章三例题4——UVA 340 Master -Mind Hints

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 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 strongwhen 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 (thelengthofthecode).Following these will be the secret code,representedas 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,inorder,onehint 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)

题意：
猜数字 1，求位置一样且位置上对应的值也相同的数的个数A
2，在两个序列中都出现过，但是位子不一样的数的个数B

先用一个数组存原序列的1-9的数字出现的个数
然后猜的序列，用另一个数组存1-9出现的个数
扫一遍，然后将位子相同且值相同的数个数记录下来。A
B的求法，就是两个记录数组比较，值最小的和就是B

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[1005];int visa[15];int b[1005];int visb[15];int main(){    int n;    int case1=0;    while(scanf("%d",&n)!=EOF&&n)    {        printf("Game %d:\n",++case1);        memset(a,0,sizeof(a));        memset(visa,0,sizeof(visa));        for(int i=0;i<n;i++)        {             scanf("%d",&a[i]);             visa[a[i]]++;        }        int q=0;       for(;;)//这里，因为我们不知道b有多少        {            memset(b,0,sizeof(b));            memset(visb,0,sizeof(visb));            for(int i=0;i<n;i++)                scanf("%d",&b[i]);            if(b[0]==0)                break;            int t=0;            for(int i=0;i<n;i++)            {                if(a[i]==b[i])                     t++;                visb[b[i]]++;            }            int ans=0;            for(int i=0;i<10;i++)            {                if(visa[i]!=0&&visb[i]!=0)                    ans+=min(visa[i],visb[i]);            }            printf("    ");            cout<<'('<<t<<','<<ans-t<<')'<<endl;        }    }    return 0;}