Chinese Mahjong】的更多相关文章

Chinese Mahjong Mahjong () is a game of Chinese origin usually played by four persons with tiles resembling dominoes and bearing various designs, which are drawn and discarded until one player wins with a hand of four combinations of three tiles each…
大白书第一章的例题,当时看起来很吃力,现如今A这道题的话怎么写都无所谓了. 思路很简单,就是枚举胡哪张牌,然后枚举一下将牌,剩下如果能找到4个顺子或者刻子就胡了. 由于粗心,34个字符串初始化写错,各种WA. #include <cstdio> #include <cstring> #include <string> #include <iostream> using namespace std; ], c[]; string s; ] = { "…
思路:得到输入得到mj[]的各个牌的数量,还差最后一张牌.直接暴力枚举34张牌就可以了. 当假设得到最后一张牌,则得到了的牌看看是不是可以胡,如果可以胡的话,就假设正确.否者假设下一张牌. 关键还是如何判断这组牌是不是可以胡.怎么判断呢?因为胡牌的条件是:nABC+mBBB+1DD, 而且n+m=4,其中n,m可以为0. 下面是一张:递归DD的牌的递归情况: 然后:依次是找顺子和刻子也是相同的算法 #include<iostream> #include<cstring> using…
先记录下每一种麻将出现的次数,然后枚举每一种可能得到的麻将,对于这个新的麻将牌,去判断可不可能胡,如果可以胡,就可以把这张牌输出出来. 因为eye只能有一张,所以这个是最好枚举的,就枚举每张牌成为eye的可能,然后对于剩下的牌去判断成为pong和chow的可能,然后判断可不可能胡牌 #include<map> #include<set> #include<ctime> #include<cmath> #include<stack> #includ…
题意:这个题意.有点麻烦,就是说给定13张牌,让你求能“听”的牌.(具体的见原题) 原题链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2151 析:看到这个题,真是麻烦啊,我又不懂麻将,看了好久才明白什么是“听”.分析一下思路. 首先对所有的牌都进行编号,然后暴力,首先的是先判断是哪个是将,然后再进一步判断, 哪一个是刻子,和顺子…
pdf传送门 直接模拟+搜索,考虑一下选哪一个是将,然后搜出顺子和刻子,最后判断一下可不可行就好了! #include<stdio.h> #include<string.h> const char* mahjong[]={"1T","2T","3T","4T","5T","6T","7T","8T","9T&quo…
简单的回溯题 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; char *mahjong[]={ "1T","2T","3T","4T","5T","6T","7T","…
OJ题号:UVa11210 思路: 首先字符串处理读入手牌,str数组将手牌和数字对应,接下来搜索,先搜对子,如果搜过对子就不搜了.由于对子有且只有一个,可以在搜到以后直接跳出.同时注意一副麻将中每种牌只有4个,所以不能作为将. #include<cstdio> #include<string> #include<vector> #include<sstream> #include<iostream> using namespace std; v…
解题报告:麻将的规则这里就不说了,这题我们可以用暴力的方法,所以我们应该这样枚举,即将34张牌的每一张牌都放到原来的十三张牌里面去,所以这时我们只要判断这十四张牌能不能胡,因为若要胡的话一定要有一个对子,所以这个时候我们应该枚举到底用哪一种牌作为对子,不过还要先判断这种牌的数量是不是大于2,如果大于2说明这种牌可以作为对子.接下来就是判断这十四张牌到底能不能胡的问题.对于这个问题我们可以先把输入的字符串先转化成0到33的对应的数字模式,即0对应1T,1对应2T ...... 33对应BAI,这样…
题目 https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2151 题意 麻将,有13张牌,问再加哪一张牌可以凑成一对,若干个三张和若干个三联. 思路 如刘书思路,明显,这是一道模拟题. 感想 1. 三倍ice cream! 代码 #include <algorithm> #include <cassert> #incl…
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题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…
背景 作为一名川娃子,怎么能不懂麻将呢= = T1 さきなに~~ [咲 -Saki-] 天才麻将少女什么编 题目 二十一世纪,世界上的麻将竞技人数超过一亿,日本每年也有大规模的全国大赛来对麻将选手进行选拔.从小学到高中,很多学校也设立了麻将部. <天才麻将少女阿知贺篇 episode of side-A>『咲 -Saki- 阿知賀編 episode of side-A』是<天才麻将少女>『咲 -Saki-』的外传,从另一个视角讲述麻将少女们的故事.故事背景设定在<天才麻将少女…
每周日更新 2016.05.29 UVa中国麻将(Chinese Mahjong,Uva 11210) UVa新汉诺塔问题(A Different Task,Uva 10795) NOIP2012同余方程 NOIP2007统计数字 NOIP2013火柴排队 NOIP2013花匠 2016.06.05 Uva 组装电脑 12124 - Assemble Uva 派 (Pie,NWERC 2006,LA 3635) 2016.06.19 Uva 网络(Network,Seoul 2007,LA 39…
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题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5379 Problem Description Little sun is an artist. Today he is playing mahjong alone. He suddenly feels that the tree in the yard doesn't look good. So he wants to decorate the tree.(The tree has n…
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题目传送门 /* 杭电一题(ACM_steps 2.2.4)的升级版,使用到高精度: 这次不是简单的猜出来的了,求的是GCD (n, k) == 1 最大的k(1, n/2): 1. 若n是奇数,则k = (n-1) / 2: 2. 若n是偶数,讨论(n-1)/2 的奇偶性,若不是奇数,则是n/2-2: 详细解释(证明):http://www.xuebuyuan.com/1552889.html */ #include <cstdio> #include <iostream> #i…
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https://vjudge.net/problem/UVA-11210 //被水题虐了一上午... #include<iostream> #include<cstdio> #include<cstring> using namespace std; const char* mahjong[]={"GZH","1T","2T","3T","4T","5T&quo…
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