#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
using namespace std; int main(){
//initialize for positive number
char h[][];
strcpy(h[], "*****..***..*...*.*****...*...*.*****.*****.***...*****.*...*");
strcpy(h[], "*.....*...*.*...*.*.......*...*.*...*...*...*..*..*...*..*.*.");
strcpy(h[], "*****.*****.*...*.***.....*****.*****...*...*...*.*...*...*..");
strcpy(h[], "....*.*...*..*.*..*.......*...*.*.*.....*...*..*..*...*...*..");
strcpy(h[], "*****.*...*...*...*****...*...*.*..**.*****.***...*****...*.."); //initialize for negative number
char v[][];
strcpy(v[], "*****");
strcpy(v[], "*....");
strcpy(v[], "*****");
strcpy(v[], "....*");
strcpy(v[], "*****");
strcpy(v[], ".....");
strcpy(v[], ".***.");
strcpy(v[], "*...*");
strcpy(v[], "*****");
strcpy(v[], "*...*");
strcpy(v[],"*...*");
strcpy(v[],".....");
strcpy(v[],"*...*");
strcpy(v[],"*...*");
strcpy(v[],"*...*");
strcpy(v[],".*.*.");
strcpy(v[],"..*..");
strcpy(v[],".....");
strcpy(v[],"*****");
strcpy(v[],"*....");
strcpy(v[],"***..");
strcpy(v[],"*....");
strcpy(v[],"*****");
strcpy(v[],".....");
strcpy(v[],".....");
strcpy(v[],".....");
strcpy(v[],"*...*");
strcpy(v[],"*...*");
strcpy(v[],"*****");
strcpy(v[],"*...*");
strcpy(v[],"*...*");
strcpy(v[],".....");
strcpy(v[],"*****");
strcpy(v[],"*...*");
strcpy(v[],"*****");
strcpy(v[],"*.*..");
strcpy(v[],"*..**");
strcpy(v[],".....");
strcpy(v[],"*****");
strcpy(v[],"..*..");
strcpy(v[],"..*..");
strcpy(v[],"..*..");
strcpy(v[],"*****");
strcpy(v[],".....");
strcpy(v[],"***..");
strcpy(v[],"*..*.");
strcpy(v[],"*...*");
strcpy(v[],"*..*.");
strcpy(v[],"***..");
strcpy(v[],".....");
strcpy(v[],"*****");
strcpy(v[],"*...*");
strcpy(v[],"*...*");
strcpy(v[],"*...*");
strcpy(v[],"*****");
strcpy(v[],".....");
strcpy(v[],"*...*");
strcpy(v[],".*.*.");
strcpy(v[],"..*..");
strcpy(v[],"..*..");
strcpy(v[],"..*.."); bool flag = false, pos;
int n, m, i, j, k;
while (cin >> n){
if (n == ) break;
if (flag)
cout << endl << endl; flag = true;
pos = true;
if (n < ){
pos = false;
n = abs(n);
}
//output for positive number
if (pos){
for (i = ; i < ; ++i){
m = ;
while (m < n){
for (j = ; h[i][j]; ++j){
k = ;
while (k < n){
cout << h[i][j];
++k;
}
}
++m;
cout << endl;
}
}
}
else{
for (i = ; i < ; ++i){
m = ;
while (m < n){
for (j = ; v[i][j]; ++j){
k = ;
while (k < n){
cout << v[i][j];
++k;
}
}
++m;
cout << endl;
}
}
}
}
cout << endl << endl;
return ;
}
 //画图题,直接打表
#include <stdio.h> char h[][] =
{
"*****..***..**.*******.*****.*****.********.**",
"*..**.**.*.**.****..*..**..*.*.",
"*****.*****.**.***..*****.********.***..",
".*.**..*.*..*.**.*.*..**..*..***..",
"*****.**********.*..**.*****.*********.."
}; char v[][] =
{
"*****",
"*.",
"*****",
".*",
"*****",
"..",
".***.",
"**",
"*****",
"**",
"**",
"..",
"**",
"**",
"**",
".*.*.",
"..*..",
"..",
"*****",
"*.",
"***..",
"*.",
"*****",
"..",
"..",
"..",
"**",
"**",
"*****",
"**",
"**",
"..",
"*****",
"**",
"*****",
"*.*..",
"*..**",
"..",
"*****",
"..*..",
"..*..",
"..*..",
"*****",
"..",
"***..",
"*..*.",
"**",
"*..*.",
"***..",
"..",
"*****",
"**",
"**",
"**",
"*****",
"..",
"**",
".*.*.",
"..*..",
"..*..",
"..*.."
}; void draw(int n)
{
if(n < )
{
n = -n;
for(int i = ; i < * n; i++)
{
for(int j = ; j < * n; j++)
{
printf("%c", v[i / n][j / n]);
}
printf("\n");
}
}
else
{
for(int i = ; i < * n; i++)
{
for(int j = ; j < * n; j++)
{
printf("%c", h[i / n][j / n]);
}
printf("\n");
}
}
printf("\n\n");
} int main()
{
int n;
while(scanf("%d", &n), n) draw(n);
return ;
}

uva 10894 - Save Hridoy的更多相关文章

  1. UVA - 11986 Save from Radiation

    Description J Save from Radiation Most of you are aware of Nuclear Power Plant Explosion at Fukushim ...

  2. UVa 12403 - Save Setu

    题目:有两种操作:1.当前数值添加.2.输出当前数值. 分析:简单题.模拟就可以. 说明:A+B. #include <iostream> #include <cstdlib> ...

  3. UVA 1291 十四 Dance Dance Revolution

    Dance Dance Revolution Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Su ...

  4. UVA 322 ships (POJ 1138)

    题目地址: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem ...

  5. UVA它11292 - Dragon of Loowater

    Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance tur ...

  6. UVA 257 - Palinwords(弦HASH)

    UVA 257 - Palinwords 题目链接 题意:输出一个文本里面的palinword,palinword的定义为.包括两个不同的回文子串,而且要求回文子串不能互相包括 思路:对于每一个单词推 ...

  7. UVA 11292 Dragon of Loowater(简单贪心)

    Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance tur ...

  8. UVA 1341 - Different Digits(数论)

    UVA 1341 - Different Digits 题目链接 题意:给定一个正整数n.求一个kn使得kn上用的数字最少.假设同样,则输出值最小的 思路: 首先利用鸽笼原理证明最多须要2个数字去组成 ...

  9. Risk UVA - 12264 拆点法+最大流+二分 最少流量的节点流量尽量多。

    /** 题目:Risk UVA - 12264 链接:https://vjudge.net/problem/UVA-12264 题意:给n个点的无权无向图(n<=100),每个点有一个非负数ai ...

随机推荐

  1. Android Framework------之Property子系统

    概述 Property是Android系统中一个重要的概念,在Android系统内,主要用于系统配置,以及不同服务间的简单信息分享.比如设备名字,蓝牙名字,编译信息,网络dns地址,以及其他的一些基本 ...

  2. 内联汇编和JMP到内联函数注意事项

    对于jmp类型的hook, 如果自己的过程没有使用_declspec(naked),那么系统会自动给添加一些额外的代码,控制堆栈平衡,但是这些额外的代码会破坏被hook函数的堆栈. 对于call类型的 ...

  3. BestCoder Round #36 [B] Gunner

    题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=5199 先对树的高度排序,然后对每次射击高度二分查找即可,打过之后数目变为0. #include< ...

  4. BZOJ1119: [POI2009]SLO

    1119: [POI2009]SLO Time Limit: 30 Sec  Memory Limit: 162 MBSubmit: 379  Solved: 181[Submit][Status] ...

  5. 自写 jQuery 大幅弹窗广告插件(不喜勿拍)

    最近写了做的两个项目都要做几乎同一件事,在首页弹出一个广告.本来是想在网上找一个的,找了几个,花了时间但都不怎么满意,尼玛呀,坑爹呀…… 最后一想,干脆自己动手了. 第一次写,在网上找一些例子来看. ...

  6. Makefile如何通过宏开关进行条件编译

    在开发中经常会遇到需要条件编译一段代码,即: #ifdef DEBUG { 如果定义了DUBUG,则执行此段代码!} #else {否则执行此段代码!} 这就需要通过宏开关来进行条件编译,也就是常说的 ...

  7. 从一个聊天信息引发的思考之Android事件分发机制

         转载请声明:http://www.cnblogs.com/courtier/p/4295235.html 起源:        我在某一天看到了下面的一条信息(如下图),我想了下(当然不是这 ...

  8. [置顶] 宏途_LCD调试流程.

    今天在调试宏途的LCD屏时,开始是开机屏幕不亮,背光都不亮,可能板子已经损坏,一般通过测试电流电压简单验证,(注:硬件引脚没焊好也会引起读lcd id出现错误!!!)出现这个问题一般是因为引脚没焊好, ...

  9. java中的“包”与C#中的“命名空间

    原文地址:http://www.cnblogs.com/lidabo/archive/2012/12/15/2819865.html Package vs. Namespace 我们知道,重用性(re ...

  10. 【转】Mac访问Windows共享文件夹

    相信大多数的用户用Windows访问Windows的共享文件夹是一件很容易的事,但是如果用Mac来访问Windows共享文件夹就会遇到很多的麻烦了,尤其是设置是比较有区别的吗,接下来的将用图文交大家怎 ...