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。