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 
and a guess 
, and are to determine the hint. A hint consists of a pair of numbers determined as follows.
A match is a pair (i,j), 
and 
, such that 
. 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)
这道题完全是虐英语渣（比如我）的，冗长的题目读完可能根本不知道题目想要说什么，还好有万能的google翻译，我总结一下题意，给还看不大懂题意得盆友：
题意：一个人写出一串数字（1-9），然后另一个人去猜这串数字，他会给出若干组数字串，我们的工作就是统计猜测的数字串和原数字串的一些信息：
1.有多少数字是完全正确的，即数字和位置都正确
2.有多少数字是不完全正确，即数字正确但是位置不正确，当然已经完全正确的数字 不算在内。
这就是我们需要计算的两个量，计算的方法也很简单：
1.将原始数字串和猜测数字串输入到数组并且统计每个数字出现的次数
2.扫描一遍两个数字串，找出完全正确的数字，记录并从统计出现次数的数组中去除他们
3.再扫描一遍统计出现次数的数组，选取最小的作为值进行累加，他们都是数字正确但是位置不正确的
4.输出两个值，注意输出格式
<span style="font-size:24px;">#include <iostream>
#include <stdio.h>
#include <memory.h>
#define MAX 1010
using namespace std;
int code[MAX];
int guess[MAX];
int code_count[10];
int guess_count[10];
int count_temp[10];
int cas=0;
int main()
{
int n,i,res1,res2;
while(scanf("%d",&n)&&n)//输入串的长度
{
printf("Game %d:\n",++cas);//打印游戏次数
memset(code_count,0,sizeof(code_count));//初始化原数字串的数字出现次数
for(i=0;i<n;i++)//输入数字串
{
scanf("%d",&code[i]);
code_count[code[i]]++;//统计次数
}
while(1)//输入猜测数字串
{
res1=res2=0;
memcpy(count_temp,code_count,sizeof(code_count));//制作原始统计信息的副本，因为需要修改
memset(guess_count,0,sizeof(guess_count));//初始化
for(i=0;i<n;i++)//输入数字串
{
scanf("%d",&guess[i]);
guess_count[guess[i]]++;//统计
}
if(guess_count[0]!=0)//如果输入的数字串中含有0，说明输入结束
break;
for(i=0;i<n;i++)//找出完全正确的
{
if(code[i]==guess[i])
{
res1++;
count_temp[code[i]]--;
guess_count[code[i]]--;
}
}
for(i=1;i<=9;i++)//计算不完全正确的
{
res2+=min(count_temp[i],guess_count[i]);
}
printf(" (%d,%d)\n",res1,res2);//输出
}
}
return 0;
}
</span>