D Digit Size

Problem Description

Digital LC-display is widely used in many different fields such as electronic calculator, electronic watch, digital instruments, etc. A simulated example of the numbers represented by digital LC-display device is shown as follows:

Each number represented by LC-display above is composed of s "-" signs for the horizontal segments and s "|" signs for the vertical ones, and each number exactly occupies s+2 columns and 2s+3 rows. Your task is to change the size of the original numbers by changing "s" — the number of signs.

 

Input

The first line of input contains a number t, which means there are t cases of the test data.
The input contains several lines, 2s + 4 for each number to be displayed. The first line of each case contains two integer s and t (1 <= s, t <= 9), where s is the original size of the numbers and t is the target size of numbers that you should output. The following 2s + 3 lines show the original number n (the digit of n will not exceed 9) you should deal with. Each digit of n will be separated by an empty column (except for the last digit).

 

Output

For each test case, output the case number and then output the target number according to the input. Output a blank line after each case.

——摘自HDOJ 3575（測試數據不好顯示）

這道題題目不難！看了後題目大家都會有想法，是的，我承認不難，但想起自己當初（也才昨天而已）實現的時候真是太痛苦了！都怪自己審題不仔細，忽略了數據的大小。所以一開始時，存儲輸入數據的數組就只開了100*100， 呃。。寫完後本地測試數據輕鬆過，但提交總是WA！這種情況是很鬱悶的！當時改了一下午，空格換行之類的細節想得都快暈了！

到後來是心灰意冷，真沒轍了。沒辦法了，就很無聊的把數據改大了點，草，竟奇蹟般地過了。。Accept！！！有木有搞錯啊……盼望了一下午啊！可還是高興不起來，當時真的是哭笑不得。。

注意審題啊，切記……切記……

#include<iostream>
#include<string>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;

int m,n;
char digit[1500][1500];

void f(int row)
{
     int i,j,k;
     int len = strlen(digit[row]);
     for(i=0; i*(m+2)+i-1<len; i++)
     {
              if(i != 0)
              printf(" ");
              printf(" ");
              for(j=0; j<n; j++)
              if(digit[row][(m+2)*i+i+1] == '-')
              printf("-");
              else printf(" ");
              printf(" ");
              }
              printf("\n");
     }
void g(int row)
{
     int i,j,k;
     int len = strlen(digit[row]);
     for(i=0; i*(m+2)+i-1<len; i++)
     {
              if(i != 0)
              printf(" ");
              printf("%c", digit[row][(m+2)*i+i]);
              for(j=0; j<n; j++)
              printf(" ");
              printf("%c", digit[row][i*(m+2)+i+m+1]);
              }
              printf("\n");
     }

int main()
{
    int cases, t;
    int i,j,k;
    
    scanf("%d", &cases);
    t = 1;
    while(t <= cases)
    {
            scanf("%d %d", &m, &n);
            
            getchar();
            for(i=0; i<2*m+3; i++)
             gets(digit[i]);

            
            printf("Case %d:\n", t);
            
            f(0);
            for(i=0; i<n; i++)
            g(1);
            f(m+1);
            for(i=0; i<n; i++)
            g(m+2);
            f(2*m+2);
            t++;
            //if(t != cases)
            printf("\n");
            }   
    return 0;
    }