题解 | #打牌#

简单粗暴解法c++

#include<iostream>
#include<map>
#include<set>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    string a, b;
    map<int, string> mp;
    mp[0] = "12345";
    mp[1] = "23456";
    mp[2] = "34567";
    mp[3] = "45678";
    mp[4] = "56789";
    while (cin >> a >> b) {
        int len = b.size();
        bool flag = false;
        switch (len) {
        case 1:
            for (int i = 0; i < a.size(); ++i)
                if (a[i] > b[0]) {
                    flag = true;
                    break;
                }
            break;
        case 2:
            for (int i = 0; i < a.size() - 1; ++i) {
                if (a[i] == a[i + 1] && (a[i] - '0') * 10 + (a[i + 1] - '0') > (b[0] - '0') * 10 + (b[1] - '0')) {
                    flag = true;
                    break;
                }
            }
            break;
        case 3:
            for (int i = 0; i < a.size() - 2; ++i)
                if (a[i] == a[i + 1] && a[i + 1]== a[i + 2] && (a[i] - '0') * 100 + (a[i + 1] - '0') * 10 + a[i + 2] - '0' > (b[0] - '0') * 100 + (b[1] - '0') * 10 + b[2] - '0') {
                    flag = true;
                    break;
                }
            break;
        case 4:
            for (int i = 0; i < a.size() - 3; ++i)
                if (a[i] == a[i + 1] && a[i + 1] == a[i + 2] && a[i + 2] == a[i + 3] && (a[i] - '0') * 1000 + (a[i + 1] - '0') * 100 + (a[i + 2] - '0') * 10 + a[i + 3] - '0' > (b[0] - '0') * 1000 + (b[1] - '0') * 100 + (b[2] - '0') * 10 + b[3] - '0') {
                    flag = true;
                    break;
                }
            break;
        default:
            int i = 0;
            set<char> Set;
            for(int i = 0;i < a.size();++i)
                Set.insert(a[i]);
            string tmp = "";
            for(auto it : Set)
                tmp = tmp + it;
            a = tmp;
            for (; i < mp.size(); ++i)
                if (mp[i] == b)
                    break;
            for (int j = i + 1; j <= 4; ++j) {
                if (a.find(mp[j]) != string::npos) {
                    flag = true;
                    break;
                }
            }
            break;
        }
        if (flag)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
}