2021-3-17-byte-hu Pai

Title Description ：
1. All in all 36 card , Each card is 1~9. Every number 4 card .
2. You have... In your hand 14 card , If this 14 Cards meet the following conditions , That is, it's a sum of cards
3.14 There's... In the cards 2 A card with the same number , It's called a bird's head .
4. Remove the above 2 card , be left over 12 Cards can make up 4 A shunzi or Kezi . Shunzi means increasing continuously 3 A number card （ for example 234,567 etc. ）, It means the same number 3 A number card （ for example 111,777）
for example ：
1 1 1 2 2 2 6 6 6 7 7 7 9 9 It can make up 1,2,6,7 Of 4 A quarter and 9 The head of a bird , You can play cards
1 1 1 1 2 2 3 3 5 6 7 7 8 9 use 1 Make a bird's head , Group 123,123,567,789 The four shunzi , You can play cards
1 1 1 2 2 2 3 3 3 5 6 7 7 9 No matter what 1 2 3 7 Which is the head of a bird , Can't form and card conditions .
Now? , Small bag from 36 Draw out of cards 13 card , He wants to know about the rest of 23 Zhang paizhong , Take another card , What kinds of number cards can be taken to match the cards .

#include <iostream>
#include <vector>
using namespace std;

/*  All in all 36 card , Each card is 1~9. Every number 4 card .  You have... In your hand 14 card , If this 14 Cards meet the following conditions , That is, it's a sum of cards  14 There's... In the cards 2 A card with the same number , It's called a bird's head .  Remove the above 2 card , be left over 12 Cards can make up 4 A shunzi or Kezi .  Shunzi means increasing continuously 3 A number card （ for example 234, 567 etc. ）, It means the same number 3 A number card （ for example 111, 777） */

vector<int> card(9);

// Check whether the remaining cards can form n A shunzi or Kezi 
bool hasTrible(int n)
{
    
	if (n == 0) return true;
	for (int i = 0; i<card.size(); ++i)
	{
    
		// Because still from 1 Start inspection , So if you check the number of cards >=3, Then it must be engraved 
		if (card[i]>2)
		{
    
			card[i] -= 3;
			if (hasTrible(n - 1))
			{
       // Check whether the remaining cards can form n-1 A shunzi or Kezi 
				card[i] += 3;
				return true;
			}
			card[i] += 3;
		}
		// Otherwise, it can only be shunzi 
		else if (i<card.size() - 2 && card[i]>0 && card[i + 1]>0 && card[i + 2]>0)
		{
    
			card[i]--;
			card[i + 1]--;
			card[i + 2]--;
			if (hasTrible(n - 1))
			{
    
				card[i]++;
				card[i + 1]++;
				card[i + 2]++;
				return true;
			}
			card[i]++;
			card[i + 1]++;
			card[i + 2]++;
		}
	}
	return false;
}

// Check 14 Whether a card can match a card 
bool isWin()
{
    
	for (int i = 0; i<9; ++i)
	{
      // In turn 1~9 Take it out as a bird's head , Check the rest 12 Whether a card is smooth or carved 
		if (card[i]<2)
			continue;
		card[i] -= 2;
		if (hasTrible(4))
		{
    
			card[i] += 2;
			return true;
		}
		card[i] += 2;
	}
	return false;
}

int main(){
    
	vector<int> res;
	int tmp;
	for (int i = 0; i<13; ++i)
	{
    
		cin >> tmp;
		card[tmp - 1]++;
	}
	for (int i = 0; i<9; ++i)
	{
      //1~9 Add... In turn , Check whether you can match the cards 
		if (card[i]>3)
			continue;
		card[i]++;
		if (isWin())           // If you add this card, you can match it , Then add it to the output 
			res.push_back(i + 1);
		card[i]--;
	}
	if (res.empty()) 
		res.push_back(0);
	for (int i = 0; i<res.size(); ++i)
		cout << res[i] << ' ';

	return 0;
}