rt

调了好久，样例都过不去。 样例三个点对两个。

题目链接：Texas hold 'em

代码如下：

#include <bits/stdc++.h>
using namespace std;

int id(string s)
{
	if (s == "#") return -1;
	int v = 0;
	if (s[0] == 'H') v += 13;
	if (s[0] == 'D') v += 26;
	if (s[0] == 'C') v += 39;
	if (s[1] == 'A') v += 12;
	else if (s[1] == 'T') v += 8;
	else if (s[1] == 'J') v += 9;
	else if (s[1] == 'Q') v += 10;
	else if (s[1] == 'K') v += 11;
	else v += s[1] - '2';
	return v;
}

int card[15];

int get()
{
	string s;
	cin >> s;
	return id(s);
}

int po[10];

int calc(int a[])
{
	int v[5];
	for (int i = 0;i < 5;i++)
		v[i] = a[i];
	sort(v,v + 5);
	if (v[0] / 13 == v[4] / 13)
	{
		for (int i = 0;i < 5;i++)
			v[i] %= 13; 
		sort(v,v + 5);
		if (v[0] == 8 && v[4] == 12) return 90000000;
		if (v[0] + 4 == v[4]) return 80000000 + v[4];
		if (v[0] + 3 == v[3] && v[0] == 0 && v[4] == 12) return 80000005;
		return 50000000 + v[4] * po[4] + v[3] * po[3] + v[2] * po[2] + v[1] * po[1] + v[0];
	}
	for (int i = 0;i < 5;i++)
		v[i] %= 13; 
	sort(v,v + 5);
	if (v[1] == v[2] && v[2] == v[3] && v[3] == v[4]) return 70000000 + v[1] * po[1] + v[0];
	if (v[1] == v[2] && v[2] == v[3] && v[3] == v[0]) return 70000000 + v[0] * po[1] + v[4];
	if (v[0] == v[1] && v[1] == v[2])
	{
		if (v[3] == v[4]) return 60000000 + v[0] * po[1] + v[3];
		return 30000000 + v[0] * po[2] + v[4] * po[1] + v[3];
	}
	if (v[2] == v[1] && v[3] == v[2]) return 30000000 + v[1] * po[2] + v[4] * po[1] + v[0];
	if (v[2] == v[3] && v[4] == v[2])
	{
		if (v[0] == v[1]) return 60000000 + v[2] * po[1] + v[0];
		return 30000000 + v[2] * po[2] + v[1] * po[1] + v[0];
	}
	if (v[0] == v[1] && v[2] == v[3]) return 20000000 + v[2] * po[2] + v[0] * po[1] + v[4];
	if (v[0] == v[1] && v[4] == v[3]) return 20000000 + v[3] * po[2] + v[0] * po[1] + v[2];
	if (v[2] == v[1] && v[4] == v[3]) return 20000000 + v[4] * po[2] + v[1] * po[1] + v[0];
	if (v[0] == v[1]) return 10000000 + v[0] * po[3] + v[4] * po[2] + v[3] * po[1] + v[2];
	if (v[1] == v[2]) return 10000000 + v[1] * po[3] + v[4] * po[2] + v[3] * po[1] + v[0];
	if (v[2] == v[3]) return 10000000 + v[2] * po[3] + v[4] * po[2] + v[1] * po[1] + v[0];
	if (v[3] == v[4]) return 10000000 + v[3] * po[3] + v[2] * po[2] + v[1] * po[1] + v[0];
	if (v[0] + 3 == v[3] && v[0] == 0 && v[4] == 12) return 40000005;
	if (v[0] + 4 == v[4]) return 40000000 + v[4];
	return v[4] * po[4] + v[3] * po[3] + v[2] * po[2] + v[1] * po[1] + v[0];
}

bool work()
{
	int me[5],opp[5];
	int ta = 0,tb = 0;
	for (int i = 5;i <= 9;i++)
		for (int j = i + 1;j <= 9;j++)
			for (int k = j + 1;k <= 9;k++)
			{
				me[0] = card[1],me[1] = card[2],me[2] = card[i],me[3] = card[j],me[4] = card[k];
				ta = max(ta,calc(me));
			}
	for (int i = 5;i <= 9;i++)
		for (int j = i + 1;j <= 9;j++)
			for (int k = j + 1;k <= 9;k++)
			{
				opp[0] = card[3],opp[1] = card[4],opp[2] = card[i],opp[3] = card[j],opp[4] = card[k];
				tb = max(tb,calc(opp));
			}
	return (ta > tb);
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	po[1] = 13;
	for (int i = 2;i <= 4;i++)
		po[i] = po[i - 1] * 13;
	while (1)
	{
		memset(card,0,sizeof(card));
		for (int i = 1;i <= 7;i++)
		{
			card[i] = get();
			if (card[i] == -1) return 0;
		}
		int sum = 0,all = 45 * 44,cnt = 0;
		for (int i = 0;i < 52;i++)
		{
			bool b = 1;
			for (int j = 1;j <= 7;j++)
				if (i == card[j]) b = 0;
			if (!b) continue;
			for (int j = 0;j < 52;j++)
			{
				bool b2 = 1;
				for (int k = 1;k <= 7;k++)
					if (j == card[k] || j == i) b2 = 0;
				if (!b2) continue;
				card[8] = i,card[9] = j;
				if (work()) sum++;
			}
		}
		cout << fixed << setprecision(20) << 1.0 * sum / all << endl;
	}
	return 0;
}