D - Master-Mind Hints UVA - 340

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 s1 . . . sn and a guess g1 . . . gn, and are to determine
the hint. A hint consists of a pair of numbers determined as follows.
A match is a pair (i, j), 1 ≤ i ≤ n and 1 ≤ j ≤ n, such that si = gj . 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 ‘0’ (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)
解题思路：就是找给出序列与其相同部分有多少个，不同部分有多少个。
注意找到后要将其归0，所以加了一个数组t将其重置。

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int n,i,j;
    int a[1000],b[1000],s[1000],c[1000];
    int cas=0;
    int p;
    int t[1000];
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)break;
        for(i=0;i<1000;i++)
        {
            b[i]=0;
            c[i]=0;
        }
        for(i=0;i<n;i++)
          scanf("%d",&s[i]);
          for(i=0;i<n;i++)
            t[i]=s[i];
          int x=0,y=0;
        while(1)
        {
            for(i=0;i<n;i++)
                scanf("%d",&a[i]);
            if(a[0]==0)break;
            for(i=0,j=0;i<n,j<n;i++,j++)
            {
                if(s[i]==a[j])
                {
                    b[x]++;
                    a[j]=0;
                    s[i]=0;
                }
            }
            for(i=0;i<n;i++)
            {
                if(a[i])
                {
                    for(p=0;p<n;p++)
                    {
                        if(s[p])
                        {
                            if(s[p]==a[i])
                            {
                                c[y]++;
                                s[p]=0;
                                a[i]=0;
                            }
                        }
                    }
                }
            }
            x++;
            y++;
            for(i=0;i<n;i++)
                s[i]=t[i];
        }
        printf("Game %d:\n",++cas);
        for(i=0,j=0;i<x,j<y;i++,j++)
        {
            printf("    (%d,%d)\n",b[i],c[j]);
        }
    }
    return 0;
}