B. Tokitsukaze and Mahjong

time limit per test1 second

memory limit per test256 megabytes

Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s.

In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand.

Do you know the minimum number of extra suited tiles she needs to draw so that she can win?

Here are some useful definitions in this game:

A mentsu, also known as meld, is formed by a koutsu or a shuntsu;

A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu;

A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu.

Some examples:

[2m, 3p, 2s, 4m, 1s, 2s, 4s] — it contains no koutsu or shuntsu, so it includes no mentsu;

[4s, 3m, 3p, 4s, 5p, 4s, 5p] — it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu;

[5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] — it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu.

Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite.

Input

The only line contains three strings — the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s.

Output

Print a single integer — the minimum number of extra suited tiles she needs to draw.

Examples

input

1s 2s 3s

output

0

input

9m 9m 9m

outputCopy

0

inputCopy

3p 9m 2p

outputCopy

1

Note

In the first example, Tokitsukaze already has a shuntsu.

In the second example, Tokitsukaze already has a koutsu.

In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile — 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p].

题意

日麻,给你三张麻将牌,问你怎么最少增加多少张牌可以凑成出一碰,或者一shuntsu

题解

答案就3种,0,1,2;0张直接check,1就加一张check,2就剩下的。

代码

#include<bits/stdc++.h>
using namespace std; pair<int,int> form(string s){
pair<int,int> p;
p.first=int(s[0]-'0');
if(s[1]=='m'){
p.second=0;
}else if(s[1]=='p'){
p.second=1;
}else{
p.second=2;
}
return p;
}
bool check(vector<pair<int,int> >v){
sort(v.begin(),v.end());
for(int i=0;i<v.size();i++){
for(int j=i+1;j<v.size();j++){
for(int k=j+1;k<v.size();k++){
if(v[i].second==v[j].second&&v[j].second==v[k].second){
if(v[i].first==v[j].first&&v[j].first==v[k].first){
return true;
}
if(v[i].first+1==v[j].first&&v[j].first+1==v[k].first){
return true;
}
}
}
}
}
return false;
}
int main(){
vector<pair<int,int> >v;
for(int i=0;i<3;i++){
string s;
cin>>s;
v.push_back(form(s));
}
if(check(v)){
cout<<"0"<<endl;
return 0;
}
for(int i=1;i<10;i++){
for(int j=0;j<3;j++){
vector<pair<int,int> >d = v;
d.push_back(make_pair(i,j));
if(check(d)){
cout<<"1"<<endl;
return 0;
}
}
}
cout<<"2"<<endl;
return 0;
}

Codeforces Round #573 (Div. 2) Tokitsukaze and Mahjong 水题的更多相关文章

  1. Codeforces Round #367 (Div. 2) A. Beru-taxi (水题)

    Beru-taxi 题目链接: http://codeforces.com/contest/706/problem/A Description Vasiliy lives at point (a, b ...

  2. Codeforces Round #334 (Div. 2) A. Uncowed Forces 水题

    A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/pro ...

  3. Codeforces Round #353 (Div. 2) A. Infinite Sequence 水题

    A. Infinite Sequence 题目连接: http://www.codeforces.com/contest/675/problem/A Description Vasya likes e ...

  4. Codeforces Round #327 (Div. 2) A. Wizards' Duel 水题

    A. Wizards' Duel Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/prob ...

  5. Codeforces Round #146 (Div. 1) A. LCM Challenge 水题

    A. LCM Challenge 题目连接: http://www.codeforces.com/contest/235/problem/A Description Some days ago, I ...

  6. Codeforces Round #335 (Div. 2) B. Testing Robots 水题

    B. Testing Robots Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/606 ...

  7. Codeforces Round #335 (Div. 2) A. Magic Spheres 水题

    A. Magic Spheres Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/606/ ...

  8. Codeforces Round #306 (Div. 2) A. Two Substrings 水题

    A. Two Substrings Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/550/pro ...

  9. Codeforces Round #188 (Div. 2) A. Even Odds 水题

    A. Even Odds Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/318/problem/ ...

随机推荐

  1. Mysql如何在删库后可以不用跑路

    我一直在想,地球上这么多程序员,应该有很多人在团队做项目的时候,出过很大的错误,比如说不小心删了库,活动福利字段多写了个零导致全服务器玩家领到数倍奖励,听了沙雕群友的话执行rm -rf命令. 记得有一 ...

  2. CMake使用总结(一)

    当我们在写CMakeLists.txt文件时,常常会搞不明白link_directories, LINK_LIBRARIES, target_link_libraries这3者的区别,下面就其详细介绍 ...

  3. Linux 部署redis集群

    下载Redis 下载网址 http://www.redis.cn/download.html Redis需要gcc环境(如果已经有该环境跳过此步骤) yum install gcc-c++ 安装Red ...

  4. 【计算机网络】如何让Ajax通信过程携带Cookie呢?

    Ajax 1. 介绍一下ajax并代码实现 1.1 基本概念 JavaScript 和XML(Asynchronous JavaScript And XML).简单点说,就是使用 XMLHttpReq ...

  5. pymongo和mongoengine安装和使用教程 包含常用命令行和代码示例 | pymongo and mongoengine tutorial on ubuntu 16.04

    本文首发于个人博客https://kezunlin.me/post/e88f04e5/,欢迎阅读最新内容! pymongo and mongoengine tutorial on ubuntu 16. ...

  6. WPF 解决多个TreeViewItem同时触发某事件的简单方法

    原理是: 因为是双击事件,也就是同时引发了选择事件,TreeViewItem的IsSelected为True.通过触发器设置某个属性为特殊值,然后在CS页面的双击事件方法内判断是否此值即可. xaml ...

  7. 零零总总遇到过的CSS 样式

    1:添加弹出框阴影 2:禁止文本域缩放 3:直接使用CSS 完成文本内容大小写(针对英文) 4: 文本框中的占位符 5:让table每列一样高 6:不使用js 让内容换行 word-break 7:曾 ...

  8. python基础(25):面向对象三大特性二(多态、封装)

    1. 多态 1.1 什么是多态 多态指的是一类事物有多种形态. 动物有多种形态:人,狗,猪. import abc class Animal(metaclass=abc.ABCMeta): #同一类事 ...

  9. python基础(18):初识模块、re模块

    1. 认识模块 常见的场景:一个模块就是一个包含了python定义和声明的文件,文件名就是模块名字加上.py的后缀. 但其实import加载的模块分为四个通用类别: 1.使用python编写的代码(. ...

  10. String与StringBuilder性能比对

    //String与StringBuilder性能比对package seday01;/** * String修改字符串带来的性能开销 * @author xingsir * */public clas ...