UVa 10189 - Minesweeper

很水的题，但是交了很多次都WA了，纠结啊。后来把输入输出格式该了才对。。。。

 

The Problem

Have you ever played Minesweeper? It's a cute little game which comes within a certain Operating System which name we can't really remember. Well, the goal of the game is to find where are all the mines within a MxN field. To help you, the game shows a number in a square which tells you how many mines there are adjacent to that square. For instance, supose the following 4x4 field with 2 mines (which are represented by an * character):

*...
....
.*..
....

If we would represent the same field placing the hint numbers described above, we would end up with:

*100
2210
1*10
1110

As you may have already noticed, each square may have at most 8 adjacent squares.

The Input

The input will consist of an arbitrary number of fields. The first line of each field contains two integers n and m (0 < n,m <= 100) which stands for the number of lines and columns of the field respectively. The next n lines contains exactly m characters and represent the field. Each safe square is represented by an "." character (without the quotes) and each mine square is represented by an "*" character (also without the quotes). The first field line where n = m = 0 represents the end of input and should not be processed.

The Output

For each field, you must print the following message in a line alone:

Field #x:

Where x stands for the number of the field (starting from 1). The next n lines should contain the field with the "." characters replaced by the number of adjacent mines to that square. There must be an empty line between field outputs.

Sample Input

 

4 4
*...
....
.*..
....
3 5
**...
.....
.*...
0 0

 

Sample Output

 

Field #1:
*100
2210
1*10
1110

Field #2:
**100
33200
1*100

#include<stdio.h>
#include<string.h>

int main()
{
    int n, m, i, j, num=0, k, l;
    int s[105][105], ch;
    while(scanf("%d%d", &n, &m)&& (n||m))
    {
        getchar();
        if(num)
            printf( "\n" );
        memset(s,0,sizeof(s));
        for (i=1; i<=n; i++)
        {
            for (j=1; j<=m; j++)
            {
                ch = getchar();

                if( ch == '*' )
                {
                    s[i][j] = -100;
                    for(k = -1 ; k <= 1 ; k++ )
                        for(l = -1 ; l <= 1 ;l++ )
                            s[i+k][j+l]++;
                }
            }
            getchar();
        }
        printf("Field #%d:\n", ++num);
        for (i=1; i<=n; i++)
        {
            for (j=1; j<=m; j++)
            {
                if (s[i][j] < 0)
                    printf("*");
                else
                    printf("%d", s[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}