UVA227

步骤：1.把输入的数字和空格保存。（这里用到gets函数读取整行）2.定位空格。3.输入指令。

#include<stdio.h>
#include<string.h>
char a[5][7];//声明数组
int main(){
    memset(a,0,sizeof(a));
    int blank_x=0;
    int blank_y=0;
    char s[100];
    char c[10];
    //输入网格
    gets(a[0]);
    gets(a[1]);
    gets(a[2]);
    gets(a[3]);
    gets(a[4]);
    //定位空格
    for(int x=0; x<5; x++)
    {
        for(int y=0; y<5; y++)
        {
            if(a[x][y] == ‘ ‘)
            {
                blank_x = x;
                blank_y = y;
            }
        }
    }
    gets(s);//输入指令
    for(int i=0;s[i] != ‘0‘;i++){
        if(s[i] == ‘A‘){
            if(!a[blank_x-1][blank_y])
                {
                 printf("1 This puzzle has no final configuration.\n");
                 return 0;}
            a[blank_x][blank_y]=a[blank_x-1][blank_y];
            a[blank_x-1][blank_y]=‘ ‘;
            blank_x=blank_x-1;
        }
        else if(s[i] == ‘B‘){
            if(!a[blank_x+1][blank_y])
                    {
                 printf("2 This puzzle has no final configuration.\n");
                 return 0;}
            a[blank_x][blank_y]=a[blank_x+1][blank_y];
            a[blank_x+1][blank_y]=‘ ‘;
            blank_x=blank_x+1;
        }
        else if(s[i] == ‘L‘){
            if(!a[blank_x][blank_y-1])
                    {
                 printf("3 This puzzle has no final configuration.\n");
                 return 0;}
            a[blank_x][blank_y]=a[blank_x][blank_y-1];
            a[blank_x][blank_y-1]=‘ ‘;
            blank_y=blank_y-1;
        }
        else if(s[i] == ‘R‘){
            if(!a[blank_x][blank_y+1])
                {
                 printf("4 This puzzle has no final configuration.\n");
                 return 0;}
            a[blank_x][blank_y]=a[blank_x][blank_y+1];
            a[blank_x][blank_y+1]=‘ ‘;
            blank_y=blank_y+1;
        }
    }
    //输出
    puts(a[0]);
    puts(a[1]);
    puts(a[2]);
    puts(a[3]);
    puts(a[4]);
    return 0;
}

OJ里还要加Puzzle #1:...