D Digit Size

Problem DescriptionDigital LC-display is widely used in your 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 abve 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 2 s + 3 rows. your task is to change the size of the original numbers
Changing "s"-the number of signs.

InputThe first line of input contains a number t, which means there are t cases of the test data.
The input contains several lines, 2 s + 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 2 s + 3 lines show the original number n (the digit of n will not exceed 9) you shoshould deal. each digit of n will be separated by an empty column (blocks t for the last digit ).

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

-- From HDOJ 3575 (poor display of test data)

This question is not difficult! After reading the questions, everyone will have an idea. Yes, I admit it is not difficult, but it is so painful to remember that I realized it yesterday! I blame myself for failing to examine the questions carefully and ignore the data size. So at the beginning, the array that stores input data only opened 100*100 .. The local test data is easy to use after writing, but the submission is always WA! This situation is depressing! At that time, I changed it for one afternoon and thought about details like space wrapping!

It was disheartened later. No way, it's boring to make the data bigger, grass, it's a miracle .. Accept !!! Something wrong ...... Hope for an afternoon! But I was still not happy. At that time, I was really crying and laughing ..

Pay attention to questions. Remember ...... Remember ......

#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;    }