讲一下最后一个条件，不能含有长度大于2的不含有公共元素的重复子串，这句话意思是含有公共元素的重复子串是OK的，暴力解法的话直接遍历两个for验证长度是3的情况，因为长度更大的是3的子集。

#include <string>
#include <map>
#include <vector>
#include <cctype>

using namespace std;

int main() {
    string str, result;
    
    while (cin >> str) {
        bool lowalpha = false, upalpha = false, digit = false, other = false;
        bool changed = false;
        if (str.size() > 8) {
            for (int i = 0; i < str.size(); i++) {
                if (islower(str[i])) lowalpha = true;
                if (isupper(str[i])) upalpha = true;
                if (isdigit(str[i])) digit = true;
                if (ispunct(str[i])) other = true;
            }
            if (lowalpha & upalpha & digit & other ||
                lowalpha & upalpha & digit || upalpha & digit & other ||
                lowalpha & digit & other) {
                for(int i =0;i<str.size()-6;i++){
                    for(int j= i+3;j<str.size()-3;j++){
                        
                        if((str[i]==str[j])&&(str[i+1]==str[j+1])&&
                           (str[i+2]==str[j+2])){
                        result ="NG";changed=true;}
                    }
                }
                if(!changed) result = "OK";
                
            }
            else result ="NG";


        }
        else result = "NG";
        cout<<result<<endl;

    }


}