Project Euler :Problem 54 Poker hands
In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way:
- High Card: Highest value card.
- One Pair: Two cards of the same value.
- Two Pairs: Two different pairs.
- Three of a Kind: Three cards of the same value.
- Straight: All cards are consecutive values.
- Flush: All cards of the same suit.
- Full House: Three of a kind and a pair.
- Four of a Kind: Four cards of the same value.
- Straight Flush: All cards are consecutive values of same suit.
- Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.
The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives (see example 1 below). But if two ranks
tie, for example, both players have a pair of queens, then highest cards in each hand are compared (see example 4 below); if the highest cards tie then the next highest cards are compared, and so on.
Consider the following five hands dealt to two players:
Hand | Player 1 | Player 2 | Winner | |||
1 | 5H 5C 6S 7S KD
Pair of Fives
|
2C 3S 8S 8D TD
Pair of Eights
|
Player 2 | |||
2 | 5D 8C 9S JS AC
Highest card Ace
|
2C 5C 7D 8S QH
Highest card Queen
|
Player 1 | |||
3 | 2D 9C AS AH AC
Three Aces
|
3D 6D 7D TD QD
Flush with Diamonds
|
Player 2 | |||
4 | 4D 6S 9H QH QC
Pair of Queens
Highest card Nine |
3D 6D 7H QD QS
Pair of Queens
Highest card Seven |
Player 1 | |||
5 | 2H 2D 4C 4D 4S
Full House
With Three Fours |
3C 3D 3S 9S 9D
Full House
with Three Threes |
Player 1 |
The file, poker.txt, contains one-thousand
random hands dealt to two players. Each line of the file contains ten cards (separated by a single space): the first five are Player 1's cards and the last five are Player 2's cards. You can assume that all hands are valid (no invalid characters or repeated
cards), each player's hand is in no specific order, and in each hand there is a clear winner.
How many hands does Player 1 win?
果然不适合做动脑子的题目。全然没有耐心(╯‵□′)╯︵┻━┻
每轮每一个人的牌都用两个数组来存,由于这题目比較的东西要么是牌值要么就是推断全部牌是不是同一个花色。
对于上面的是十个较项目,从下往上进行比較。
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
using namespace std; bool same_suit(char s[5])
{
if (s[0] == s[1] && s[1] == s[2] && s[2] == s[3] && s[3] == s[4])
return true;
else
return false;
} int cv(char a)
{
if (a >= '2'&&a <= '9')
return a - '2';
else if (a == 'T')
return 8;
else if (a == 'J')
return 9;
else if (a == 'Q')
return 10;
else if (a == 'K')
return 11;
else
return 12;
} void ch_int(char a[5][2],int ac[5])
{
for (int i = 0; i < 5; i++)
{
int tmp = cv(a[i][0]);
ac[i] = tmp;
}
} int high_card(int ac[5],char s[5])
{
return ac[4];
} int one_pair(int ac[5],char s[5])
{
int count = 0;
for (int i = 5; i >= 1; i--)
{
if (ac[i] == ac[i - 1])
return ac[i];
}
return -1;
} int two_pair(int ac[5],char s[5])
{
vector<int>res;
int count = 0;
for (int i = 1; i < 5; i++)
{
if (ac[i] == ac[i - 1])
{
res.push_back(ac[i]);
i++;
}
}
if (res.size() == 2)
return res[0] + res[1] * 13;
else
return -1;
} int three_kind(int ac[5],char s[5])
{
int count = 0;
for (int i = 2; i < 5; i++)
{
if (ac[i - 2] == ac[i - 1] && ac[i - 1] == ac[i])
return ac[i - 2];
}
return -1;
} int straight(int ac[5],char s[5])
{
if (ac[0] == 0 && ac[1] == 1 && ac[2] == 2 && ac[3] == 3 && ac[4] == 12)
return ac[3];
for (int i = 1; i < 5; i++)
{
if (ac[i] != ac[i - 1] + 1)
return -1;
}
return ac[4];
} int flush(int ac[5],char s[5])
{
if (same_suit(s))
return ac[4];
return -1;
} int full_house(int ac[5],char s[5])
{
if (ac[0] == ac[1] && ac[2] == ac[3] && ac[3] == ac[4])
return ac[0] + ac[4] * 13;
if (ac[0] == ac[1] && ac[1] == ac[2] && ac[3] == ac[4])
return ac[4] + ac[0] * 13;
return -1;
} int four_kind(int ac[5],char s[5])
{
if ((ac[0] == ac[1] && ac[1] == ac[2] && ac[2] == ac[3]) || (ac[1] == ac[2] && ac[2] == ac[3] && ac[3] == ac[4]))
return ac[2];
return -1;
} int strai_flush(int ac[5],char s[5])
{
int tmp = straight(ac,s);
if (same_suit(s) && tmp >= 0)
return ac[4];
return -1;
} int royal_flush(int ac[5], char s[5])
{
int tmp = strai_flush(ac, s);
if (tmp >= 0 && ac[4] == 12)
return ac[4];
return -1;
} int compareHighest(int ac[5], int bc[5])
{
for (int i = 4; i >= 0; i--)
{
if (ac[i] > bc[i])
return 1;
if (ac[i] < bc[i])
return 2;
}
} int comp(int ac[5], int bc[5], char as[5], char bs[5])
{
int(*compareList[10])(int *, char *) = { high_card, one_pair, two_pair, three_kind, straight, flush, full_house, four_kind, strai_flush, royal_flush };
for (int i = 9; i >= 0; i--)
{
int pa = (*compareList[i])(ac, as);
int pb = (*compareList[i])(bc, bs); if (pa != -1 || pb != -1)
{
if (pa == -1)
return 2;
if (pb == -1)
return 1;
if (pa > pb)
return 1;
if (pa < pb)
return 2;
if (pa == pb)
return compareHighest(ac, bc);
} }
} int main()
{
ifstream input;
input.open("poker.txt");
string s;
int ct = 0;
while (getline(input, s))
{
char a[5][2];
char b[5][2];
int count = 0;
int flag = 0;
for (int i = 0; i < s.length(); i++)
{
if (s[i] == ' ')
{
count++;
if (count>4)
{
if (flag == 0)
flag = 1;
count -= 5;
}
}
else
{
if (flag == 0)
{
a[count][0] = s[i++];
a[count][1] = s[i];
}
else
{
b[count][0] = s[i++];
b[count][1] = s[i];
}
}
} int ac[5], bc[5];
char as[5], bs[5];
for (int i = 0; i < 5; i++)
{
a[i][1] = as[i];
b[i][1] = bs[i];
}
ch_int(a, ac);
ch_int(b, bc);
sort(ac, ac + 5);
sort(bc, bc + 5); if (comp(ac, bc, as, bs) == 1)
ct++;
}
cout << ct << endl;
system("pause");
return 0;
}
然后学C++都好几年了才知道有函数指针数组这样的奇妙好用的东东,感动哭。
Project Euler :Problem 54 Poker hands的更多相关文章
- Project Euler:Problem 55 Lychrel numbers
If we take 47, reverse and add, 47 + 74 = 121, which is palindromic. Not all numbers produce palindr ...
- Project Euler:Problem 63 Powerful digit counts
The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is ...
- Project Euler:Problem 86 Cuboid route
A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3, and a fly, F, sits in the o ...
- Project Euler:Problem 76 Counting summations
It is possible to write five as a sum in exactly six different ways: 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 ...
- Project Euler:Problem 87 Prime power triples
The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is ...
- Project Euler:Problem 89 Roman numerals
For a number written in Roman numerals to be considered valid there are basic rules which must be fo ...
- Project Euler:Problem 93 Arithmetic expressions
By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and making use of the four ari ...
- Project Euler:Problem 39 Integer right triangles
If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exact ...
- Project Euler:Problem 28 Number spiral diagonals
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is forme ...
随机推荐
- UVa 10954 Add All 贪心
贪心 每一次取最小的两个数,注意相加的数也要算' #include<cstring> #include<iostream> #include<cstdio> # ...
- BZOJ2179: FFT快速傅立叶 & caioj1450:【快速傅里叶变换】大整数乘法
[传送门:BZOJ2179&caioj1450] 简要题意: 给出两个超级大的整数,求出a*b 题解: Rose_max出的一道FFT例题,卡掉高精度 = =(没想到BZOJ也有) 只要把a和 ...
- BAPC 2014 Preliminary(第一场)
D:Lift Problems On the ground floor (floor zero) of a large university building a number of students ...
- 排序算法(Apex 语言)
/* Code function : 冒泡排序算法 冒泡排序的优点:每进行一趟排序,就会少比较一次,因为每进行一趟排序都会找出一个较大值 时间复杂度:O(n*n) 空间复杂度:1 */ List< ...
- bash编程,while嵌套case语句, file不能判断文件存在与否
写一个脚本, 完成如下要求 (1)脚本可接受参数 : start, stop, restart, status, (2)如果参数非非法, 提示使用格式后报错退出; (3)如果是start, 则创建/t ...
- ES6学习5 字符串的扩展
1.ES6 为字符串添加了遍历器接口,使得字符串可以被for...of循环遍历. for (let codePoint of 'foo') { console.log(codePoint) } // ...
- Hexo 与 Git 集成
git初始化项目 登录Github,初始化GitHub Pages项目.即是添加一个Git Project. 点击New repository创建一个新的Project.需要填写选项如下: - ...
- ubuntu18.04安装dash-to-dock出错的问题
之前在安装dash-to-dock出现了这种错误: TypeError: ExtensionUtils.initTranslations is not a function Stack trace:i ...
- vsCode 快捷键、插件
插件 参考链接:https://blog.csdn.net/shunfa888/article/details/79606277 快捷键及常用插件:https://www.jianshu.com/p/ ...
- oracle基础入门(四)
一:其实oracle的语法跟sql servce 挺像的只有一些个别的差异而已 1):安装Oracle的数据库一般它的表中会自带了两站表: 是 emp(员工表) , dept(部门) 单表查询 sel ...