UVa 340 Master-Mind Hints

UVa 340 Master-Mind Hints

E文不够好，题目意思没有看明白。看了别人题解上讲述的题目大意加上样例分析，大致理解了。
题目意思是这样的，给出两个序列A、B，对于A[i]==B[j]，如果i==j，那么这是一个strong对，否则这是一个weak对。题目要求先找出所有的strong对，然后找出weak对（一对数字只能用一次）。
这样的模拟题还是不能急，心里要对过程很明白，一步一步来，不要怕代码长。

以下是我的代码：

#include < cstdio >
#include < cstring >
using   namespace  std;
const   int  kMaxn( 1007 );

int  main()
{
     /*
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    // */

     int  T( 0 ),n;
     while (scanf( " %d " , & n) == 1   &&  n)
    {
         int  code[kMaxn];
         for ( int  i = 1 ;i <= n;i ++ )
            scanf( " %d " , & code[i]);
        T ++ ;
        printf( " Game %d:\n " ,T);
         int  guess[kMaxn];
         for ( int  i = 1 ;i <= n;i ++ )
            scanf( " %d " , & guess[i]);
         while (guess[ 1 ])
        {
             int  t[kMaxn];
            memcpy(t,code,kMaxn * sizeof ( int ));
             int  strong( 0 ),weak( 0 );
             for ( int  i = 1 ;i <= n;i ++ )
                 if (t[i] == guess[i])
                {
                    strong ++ ;
                    t[i] = guess[i] = 0 ;
                }
             for ( int  i = 1 ;i <= n;i ++ )
            {
                 if ( ! t[i])
                     continue ;
                 for ( int  j = 1 ;j <= n;j ++ )
                     if (t[i] == guess[j])
                    {
                        weak ++ ;
                        t[i] = guess[j] = 0 ;
                         break ;
                    }
            }

            printf( "     (%d,%d)\n " ,strong,weak);

             for ( int  i = 1 ;i <= n;i ++ )
                scanf( " %d " , & guess[i]);
        }
    }

     return   0 ;
}