CF581D Three Logos 暴力

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 x_i and y_i 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;

}

CF581D Three Logos 暴力的更多相关文章

cf581D Three Logos
Three companies decided to order a billboard with pictures of their logos. A billboard is a big squa ...
Codeforces Round #322 (Div. 2) D. Three Logos 暴力
D. Three Logos Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/581/problem ...
hdu2328 Corporate Identity【string库使用】【暴力】【KMP】
Corporate Identity Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
（ＫＭＰ　暴力）Corporate Identity　－－　ｈｄｕ　－－　２３２８
http://acm.hdu.edu.cn/showproblem.php?pid=2328 Corporate Identity Time Limit: 9000/3000 MS (Java/Oth ...
zone.js - 暴力之美
在ng2的开发过程中,Angular团队为我们带来了一个新的库 – zone.js.zone.js的设计灵感来源于Dart语言,它描述JavaScript执行过程的上下文,可以在异步任务之间进行持久性 ...
[bzoj3123][sdoi2013森林] (树上主席树+lca+并查集启发式合并+暴力重构森林)
Description Input 第一行包含一个正整数testcase,表示当前测试数据的测试点编号.保证1≤testcase≤20. 第二行包含三个整数N,M,T,分别表示节点数.初始边数.操作数 ...
HDU 5944 Fxx and string（暴力/枚举）
传送门 Fxx and string Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Othe ...
1250 Super Fast Fourier Transform(湘潭邀请赛暴力思维)
湘潭邀请赛的一题,名字叫"超级FFT"最终暴力就行,还是思维不够灵活,要吸取教训. 由于每组数据总量只有1e5这个级别,和不超过1e6,故先预处理再暴力即可. #include&l ...
fragment+viepager 的简单暴力的切换方式
这里是自定义了一个方法来获取viewpager private static ViewPager viewPager; public static ViewPager getMyViewPager() ...

随机推荐

Solaris10 如何设置空闲ssh连接超时断开
在ssh的配置文件中有2个参数可以控制空闲连接超时断开.这2个参数是ClientAliveCountMax和ClientAliveInterval. Solaris10上设置空闲ssh连接超时断开的方 ...
Delphi Cookie
Cookie IdHTTP1.CookieManager.AddCookies(); IdHTTP1.Post(); IdHTTP1.Get('http://1.1.1.1:9000/'); for ...
logstash日志写入kafka
安装kafka curl -L -O https://mirrors.cnnic.cn/apache/kafka/0.10.2.1/kafka_2.10-0.10.2.1.tgz tar xf kaf ...
LAMP 3.3 mysql常用操作-1
有一个图形化管理 mysql 的工具叫做 phpmyadmin,如何在命令行下面来管理和操作 mysql. 首先进入mysql mysql -uroot -pwangshaojun 查看有那些库 &g ...
setAttribute这个方法
setAttribute这个方法,在JSP内置对象session和request都有这个方法,这个方法作用就是保存数据,然后还可以用getAttribute方法来取出.比如现在又个User对象,Use ...
java中sleep和join和yield和wait和notify的区别
1.sleep() 使当前线程(即调用该方法的线程)暂停执行一段时间,让其他线程有机会继续执行,但它并不释放对象锁.也就是说如果有synchronized同步快,其他线程仍然不能访问共享数据.注意该方 ...
【oracle】首次启动SQL Developer配置java.exe出错（Could not find jvm.cfg! ）
1.环境 win7/8/8.1 x64,Oracle 11g r2,jdk7 x64 2.问题第一次启动Oracle SQL Developer的时候会让我们填写Java.exe的路径,我在jdk ...
解析IFC数据并转成json格式
{ "com.bim.ifc.ifc2x3.ifc2x3tc1.IfcBuilding (#104)-": [{ "objKey": "GlobalI ...
PCL—点云分割（基于凹凸性）低层次点云处理
博客转载自:http://www.cnblogs.com/ironstark/p/5027269.html 1.图像分割的两条思路场景分割时机器视觉中的重要任务,尤其对家庭机器人而言,优秀的场景分割 ...
Paxos在大型系统中的应用场景
https://timyang.net/distributed/paxos-scenarios/ 在分布式算法领域,有位非常重要的短发叫Paxos,它的重要性有多高呢?Google的Chubby[1] ...

CF581D Three Logos 暴力

CF581D Three Logos 暴力的更多相关文章

随机推荐

热门专题