• uva-340


     Master-Mind Hints 

    MasterMind is a game for two players. One of them, Designer, selects a secret code. The other, Breaker, tries to break it. A code is no more than a row of colored dots. At the beginning of a game, the players agree upon the length N that a code must have and upon the colors that may occur in a code.

    In order to break the code, Breaker makes a number of guesses, each guess itself being a code. After each guess Designer gives a hint, stating to what extent the guess matches his secret code.

     

    In this problem you will be given a secret code tex2html_wrap_inline35 and a guess tex2html_wrap_inline37 , and are to determine the hint. A hint consists of a pair of numbers determined as follows.

    match is a pair (i,j), tex2html_wrap_inline41 and tex2html_wrap_inline43 , such that tex2html_wrap_inline45 . Match (i,j) is called strong when i =j, and is called weak otherwise. Two matches (i,j) and (p,q) are called independent when i = p if and only if j = q. A set of matches is called independent when all of its members are pairwise independent.

     

    Designer chooses an independent set M of matches for which the total number of matches and the number of strong matches are both maximal. The hint then consists of the number of strong followed by the number of weak matches in M. Note that these numbers are uniquely determined by the secret code and the guess. If the hint turns out to be (n,0), then the guess is identical to the secret code.

     

    Input

    The input will consist of data for a number of games. The input for each game begins with an integer specifying N (the length of the code). Following these will be the secret code, represented as N integers, which we will limit to the range 1 to 9. There will then follow an arbitrary number of guesses, each also represented as N integers, each in the range 1 to 9. Following the last guess in each game will be N zeroes; these zeroes are not to be considered as a guess.

     

    Following the data for the first game will appear data for the second game (if any) beginning with a new value for N. The last game in the input will be followed by a single zero (when a value for N would normally be specified). The maximum value for N will be 1000.

     

    Output

    The output for each game should list the hints that would be generated for each guess, in order, one hint per line. Each hint should be represented as a pair of integers enclosed in parentheses and separated by a comma. The entire list of hints for each game should be prefixed by a heading indicating the game number; games are numbered sequentially starting with 1. Look at the samples below for the exact format.

     

    Sample Input

     

    4
    1 3 5 5
    1 1 2 3
    4 3 3 5
    6 5 5 1
    6 1 3 5
    1 3 5 5
    0 0 0 0
    10
    1 2 2 2 4 5 6 6 6 9
    1 2 3 4 5 6 7 8 9 1
    1 1 2 2 3 3 4 4 5 5
    1 2 1 3 1 5 1 6 1 9
    1 2 2 5 5 5 6 6 6 7
    0 0 0 0 0 0 0 0 0 0
    0

     

    Sample Output

     

    Game 1:
        (1,1)
        (2,0)
        (1,2)
        (1,2)
        (4,0)
    Game 2:
        (2,4)
        (3,2)
        (5,0)
        (7,0)


    明白题目大意很好A。
    题目大意:在游戏开始之前,双方先协议密码的长度,假设是n。 在出题者设定密码(s1,s2,...sn) 后,由解题者来猜(g1,g2,...gn)。 如 果同样位置gi=si, 那解题者得到一个A。 如果gi=sj,但i不等于j,那解题者得到一个B。 请注意:不管是得A或得B,每个 gi最多只能对应到一个sj,而且得A比得B优先。 举例说明:密码为1 2 3 4,若猜题者猜1 1 5 6,那出题者应该回答1A0B,而不是0A1B。 如果某个猜测得到 nA0B,那就代表猜题者完全猜中密码了。

     1 #include<cstdio>
     2 
     3 #define maxn 1005
     4 
     5 struct secret
     6 {
     7     int num;
     8     int flag;
     9 };
    10 
    11 struct guess
    12 {
    13     int num;
    14     int flag;
    15 };
    16 
    17 int main()
    18 {
    19     struct secret str[maxn];
    20     struct secret guess[maxn];
    21     int n,i,j,count,k=1;
    22     int a,b;
    23     while(scanf("%d",&n) != EOF && n)
    24     {
    25         for(i=0;i<n;i++)
    26         {
    27             scanf("%d",&str[i].num);
    28         }
    29         printf("Game %d:
    ",k++);
    30         while(1)
    31         {
    32             count=0;
    33             a=0;
    34             b=0;
    35             for(i=0;i<n;i++)
    36             {
    37                 str[i].flag=0;
    38             }
    39             for(i=0;i<n;i++)
    40             {
    41                 scanf("%d",&guess[i].num);
    42                 guess[i].flag=0;
    43                 if(guess[i].num==0)
    44                     count++;
    45             }
    46             if(count==n)
    47                 break;
    48             for(i=0;i<n;i++)
    49             {
    50                 if(str[i].num==guess[i].num)
    51                 {
    52                     a++;
    53                     str[i].flag=1;
    54                     guess[i].flag=1;
    55                 }
    56             }
    57             for(i=0;i<n;i++)
    58             {
    59                 for(j=0;j<n;j++)
    60                 {
    61                     if(i!=j&&str[i].flag==0&&guess[j].flag==0&&str[i].num==guess[j].num)
    62                     {
    63                         b++;
    64                         str[i].flag=1;
    65                         guess[j].flag=1;
    66                     }
    67                 }
    68             }
    69             printf("    (%d,%d)
    ",a,b);
    70         }
    71     }
    72 }
    
    
    

         
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  • 原文地址:https://www.cnblogs.com/youdiankun/p/3676109.html
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