题解 | #密码验证合格程序#

前面两个要求都很简单，第三个其实也不难，长度大于2字串不能重复那么只需要判断长度为3的字串是否重复即可。看到别人用find找重复字串确实更方便。。。

#include <iostream>
#include <string>
using namespace std;
int main()
{
    string str;
    while (cin >> str) {
        int count[4] = {0};
        bool flag = true;
        if (str.length() < 8) {
            cout << "NG" << endl;
            continue;
        }
        for (int i = 0; i < str.length(); i++) {
            if (isdigit(str[i]))
                count[0] = 1;
            else if (isupper(str[i]))
                count[1] = 1;
            else if (islower(str[i]))
                count[2] = 1;
            else 
                count[3] = 1;
        }
        if (count[0] + count[1] + count[2] + count[3] < 3) {
            cout << "NG" << endl;
            continue;
        }
        //判断长度大于2的字串是否重复只需要判断有三个字符串的字串重复就行了
        for (int i = 0; i < str.size() - 3; i++) {
            for (int j = i + 1; j < str.size() - 3; j++) {
                if (str.substr(i, 3) == str.substr(j, 3)) {
                    cout << "NG" << endl;
                    flag = false;
                    break;
                }
            }
        }
        if (flag == true)
            cout << "OK" << endl;
    }
}