Poker Face


Time Limit: 2 Seconds      Memory Limit: 65536 KB

As is known to all, coders are lack of exercise and Kato is one of them. In order to get fit, he decides to go mining in MINECRAFT. However, it is always dangerous in the mine because there are a lot of ZOMBIEs in MINECRAFT. They kill Kato time and time again when he is mining. But this time, Kato is in invincible mode and wants to observe what the ZOMBIEs' faces are like. After his obervation, he found that the ZOMBIEs' faces are regular. Now he provides you the ZOMBIEs' faces of size 8, 16 and 32, it is your task to draw the other ZOMBIEs. You should find the recursivepattern from the small cases and learn to draw bigger cases.

Input

There are multiple test cases. For each case there is only a number N (8 ≤ N≤ 1024, N is a power of 2), which is size of the ZOMBIE's face. The input ends with an integer less than eight.

Output

For each case, output the image of the ZOMBIE's face. And print an empty line after each case.

Sample Input

8
16
32
0

Sample Output

********
*** ***
*** ***
*** ***
* **** *
* * * *
* * * *
******** ****************
* *
* **** **** *
* * * * * *
* * * * * *
* * * * * *
* **** **** *
* *
* ******** *
* * * * * *
* * * * * *
* * **** * *
* *** *** *
* *** *** *
* *** *** *
**************** ********************************
* *
* *
* *
* ******** ******** *
* * * * * *
* * * * * *
* * * * * *
* * * * * *
* * * * * *
* * * * * *
* * * * * *
* ******** ******** *
* *
* *
* *
* **************** *
* * *** *** * *
* * *** *** * *
* * *** *** * *
* * * **** * * *
* * * * * * * *
* * * * * * * *
* * ******** * *
* * * *
* * **** **** * *
* * * * * * * *
* * * * * * * *
* * * * * * * *
* * **** **** * *
* * * *
********************************

Hint

The sample out put is the ZOMBIE's face of size 8, 16, 32.

Author: ZHU, Heming

解题:递推打印就是了。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define INF 0x3f3f3f3f
#define pii pair<int,int>
using namespace std;
const int maxn = ;
int n;
char table[][maxn][maxn];
void printLine(int m,int n,int x,int y) {
for(int i = ; i < n; ++i)
table[m][x][y+i] = '*';
}
void printR(int m,int n,int x,int y) {
for(int i = ; i < n-; ++i) {
table[m][x+i][y] = '*';
for(int j = ; j < n-; ++j)
table[m][x+i][y+j] = ' ';
table[m][x+i][y+n-] = '*';
}
}
void printG(int m,int n) {
for(int i = ; i < n; ++i) {
for(int j = ; j < n; ++j)
putchar(table[m][i][j]);
putchar('\n');
}
}
void printB(int n,int x,int y,int m) {
for(int i = n-; i >= ; --i) {
for(int j = ; j < n; ++j) {
table[m][x+i][y+j] = table[m-][i][j];
}
}
}
void draw(int m,int n) {
printLine(m,n,,);
printR(m,n,,);
printLine(m,n,n-,); printLine(m,n>>,n>>,n>>);
printR(m,n>>,(n>>)+,n>>);
printLine(m,n>>,(n>>)+(n>>),n>>); printLine(m,n>>,n>>,n-(n>>)-(n>>));
printR(m,n>>,(n>>)+,n-(n>>)-(n>>));
printLine(m,n>>,(n>>)+(n>>),n-(n>>)-(n>>));
}
void draw2(int m,int n) {
int y = n>>;
int x = (n>>);
for(int i = (n>>)-; i >= ; --i,x++) {
for(int j = ; j < (n>>); ++j)
table[m][x][y+j] = table[m-][i][j];
}
}
int main() {
memset(table,' ',sizeof(table));
draw(,<<);
printLine(,,,);
table[][][] = table[][][] = '*';
table[][][] = table[][][] = '*';
for(int i = ; i <= ; ++i) {
draw(i,<<i);
draw2(i,<<i);
}
while(scanf("%d",&n),n >= ) {
int k = log2(n)+0.5;
printG(k,n);
putchar('\n');
}
return ;
}

137 - ZOJ Monthly, November 2014 - J Poker Face的更多相关文章

  1. ZOJ Monthly, November 2014

    做了一次月赛,没想到这么难,加上后来补上的题目也只有3个题.第一名也只有4个题啊啊啊啊~.其中两道还是水题.留坑慢慢补上来. 3832 Tilt Cylinder 给定如图所示有盖圆柱体,R,H,水面 ...

  2. 浙大月赛ZOJ Monthly, August 2014

    Abs Problem Time Limit: 2 Seconds Memory Limit: 65536 KB Special Judge Alice and Bob is playing a ga ...

  3. 135 - ZOJ Monthly, August 2014

    135 - ZOJ Monthly, August 2014 A:构造问题,推断序列奇偶性.非常easy发现最小值不是1就是0.最大值不是n就是n - 1,注意细节去构造就可以 E:dp.dp[i][ ...

  4. ZOJ Monthly, November 2012

    A.ZOJ 3666 Alice and Bob 组合博弈,SG函数应用 #include<vector> #include<cstdio> #include<cstri ...

  5. ZOJ Monthly, August 2014

    A Abs Problem http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5330 找规律题,构造出解.copyright@ts ...

  6. ZOJ Monthly, June 2014 月赛BCDEFGH题题解

    比赛链接:点击打开链接 上来先搞了f.c,,然后发现状态不正确,一下午都是脑洞大开,, 无脑wa,无脑ce...一样的错犯2次.. 硬着头皮搞了几发,最后20分钟码了一下G,不知道为什么把1直接当成不 ...

  7. 记次浙大月赛 134 - ZOJ Monthly, June 2014

    链接 虽做出的很少,也记录下来,留着以后来补..浙大题目质量还是很高的 B 并查集的一些操作,同类和不同类我是根据到根节点距离的奇偶判断的,删点是直接新加一个点,记得福大月赛也做过类似的,并差集的这类 ...

  8. ZOJ Monthly, June 2014 解题报告

    A.Another Recurrence Sequence problemId=5287">B.Gears 题目大意:有n个齿轮,一開始各自为一组.之后进行m次操作,包含下面4种类型: ...

  9. ZOJ 4010 Neighboring Characters(ZOJ Monthly, March 2018 Problem G,字符串匹配)

    题目链接  ZOJ Monthly, March 2018 Problem G 题意  给定一个字符串.现在求一个下标范围$[0, n - 1]$的$01$序列$f$.$f[x] = 1$表示存在一种 ...

随机推荐

  1. STM8S103 解决Rom空间不足 & Map文件分析

    STM8S103只有8KRom,很容易造成空间不足.对于空间不足,我们就要从map文件着手分析,究竟哪些函数占了多少空间,map文件分为几部分:Segments(总括了各个段所占的空间), Modul ...

  2. ADO.NET之断开数据连接的数据库操作

    在ADO.NET对数据库操作时有两种方式一种时与数据库实时连接,第二种时断开连接的操作. 断开连接的操作使用SqlDataAdapter来实现,我们要把数据库中的表数据加载到winform中的data ...

  3. 连类比事-category和extension

    extension看起来很像一个匿名的category,但是extension和有名字的category几乎完全是两个东西. extension在编译期决议,它就是类的一部分,在编译期和头文件里的@i ...

  4. python 之 MRO 异常

    今天突然遇到这个异常,先贴两个地址,待有时间写博客 https://www.jianshu.com/p/fea6e0a0cc14 https://makina-corpus.com/blog/meti ...

  5. 由Request Method:OPTIONS初窥CORS(转)

    刚接触前端的时候,以为HTTP的Request Method只有GET与POST两种,后来才了解到,原来还有HEAD.PUT.DELETE.OPTIONS…… 目前的工作中,HEAD.PUT.DELE ...

  6. [NOI2014]动物园(KMP)

    题意 题解 因为,一直用j=nxt[j]来遍历,可以遍历前i个字符所有相等的前后缀长度,所以有一个暴力的想法,就是对于每一个长度,开始遍历,记录长度小于i/2的相等的前后缀数量,最后累加即可. 但显然 ...

  7. SFTP使用key文件登录

    命令: sftp -oPort= -oIdentityFile=/root/.ssh/user.priv user@39.39.100.100 命令选项: -o ssh_option Can be u ...

  8. virtualbox挂载目录失败mount: unknown filesystem type ‘vboxsf’

    转自[https://blog.csdn.net/u011486871/article/details/79443375] [有小修改] 错误提示:Vagrant was unable to moun ...

  9. [terry笔记]data guard基础知识

    如下介绍了data guard的基础知识,整理自网络: Data Gurad 通过冗余数据来提供数据保护,Data Gurad 通过日志同步机制保证冗余数据和主数据之前的同步,这种同步可以是实时,延时 ...

  10. HDU 3108 Ant Trip

    Ant Trip Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Su ...