烟大 Contest1024 - 《挑战编程》第一章：入门 Problem D: LC-Display（模拟计算器显示数字）

Problem D: LC-Display

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 14  Solved: 3
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Description

A friend of yours has just bought a new computer. Before this, the most powerful machine he ever used was a pocket calculator. He is a little disappointed because he liked the LCD display of his calculator more than the screen on his new computer! To make him happy, write a program that prints numbers in LCD display style.

Input

The input file contains several lines, one for each number to be displayed. Each line contains integers s and n, where n is the number to be displayed ( 0n99, 999, 999) and s is the size in which it shall be displayed ( 1s10). The input will be terminated by a line containing two zeros, which should not be processed

Output

Print the numbers specified in the input file in an LCD display-style using s ``-'' signs for the horizontal segments and s ``|'' signs for the vertical ones. Each digit occupies exactly s + 2 columns and 2s + 3 rows. Be sure to fill all the white space occupied by the digits with blanks, including the last digit. There must be exactly one column of blanks between two digits. Output a blank line after each number. You will find an example of each digit in the sample output below.

Sample Input

2 12345

3 67890

0 0

Sample Output

      --   --        --
   |    |    | |  | |
   |    |    | |  | |
      --   --   --   --
   | |       |    |    |
   | |       |    |    |
      --   --        -- 

 ---   ---   ---   ---   ---
|         | |   | |   | |   |
|         | |   | |   | |   |
|         | |   | |   | |   |
 ---         ---   ---
|   |     | |   |     | |   |
|   |     | |   |     | |   |
|   |     | |   |     | |   |
 ---         ---   ---   ---

HINT

杭电1332

---

　　模拟题。这道题是在控制台中以计算器的显示形式显示指定尺寸大小的一串数字。

　　我的做法是提前写好数字的模板（因为尺寸不确定，所以要把数字做成模板），存放在一个三维数组中（数字，行，列）。输入尺寸和数字串后，将数字的模板组合起来，然后解析成指定尺寸的数字输出出来。

My code：

 //这是一道模拟题
 //模拟计算器的形式在控制台中输出数字

 #include <iostream>
 #include <string.h>
 using namespace std;

 int a[][][]=
 {
 {//
 ,,,
 ,,,
 ,,,
 ,,,
 ,,
 }
 ,{//
 ,,,
 ,,,
 ,,,
 ,,,
 ,,}
 ,{//
 ,,,
 ,,,
 ,,,
 ,,,
 ,,
 }
 ,{//3
 ,,,
 ,,,
 ,,,
 ,,,
 ,,
 }
 ,{//4
 ,,,
 ,,,
 ,,,
 ,,,
 ,,
 }
 ,{//5
 ,,,
 ,,,
 ,,,
 ,,,
 ,,
 }
 ,{//6
 ,,,
 ,,,
 ,,,
 ,,,
 ,,
 }
 ,{//7
 ,,,
 ,,,
 ,,,
 ,,,
 ,,
 }
 ,{//8
 ,,,
 ,,,
 ,,,
 ,,,
 ,,
 }
 ,{//9
 ,,,
 ,,,
 ,,,
 ,,,
 ,,
 }
 };
 /*
 0是一个空格
 1是size个竖杠
 2是size个横杠
 3是size个虚横杠
 */
 int b[][];
 int main()
 {
     string l;
     int size;
     while(cin>>size){
         cin>>l;
         if(size== && l=="") break;
         int length=l.length();
         //组合数字模板在数组b中
         for(int i=;i<=;i++){   //循环行
             int f=;    //组合数组b的第一个开始赋值
             for(int j=;j<=length;j++){ //循环数字
                 int num=l[j-]-'';
                 for(int k=;k<;k++){
                     b[i][f]=a[num][i-][k];
                     f++;
                 }
                 b[i][f]=;
                 f++;
             }
         }
         /*输出组合好的数字模板
         for(int i=1;i<=5;i++){
             for(int j=1;j<=3*length+length-1;j++)
                 cout<<b[i][j];
             cout<<endl;
         }
         */
         //将数字模板解析成数字
         //0输出一个空格
         //1输出size行竖杠
         //2输出size个横杠
         //3输出size行空格
         int count = ;
         for(int i=;i<=;i++){//循环行
             for(int j=;j<=*length+length-;j++){
                 //if(j==3*length+length-1 && b[i][j]==0)
                 //    continue;
                 if(i%!=){
                     switch(b[i][j]){
                     case :
                         cout<<' ';
                         break;
                     case :
                         for(int k=;k<size;k++)
                             cout<<'-';
                         break;
                     case :
                         for(int k=;k<size;k++)
                             cout<<' ';
                         break;
                     }
                 }
                 else{
                     switch(b[i][j]){
                     case :
                         cout<<' ';break;
                     case :
                         for(int k=;k<size;k++)
                             cout<<' ';
                         break;
                     case :
                         cout<<'|';break;
                     }
                 }
             }
             cout<<endl;
             if(i%==){  //如果改行是偶数行，且输出次数再输出一遍
                 ++count;    //如果是偶数行，已经输出一遍了，count+1
                 if(count<size)
                     --i;
             }
             else
                 count=;
         }
         cout<<endl;
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013