]; i++) {
					for (int j = 1; j <= h[3]; j++)mp[i][j] = 'C';
					//	cout << endl;
				}
				for (int i = 1; i <= h[1]; i++) {
					for (int j = 1 + h[2]; j <= w[3]; j++)mp[i][j] = 'A';
					//	cout << endl;
				}
				for (int i = 1 + h[1]; i <= w[3]; i++)
					for (int j = 1 + h[3]; j <= w[3]; j++)mp[i][j] = 'B';
				for (int i = 1; i <= w[3]; i++) {
					for (int j = 1; j <= w[3]; j++)cout << mp[i][j];
					cout << endl;
				}
				return 0;
			}
		}
		if (w[1] + h[2] == w[3]) {
			if (w[2] == h[1] && h[1] + h[3] == w[3]) {
				cout << w[3] << endl;
				for (int i = 1; i <= w[3]; i++) {
					for (int j = 1; j <= h[3]; j++)mp[i][j] = 'C';
					//	cout << endl;
				}
				for (int i = 1; i <= w[1]; i++) {
					for (int j = 1 + h[3]; j <= w[3]; j++)mp[i][j] = 'A';
					//	cout << endl;
				}
				for (int i = 1 + w[1]; i <= w[3]; i++)
					for (int j = 1 + h[3]; j <= w[3]; j++)mp[i][j] = 'B';
				for (int i = 1; i <= w[3]; i++) {
					for (int j = 1; j <= w[3]; j++)cout << mp[i][j];
					cout << endl;
				}
				return 0;
			}
		}
		if (w[1] + w[2] == w[3]) {
			if (h[2] == h[1] && h[1] + h[3] == w[3]) {
				cout << w[3] << endl;
				for (int i = 1; i <= w[3]; i++) {
					for (int j = 1; j <= h[3]; j++)mp[i][j] = 'C';
					//	cout << endl;
				}
				for (int i = 1; i <= w[1]; i++) {
					for (int j = 1 + h[3]; j <= w[3]; j++)mp[i][j] = 'A';
					//	cout << endl;
				}
				for (int i = 1 + w[1]; i <= w[3]; i++)
					for (int j = 1 + h[3]; j <= w[3]; j++)mp[i][j] = 'B';
				for (int i = 1; i <= w[3]; i++) {
					for (int j = 1; j <= w[3]; j++)cout << mp[i][j];
					cout << endl;
				}
				return 0;
			}
		}
		}

	}
	cout << -1 << endl;
	return 0;
}

 









转载于:https://www.cnblogs.com/zxyqzy/p/10293805.html

                

        




    

    
  



    



                



	

		

			

				
					
logos

				
			

			

				

					03-19
				

			

		

		

			
				
在IT行业中，"logos"通常指的是标志或标识，它是公司、品牌或产品视觉识别系统的核心元素。一个精心设计的logo能够传达出品牌的价值观、理念以及与目标受众的关联。下面，我们将深入探讨logo设计的相关知识点，以及...

			
		

	



                

            
              



	

		

			

				
					
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"Logos"通常指的是徽标或标志，是公司、品牌、产品或组织视觉识别系统的重要组成部分。在IT行业中，设计一个有效的Logo对于建立品牌形象、传达公司理念和吸引目标受众至关重要。下面将详细介绍Logo设计的相关知识点...

			
		

	



              


  
              

                


	

		

			

				
					
Codeforces 581D  Three Logos

				
			

			

				

					weixin_30341745的博客
				

				

					06-19
					
					119
					
				

			

		

		

			
				
Codeforces Tutorial
D. Three Logos
Problem Analysis
从结果出发进行构造
从下面的图可以看出每添加一个logo，剩余的一定是一个矩形，否则无解
维护剩余矩形的左上角点的坐标\((px,py)\)，右下角点的坐标为\((n,n)\)，\(n\)为最大边长
每加入一个矩形，更新状态。
\[
\begin{matrix}
B&B&B&am...

			
		

	





	

		

			

				
					
[Codeforces 581D]Three Logos

				
			

			

				

					
				

				

					10-20
					
					134
					
				

			

		

		

			
				
题目大意：给你三个矩阵，分别是A、B、C。现在要你贴在一个正方形上，要求不重叠且叠满，问你正方形边长是多少，并输出一种张贴方式。如果无法满足要求，输出-1。
解题思路：一道模拟题。打比赛的时候由于时间不够，代码打得很急，所以比较乱，比较浪费时空。好在数据范围小，乱搞也可以AC。
由于矩阵可以旋转，所以实际上只可能是这两种摆法：
①；②
其他可能方案均可通过旋转或交换位置得到。
...

			
		

	



		

			
		



	

		

			

				
					
Codeforces 581D Three Logos

				
			

			

				

					
				

				

					10-03
					
					813
					
				

			

		

		

			
				
D. Three Logos


time limit per test
1 second


memory limit per test
256 megabytes


input
standard input


output
standard output



Three companies decided to order a billboard 

			
		

	





	

		

			

				
					
CF_#322(Div.2) D. Three Logos(greedy)

				
			

			

				

					EaShion1994的专栏
				

				

					10-06
					
					716
					
				

			

		

		

			
				
Three companies decided to order a billboard with pictures of their logos. A billboard is a big square board. A logo of each company is a rectangle of a non-zero area.

Advertisers will put up the a

			
		

	





	

		

			

				
					
《Logos 系列产品 HMIC-H IP用户指南》


					
最新发布

				
			

			

				

					09-08
				

			

		

		

			
				
《Logos 系列产品 HMIC-H IP用户指南》是一份由深圳市紫光同创电子有限公司编写的详细技术文档，主要针对其SDRAM/DDR/DDR2/DDR3 IP核的使用进行指导。这份用户指南在多个版本中进行了更新，以适应产品的发展和改进，...

			
		

	





	

		

			

				
					
D. Three Logos （CF Round #322 (Div.2) 瞎搞 分情况）

				
			

			

				

					@you！
				

				

					09-28
					
					610
					
				

			

		

		

			
				
题意：三个logo，给出他们的长宽，问他们能不能无缝放在一个正方形内。
思路：5种情况在纸上画一下就知道了。

			
		

	





	

		

			

				
					
cf581D Three Logos

				
			

			

				

					weixin_30722589的博客
				

				

					09-28
					
					120
					
				

			

		

		

			
				
Three companies decided to order a billboard with pictures of their logos. A billboard is a bigsquareboard. A logo of each company is a rectangle of a non-zero area.
Advertisers will put ...

			
		

	





	

		

			

				
					
CF 591D Three Logos 暴力,模拟

				
			

			

				

					noone0的博客
				

				

					04-06
					
					551
					
				

			

		

		

			
				
题意:给出三个矩形的长和宽(x[i],y[i]),问用这三个矩形是否能拼成一个正方形(矩形可以旋转)
输出正方形的边长和拼成的方案.


若sqrt(矩形面积和)正好为某个整数n 则边长为n.
因为只有3个矩形,在左上角放一个以后,剩下两个部分必须填满 暴力判断即可.
#include 
using namespace std;
const int N=2e3+5;
int n=-1,

			
		

	





	

		

			

				
					
CodeForces 581 D.Three Logos（计算几何）

				
			

			

				

					ZSQ
				

				

					10-09
					
					726
					
				

			

		

		

			
				
Description 
给出三种海报A,B,C的长和宽，问能否将这三张海报毫无缝隙的放在一个正方形的框中，如果可以则输出这个正方形框的边长并将贴完海报的这个正方形矩阵输出，否则输出-1 
Input 
六个整数分别表示这三张海报的长和宽 
Output 
如果存在合法的贴海报方案则输出，否则输出-1 
Sample Input 
5 1 2 5 5 2 
Sample Output 
5 
AA...

			
		

	





	

		

			

				
					
Three Logos（模拟）

				
			

			

				

					梧桐下的四叶草
				

				

					09-29
					
					571
					
				

			

		

		

			
				
D. Three Logos


time limit per test
1 second


memory limit per test
256 megabytes


input
standard input


output
standard output



Three companies decided to order a billboard

			
		

	





	

		

			

				
					
Three Logos CodeForces - 581D

				
			

			

				

					sunyutian1998的博客
				

				

					09-06
					
					304
					
				

			

		

		

			
				
点击打开链接

模拟题 写的很难受 提心吊胆的看着status上跑过最后几个样例。。

若三个矩形符合题意 三者中必有一个的长或宽等于正方形边长 将该矩形对准放到最上面即可 六种可能分类讨论

剩下两个矩形必然各自有一边相等 对准拼成一个大矩形 看能不能和第一个矩形凑成正方形

思路应该很简单 感觉实现起来比较棘手 还是太渣了

 


#include&lt;bits/stdc++.h&gt;
...

			
		

	





	

		

			

				
					
Codeforces 581D - Three Logos (枚举)

				
			

			

				

					一只冰熊
				

				

					09-28
					
					875
					
				

			

		

		

			
				
RT

			
		

	





	

		

			

				
					
Codeforces Round #322 D.Three Logos

				
			

			

				

					_zidaoziyan的博客
				

				

					09-29
					
					566
					
				

			

		

		

			
				
D. Three Logos

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output


Three companies decided to order a billboard with pictu

			
		

	





	

		

			

				
					
cf581d

				
			

			

				

					渣渣日常
				

				

					09-29
					
					1411
					
				

			

		

		

			
				
http://codeforces.com/contest/581/problem/D
两种可能  3横或者2横1竖  三横直接判断  两横一竖先找横 在分别枚举竖
代码好丑。
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include

			
		

	





	

		

			

				
					
CallFavorites：Logos功能的详细介绍

				
			

			

				

					
				

			

		

		

			
				
同时，由于标签是“Logos”，可能暗示该主题与品牌或身份标志相关。文件名称“CallFavorites-main”可能表明这是一个主要文件或程序的主文件。  由于信息有限，我们将尝试通过以下假设进行扩展：  1. ...

			
		

	



              




          




        



    





    



      

      

        
      

      





	

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