在C＆＃x2B;＆＃x2b;中自动嵌套回路

Question

我刚刚开始使用C ++，目前我正在尝试构建一种破解密码的算法。我遇到了嵌套循环的问题，以获取更长的密码。目前，它适用于4个字符密码，我必须等待一段时间，但是它可以使用。是否有一种方法可以摆脱硬编码的嵌套循环，使该算法比对于角色而不必一遍又一遍地复制和粘贴if语句的算法更重要？

#include <iostream>
#include <string>
#include <ctime>  


using namespace std;

const char alphanum[] = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&* ";
const string pw = "0000";
bool cracked = false;
int tries = 0;
const int pwLength = pw.length();
string crack();
string guess="0";
const int maxLen = 30;

int main()
{
    clock_t begin, end;
    double time;
    begin = clock();

    string crackedPassword = crack();
        
    end = clock();

    time = double(end - begin) / (double)CLOCKS_PER_SEC;
    cout << "The password is: " << crackedPassword << "   The program checked a total of " << tries << " possibilites and took " << time << " seconds to crack the password!";

    return 0;
}


string crack() {
    int index = 0;
    int repeat = 0;
    while (!cracked) {
        for (int i = 0; i < 71; i++) {
            tries++;
            guess[index] = alphanum[i];
            cout << "Trying: " << guess << endl;
            if (repeat > 0) {
                for (int i = 0; i < 71; i++) {
                    tries++;
                    guess.push_back(alphanum[i]);
                    cout << "Trying: " << guess << endl;
                    if (repeat > 1) {
                        for (int i = 0; i < 71; i++) {
                            tries++;
                            guess.push_back(alphanum[i]);
                            cout << "Trying: " << guess << endl;
                            if (repeat > 2) {
                                for (int i = 0; i < 71; i++) {
                                    tries++;
                                    guess.push_back(alphanum[i]);
                                    cout << "Trying: " << guess << endl;
                                    if (guess == pw) {
                                        cracked = true;
                                        break;
                                    }
                                    else {
                                        guess.pop_back();
                                    }
                                }
                            }
                            if (guess == pw) {
                                cracked = true;
                                break;
                            }
                            else {
                                guess.pop_back();
                            }
                        }
                    }
                    if (guess == pw) {
                    cracked = true;
                    break;
                }
                    else {
                        guess.pop_back();
                    }
                }
            }
            if (guess == pw) {
                cracked = true;
                break;
            }
        }
        repeat++;
    }
    return guess;
}