The United Nations has decided to build a new headquarters in Saint Petersburg, Russia. It will have a
form of a rectangular parallelepiped and will consist of several rectangular
oors, one on top of another.
Each
oor is a rectangular grid of the same dimensions, each cell of this grid is an office.
Two offices are considered adjacent if they are located on the same
oor and share a common wall,
or if one's
oor is the other's ceiling.
The St. Petersburg building will host n national missions. Each country gets several offices that
form a connected set.
Moreover, modern political situation shows that countries might want to form secret coalitions. For
that to be possible, each pair of countries must have at least one pair of adjacent offices, so that they
can raise the wall or the ceiling they share to perform secret pair-wise negotiations just in case they
need to.
You are hired to design an appropriate building for the UN.
Input
Input consists of several datasets. Each of them has a single integer number n (1 n 50) | the
number of countries that are hosted in the building.
Output
On the rst line of the output for each dataset write three integer numbers h, w, and l | height, width
and length of the building respectively.
h descriptions of
oors should follow. Each
oor description consists of l lines with w characters on
each line. Separate descriptions of adjacent
oors with an empty line.
Use capital and small Latin letters to denote offices of different countries. There should be at most
1 000 000 offices in the building. Each office should be occupied by a country. There should be exactly
n different countries in the building. In this problem the required building design always exists.
Print a blank line between test cases.

 #include<cstdio>
#include<cstring>
char turn(int x)
{
if (x<=) return x+'a'-;
return x-+'A';
}
int main()
{
int i,j,k,m,n,p,q,x,y,z,l,h;
char c;
bool b=;
while (scanf("%d",&n)==)
{
if (b==) b=;
else printf("\n");
printf("2 %d %d\n",n,n);
for (i=;i<=n;i++)
{
for (j=;j<=n;j++)
printf("%c",turn(i));
printf("\n");
}
printf("\n");
for (i=;i<=n;i++)
{
for (j=;j<=n;j++)
printf("%c",turn(j));
printf("\n");
}
}
}

和http://www.cnblogs.com/AwesomeOrion/p/5380752.html这道题一样,只是让你找到一组解。那我只要存心构造万能解即可。

我的方法是:建两层,每一层n*n,第一层横着按顺序排列1..n,第二层竖着按顺序排列1..n,这样每两个国家都会交叉。

uva 1605 building for UN ——yhx的更多相关文章

  1. UVA 1605 Building for UN

    题意: 有n个国家,要求你设计一栋楼并为这n个国家划分房间,要求国家的房间必须连通,且每两个国家之间必须有一间房间是相邻的 分析: 其实非常简单,完全被样例误导了.只需要设计两层就可以了,每个国家占第 ...

  2. UVA 1605 Building for UN(思维)

    题目链接: https://cn.vjudge.net/problem/UVA-1605#author=0 /* 问题 设计一个包含若干层的联合国大厦,其中每一层都是等大的网格,每个格子分配给一个国家 ...

  3. Uva 1605 Building for UN【构造法】

    题意:给出n个国家,给它们分配办公室,使得任意两个国家都有一对相邻的格子 看的紫书,最开始看的时候不理解 后来还是搜了题解--- 发现是这样的 比如说5个国家 应该输出 AAAA BBBB CCCC ...

  4. UVA - 1605 Building for UN (联合国大楼)

    题意:一个联合国大楼每层都有数量相等大小相同的格子,将其分配给n个国家,使任意两个不同的国家都相邻(同层有公共边或相邻层的同一个格子). 分析:可以设计一个只有两层的大楼,第一层每个国家占一行,第二层 ...

  5. 贪心水题。UVA 11636 Hello World,LA 3602 DNA Consensus String,UVA 10970 Big Chocolate,UVA 10340 All in All,UVA 11039 Building Designing

    UVA 11636 Hello World 二的幂答案就是二进制长度减1,不是二的幂答案就是是二进制长度. #include<cstdio> int main() { ; ){ ; ) r ...

  6. UVa 1605 (构造) Building for UN

    题意: 有n个国家,要设计一栋长方体的大楼,使得每个单位方格都属于其中一个国家,而且每个国家都要和其他国家相邻. 分析: 紫书上有一种很巧妙的构造方法: 一共有2层,每层n×n.一层是每行一个国家,另 ...

  7. UVa 11039 - Building designing 贪心,水题 难度: 0

    题目 https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&a ...

  8. UVA 11039 Building designing 贪心

    题目链接:UVA - 11039 题意描述:建筑师设计房子有两条要求:第一,每一层楼的大小一定比此层楼以上的房子尺寸要大:第二,用蓝色和红色为建筑染色,每相邻的两层楼不能染同一种颜色.现在给出楼层数量 ...

  9. UVA 11039 - Building designing(DP)

    题目链接 本质上是DP,但是俩变量就搞定了. #include <cstdio> #include <cstring> #include <algorithm> u ...

随机推荐

  1. 将C1Chart数据导出到Excel

    大多数情况下,当我们说将图表导出到Excel时,意思是将Chart当成图片导出到Excel中.如果是这样,你可以参考帮助文档中保存和导出C1Chart章节. 不过,也有另一种情况,当你想把图表中的数据 ...

  2. wrong requestcode when using startActivityForResult

    You are calling startActivityForResult() from your Fragment. When you do this, the requestCode is ch ...

  3. 泛函编程(13)-无穷数据流-Infinite Stream

    上节我们提到Stream和List的主要分别是在于Stream的“延后计算“(lazy evaluation)特性.我们还讨论过在处理大规模排列数据集时,Stream可以一个一个把数据元素搬进内存并且 ...

  4. 选择使用c语言编写的phalcon框架

    使用这个框架,我总结了如下几点考虑 1.这个框架速度快.纯c语言编写的框架,速度都比php框架快,省去了中间环节.当然,使用它不仅仅是性能考虑.因为如果为了解决php性能问题,完全可以有很多种方式,不 ...

  5. 【iOS】Quartz2D图形上下文

      一.绘图的完整过程 程序启动,显示自定义的view.当程序第一次显示在我们眼前的时候,程序会调用drawRect:方法,在里面获取了图形上下文(在内存中拥有了),然后利用图形上下文保存绘图信息,可 ...

  6. 【翻译】配置RSVP-signaled LSP

    源地址: https://www.juniper.net/techpubs/software/junos-security/junos-security10.2/junos-security-swco ...

  7. 【使用 DOM】使用 Document 对象

    Document 对象时通往DOM功能的入口,它向你提供了当前文档的信息,以及一组可供探索.导航.搜索或操作结构与内容的功能. 我们通过全局变量document访问Document对象,它是浏览器为我 ...

  8. CSS选择器特殊性与重要性

    特殊性 在编写CSS代码的时候,我们会出现多个样式规则作用于同一个元素的情况,例如 <!-- HTML --> <header> <nav class="nav ...

  9. andriod VideoView

    package com.example.yanlei.myyk; import android.media.MediaPlayer; import android.net.Uri; import an ...

  10. android 数据文件存取至储存卡

    来自:http://blog.csdn.net/jianghuiquan/article/details/8569233 <?xml version="1.0" encodi ...