uva Minesweeper 10189

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:

*10022101*101110

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:*10022101*101110Field #2:**100332001*100

思路：就注意一定 最后一个数据后面没有换行!!!!!!!!

#include<stdio.h>#include<string.h>int a[8]= {-1,-1,-1,0,1,1,1,0};//控制xint b[8]= {-1,0,1,1,1,0,-1,-1};//控制yint main(){    int n,m;    int d=1;    char map[1005][1005],step[1005][1005];    while(~scanf("%d%d",&n,&m),(n&&m))    {        int i,j,k;        if(d>1) printf("\n");        memset(map,'\0',sizeof(map));        for(i=1; i<=n+1; i++)            for(j=1; j<=m+1; j++)                step[i][j]='0';        getchar();        for(i=1; i<=n; i++)        {            for(j=1; j<=m; j++)                scanf("%c",&map[i][j]);            getchar();        }        for(i=1; i<=n; i++)        {            for(j=1; j<=m; j++)            {                if(map[i][j]=='*')                    step[i][j]='*';                else                {                    for(k=0; k<8; k++)                    {                        if(i+a[k]<=0||i+a[k]>n||j+b[k]<=0||j+b[k]>m)                            continue;                        if(map[i+a[k]][j+b[k]]=='*')                            step[i][j]++;                    }                }            }        }        printf("Field #%d:\n",d++);        for(i=1; i<=n; i++)        {            for(j=1; j<=m; j++)                printf("%c",step[i][j]);            printf("\n");        }    }    return 0;}