小丑牌｜超级大模拟

T4 - 小丑牌

题目链接跳转：点击跳转

解题思路

也是一道模拟题，但需要注意以下几点：

1. 同一个点数可能会出现五次，那么此时应该输出 High Card（如题意）。
2. 如果有一个牌型符合上述多条描述，请输出符合描述的牌型中在规则中最后描述的牌型。
3. 牌的数量不局限于传统的扑克牌，每张牌可以题目中四种花色的任意之一。

代码实现

本题的 C++ 代码如下：

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

int t;
struct card{
    int rank;
    string suit;
} cards[10];

bool cmp(card a, card b){
    return a.rank < b.rank;
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
	cin >> t;
    while(t--){
        for (int i=1; i<=5; i++){
            string rank; string suit;
            cin >> rank >> suit;
            int act;
            if (rank == "J") act = 11;
            else if (rank == "Q") act = 12;
            else if (rank == "K") act = 13;
            else if (rank == "A") act = 14;
            else if (rank == "10") act = 10;
            else act = rank[0] - '0';
            cards[i] = (card){act, suit};
        }
        sort(cards+1, cards+6, cmp);
        int cnt[20] = {};
        bool isSameSuit = true;
        bool isRanked = true;
        int pairs = 0; int greatest = 0;
        for (int i=1; i<=5; i++){
            cnt[cards[i].rank] += 1;
            if (i > 1 && cards[i].suit != cards[i-1].suit)
                isSameSuit = false;
            if (i > 1 && cards[i-1].rank + 1 != cards[i].rank)
				isRanked = false;
        }
        for (int i=1; i<=15; i++){
            if (cnt[i] == 2) pairs++;
            greatest = max(greatest, cnt[i]);
        }
        if (isRanked && isSameSuit){
            if (cards[5].rank == 14) cout << "Royal Flush" << endl;
            else cout << "Straight Flush" << endl;
        } else if (isRanked) cout << "Straight" << endl;
        else if (pairs == 1 && greatest == 3) cout << "Full House" << endl;
        else if (greatest == 4) cout << "Four of a Kind" << endl;
        else if (greatest == 3) cout << "Three of a Kind" << endl;
        else if (pairs == 2) cout << "Two Pairs" << endl;
        else if (pairs == 1) cout << "One Pair" << endl;
        else cout << "High Card" << endl;
    }
    return 0;
}