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)
#include<stdio.h>
#define maxn 1000 + 10 int main() {
int n, a[maxn], b[maxn];
int kase = 0;
while(scanf("%d", &n) == 1 && n) { // n=0时输入结束
printf("Game %d:\n", ++kase);
for(int i = 0; i < n; i++) scanf("%d", &a[i]);
for(;;) {
int A = 0, B = 0;
for(int i = 0; i < n; i++) {
scanf("%d", &b[i]);
if(a[i] == b[i]) A++;
}
if(b[0] == 0) break; // 正常的猜测序列不会有0,所以只判断第一个数是否为0即可
for(int d = 1; d <= 9; d++) {
int c1 = 0, c2 = 0; // 统计数字d在答案序列和猜测序列中各出现多少次
for(int i = 0; i < n; i++) {
if(a[i] == d) c1++;
if(b[i] == d) c2++;
}
if(c1 < c2) B += c1; else B += c2;
}
printf(" (%d,%d)\n", A, B-A);
}
}
return 0;
}

UVA340-Master-Mind Hints(紫书例题3.4)的更多相关文章

  1. 紫书 例题 11-13 UVa 10735(混合图的欧拉回路)(最大流)

    这道题写了两个多小时-- 首先讲一下怎么建模 我们的目的是让所有点的出度等于入度 那么我们可以把点分为两部分, 一部分出度大于入度, 一部分入度大于出度 那么显然, 按照书里的思路,将边方向后,就相当 ...

  2. 紫书 例题8-3 UVa 1152(中途相遇法)

    这道题要逆向思维, 就是求出答案的一部分, 然后反过去去寻找答案存不存在. 其实很多其他题都用了这道题目的方法, 自己以前都没有发现, 这道题专门考这个方法.这个方法可以没有一直往下求, 可以省去很多 ...

  3. 紫书 例题8-12 UVa 12627 (找规律 + 递归)

    紫书上有很明显的笔误, 公式写错了.g(k, i)的那个公式应该加上c(k-1)而不是c(k).如果加上c(k-1)那就是这一次 所有的红气球的数目, 肯定大于最下面i行的红气球数 我用的是f的公式, ...

  4. 紫书 例题8-4 UVa 11134(问题分解 + 贪心)

     这道题目可以把问题分解, 因为x坐标和y坐标的答案之间没有联系, 所以可以单独求两个坐标的答案 我一开始想的是按照左区间从小到大, 相同的时候从右区间从小到大排序, 然后WA 去uDebug找了数据 ...

  5. 紫书 例题8-17 UVa 1609 (构造法)(详细注释)

    这道题用构造法, 就是自己依据题目想出一种可以得到解的方法, 没有什么规律可言, 只能根据题目本身来思考. 这道题的构造法比较复杂, 不知道刘汝佳是怎么想出来的, 我想的话肯定想不到. 具体思路紫书上 ...

  6. 紫书 例题 9-5 UVa 12563 ( 01背包变形)

    总的来说就是价值为1,时间因物品而变,同时注意要刚好取到的01背包 (1)时间方面.按照题意,每首歌的时间最多为t + w - 1,这里要注意. 同时记得最后要加入时间为678的一首歌曲 (2)这里因 ...

  7. UVA489 - Hangman Judge【紫书例题4.2】

    题意:就是给出一个字符串,让你去一个一个猜测,相同字母算一次,如果是之前猜过的也算错,如果你在错7次前猜对就算你赢,文章中是LRJ的例题代码. #include<stdio.h> #inc ...

  8. 紫书 例题 10-2 UVa 12169 (暴力枚举)

    就是暴力枚举a, b然后和题目给的数据比较就ok了. 刘汝佳这道题的讲解有点迷,书上讲有x1和a可以算出x2, 但是很明显x2 = (a * x1 +b) 没有b怎么算x2?然后我就思考了很久,最后去 ...

  9. 紫书 例题 10-26 UVa 11440(欧拉函数+数论)

    这里用到了一些数论知识 首先素因子都大于M等价与M! 互质 然后又因为当k与M!互质且k>M!时当且仅当k mod M! 与M!互质(欧几里得算法的原理) 又因为N>=M, 所以N!为M! ...

随机推荐

  1. HDU2149 - Public Sale【巴什博弈】

    虽然不想,但是现实总归是现实,Lele始终没有逃过退学的命运,因为他没有拿到奖学金.现在等待他的,就是像FarmJohn一样的农田生涯.  要种田得有田才行,Lele听说街上正在举行一场别开生面的拍卖 ...

  2. [bzoj 2726] 任务安排 (斜率优化 线性dp)

    3月14日第三题!!!(虽然是15号发的qwq) Description 机器上有N个需要处理的任务,它们构成了一个序列.这些任务被标号为1到N,因此序列的排列为1,2,3-N.这N个任务被分成若干批 ...

  3. mysql数据库增量恢复

    mysqldump -uroot -p -B discuzx -F -x --master-data=2 --events|gzip >/root/discuzx.sql.gz 写入数据 删除数 ...

  4. CentOS中防火墙相关的命令(CentOS7中演示)

    CentOS中防火墙程序主要是firewall和iptables,CentOS7中firewall服务已经默认安装好了,而iptables服务需要自己用yum  install  iptabes-se ...

  5. ZOJ 3229 Shoot the Bullet

    Shoot the Bullet Time Limit: 2000ms Memory Limit: 32768KB This problem will be judged on ZJU. Origin ...

  6. ASP.NET中的Webconfig 和 Global.asax区别

    Web.Config与Global.asax的区别: Config可以根据不同的错误类型定义不同的错误页,网站重定义转向新的错误页面. Global,在全局错误中写入应用程序事件错误信息,并在当前页输 ...

  7. 使用Eclipse进行远程调控

    什么是远程调试,就是在A机器上利用Eclipse单步跟踪调试B机器上的Web应用,当然调试A机器上Web应用也是没有问题的,90%我都是调试本机的Web应用,远程调试的意义我想我不用说了,大家都会想到 ...

  8. 关于sql连接查询(内联、左联、右联、全联)

    内连接(INNER JOIN)(典型的连接运算,使用像   =   或   <>   之类的比较运算符).包括相等连接和自然连接. 内连接使用比较运算符根据每个表共有的列的值匹配两个表中的 ...

  9. 沃通SSL精灵,让站点HTTPS永只是期

    告别HTTP明文"裸奔"时代 百度.阿里巴巴.必应等越来越多的互联网巨头相继启用全站HTTPS加密,保护用户数据和隐私安全.逐步告别HTTP明文"裸奔"时代. ...

  10. 启用QNX系统,海尔智能冰箱或成业界“宝马”

        智能家电正处于迅猛发展的态势,国内眼下有非常多企业都在积极布局智能家电,当中又以海尔最为典型.作为家电领域的领头羊,海尔近年来在智能家电领域的动作不小.近期有消息透露.海尔也许会在IFA展会上 ...