poj1102 LC-Display

LC-Display

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13320 Accepted: 5247

Description

A friend of you has just bought a new computer. Until now, the most powerful computer he ever used has been a pocket calculator. Now, looking at his new computer, he is a bit disappointed, because he liked the LC-display of his calculator so much. So you decide to write a program that displays numbers in an LC-display-like style on his computer.

Input

The input contains several lines, one for each number to be displayed. Each line contains two integers s, n (1 <= s <= 10, 0 <= n <= 99 999 999), where n is the number to be displayed and s is the size in which it shall be displayed.

The input file will be terminated by a line containing two zeros. This line should not be processed.

Output

Output the numbers given in the input file in an LC-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, also for the last digit.) There has to be exactly one column of blanks between two digits.

Output a blank line after each number. (You will find a sample of each digit in the sample output.)

Sample Input

2 123453 678900 0

Sample Output

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

Source

Mid-Central European Regional Contest 1999


话说很简单，就是打印大小为n的数字，但是各种PE卡的蛋疼
首先末尾要多个空格，但是他只说两个数字之间要有个空格啊。。。。
然后每行的数字后面空格要填完
最后每个测试数据之后有个空行。。。。
代码

#include <stdio.h>#include <string.h>char zm[10][10][10]={    {       " - ",       "| |",       "   ",       "| |",       " - "    },    {       "   ",       "  |",       "   ",       "  |",       "   "    },    {       " - ",       "  |",       " - ",       "|  ",       " - "    },    {       " - ",       "  |",       " - ",       "  |",       " - "    },    {       "   ",       "| |",       " - ",       "  |",       "   "    },    {       " - ",       "|  ",       " - ",       "  |",       " - "    },    {       " - ",       "|  ",       " - ",       "| |",       " - "    },    {       " - ",       "  |",       "   ",       "  |",       "   "    },    {       " - ",       "| |",       " - ",       "| |",       " - "    },    {       " - ",       "| |",       " - ",       "  |",       " - "    }};char str[100];void Expend(char *s,int re){    int i;    printf("%c",s[0]);    for (i=0;i<re;i++)    {        printf("%c",s[1]);    }    printf("%c",s[2]);}int main(){    int i,j,n;    while(1)    {        scanf("%d",&n);        scanf("%s",str);        if (n==0) break;        for (i=0;i<2*n+3;i++)        {            for (j=0;j<strlen(str);j++)            {                if (i==0)                {                    Expend(zm[str[j]-'0'][0],n);                }                else if (i==n+1)                {                    Expend(zm[str[j]-'0'][2],n);                }                else if (i==2*n+2)                {                    Expend(zm[str[j]-'0'][4],n);                }                else if (i<n+1)                {                    Expend(zm[str[j]-'0'][1],n);                }                else                {                    Expend(zm[str[j]-'0'][3],n);                }                printf(" ");            }            printf("\n");        }        printf("\n");    }    return 0;}