UVa 120 Stacks of Flapjacks

UVa 120 Stacks of Flapjacks

题目大意：通过一系列的反转操作，给一个序列排序。
类似于选择排序的思路，每次找当前序列的最大值放在合适的位置即可。
以下是我的代码：

#include < stdio.h >
#include < string .h >
void  deal( char   * s, long   * a, long   & n)
{
     long  begin,end,len = strlen(s);
    n = 0 ;begin = 0 ,end = begin;
     while (end < len)
    {
        while (s[end] != '   ' && s[end] != 0 ) end ++ ;
        long  t = 0 ;
        while (begin < end)
       {
          t *= 10 ;
          t += s[begin] - ' 0 ' ;
          begin ++ ;
       }
       n ++ ;a[n] = t;
       begin = end + 1 ;end = begin;
    }
}
void  flip( long   * a, long  n, long  pos)
{
     long  begin = pos,end = n;
     while (begin <= end)
    {
        long  t = a[begin];a[begin] = a[end];a[end] = t;
       begin ++ ;end -- ;
    }
}
int  main()
{
     /*
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    // */
     const   long  maxn = 37 ;
     char  s[maxn * 10 ];
     while (gets(s) != NULL)
    {
        long  n = 0 ,a[maxn],tmp[maxn];
       deal(s,a,n);
        for ( long  i = 1 ;i <= n;i ++ ) tmp[n - i + 1 ] = a[i];
        //   Init
        for ( long  i = 1 ,first = 1 ;i <= n;i ++ )
       {
           if (first) first = 0 ;
           else  putchar( '   ' );
          printf( " %ld " ,a[i]);
       }
       putchar( ' \n ' );
        //   First Line
        for ( long  i = 1 ;i <= n;i ++ )
       {
           long  max = i;
           for ( long  j = n;j > i;j -- )
             if (tmp[max] < tmp[j]) max = j;
           if (max != i)
          {
              if (max != n)
             {
                flip(tmp,n,max);
                printf( " %ld  " ,max);
             }
              if (max != i)
             {
                flip(tmp,n,i);
                printf( " %ld  " ,i);
             }
          }
       }
       printf( " %ld\n " , 0 );
    }
return   0 ;
}