我是攻读计算机工程专业的一年级学生，我们的任务是创建一个蛮力算法，该算法将破解用户提供的密码，我决定再加倍努力并使用并行编程，现在这是代码没有并行编程：
EDIT:OLD CODE WAS HERE
它可以工作，但是我已经尝试过以多种方式实施OpenMP，要么我遇到了无法解决的非常混乱的比赛条件，要么就无法正常工作，我只是想问一些提示，我知道这是由我来完成任务。
编辑：这是新代码
#include <omp.h>
#include <iostream>
#include <ctime>
#include <string>
#include <stdio.h>
using namespace std;
void crackPassword(string pass);
long long int attempt;
clock_t start_t, end_t;
string test2[3];
string alphabet;
static int digit[7], alphabetSet = 1;
string test;
int main() {
string password;
std::cout << "Enter the password to crack : ";
cin >> password;
crackPassword(password);
std::cout << "The number of attempts : " << attempt << endl;
return 0;
}
void alphabets(int alphabetSet) {
switch (alphabetSet) {
case 1: alphabet = "-0123456789";
break;
case 2: alphabet = "-abcdefghijklmnopqrstuvwxyz";
break;
case 3: alphabet = "-ABCDEFGHIJKLMNOPQRSTUVWXYZ";
break;
case 4: alphabet = "-0123456789abcdefghijklmnopqrstuvwxyz";
break;
case 5: alphabet = "-0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
break;
case 6: alphabet = "-abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
break;
case 7: alphabet = "-0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
break;
}
}
void cases7( int alphabetSet, string pass, int passwordLength) {
while (alphabetSet < 8) {
alphabets(alphabetSet);
for (digit[6] = 0; digit[6] < alphabet.length() ; digit[6]++)
for (digit[5] = 0; digit[5] < alphabet.length() ; digit[5]++)
for (digit[4] = 0; digit[4] < alphabet.length() ; digit[4]++)
for (digit[3] = 0; digit[3] < alphabet.length(); digit[3]++)
for (digit[2] = 0; digit[2] < alphabet.length() ; digit[2]++)
for (digit[1] = 0; digit[1] < alphabet.length() ; digit[1]++)
for (digit[0] = 1; digit[0] < alphabet.length(); digit[0]++) {
attempt++;
if (attempt % 2500000 == 0) std::cout << ".";
test = alphabet[digit[0]];
for (int i = 1; i < passwordLength; i++)
if (alphabet[digit[i]] != '-')test += alphabet[digit[i]];
if (pass.compare(test) == 0) {
end_t = clock(); std::cout << endl << endl << endl << ">\n>> CRACKED THE PASSWORD! >>\n>" << endl << endl << "The password : " << pass;
std::cout << "The time duration passed : " << (double)(end_t - start_t) / 1000 << " seconds" << endl << endl;
system("pause");
}
}
alphabetSet++;
}
}
void cases6(int alphabetSet, string pass, int passwordLength) {
while (alphabetSet < 8) {
alphabets(alphabetSet);
for (digit[5] = 0; digit[5] < alphabet.length() ; digit[5]++)
for (digit[4] = 0; digit[4] < alphabet.length(); digit[4]++)
for (digit[3] = 0; digit[3] < alphabet.length() ; digit[3]++)
for (digit[2] = 0; digit[2] < alphabet.length(); digit[2]++)
for (digit[1] = 0; digit[1] < alphabet.length() ; digit[1]++)
for (digit[0] = 1; digit[0] < alphabet.length(); digit[0]++) {
attempt++;
if (attempt % 2500000 == 0) std::cout << ".";
test = alphabet[digit[0]];
for (int i = 1; i < passwordLength; i++)
if (alphabet[digit[i]] != '-')test += alphabet[digit[i]];
if (pass.compare(test) == 0) {
end_t = clock(); std::cout << endl << endl << endl << ">\n>> CRACKED THE PASSWORD! >>\n>" << endl << endl << "The password : " << pass;
std::cout << "The time duration passed : " << (double)(end_t - start_t) / 1000 << " seconds" << endl << endl;
system("pause");
}
}
alphabetSet++;
}
}
void cases5(int alphabetSet, string pass, int passwordLength) {
while (alphabetSet < 8) {
alphabets(alphabetSet);
for (digit[4] = 0; digit[4] < alphabet.length(); digit[4]++)
for (digit[3] = 0; digit[3] < alphabet.length(); digit[3]++)
for (digit[2] = 0; digit[2] < alphabet.length(); digit[2]++)
for (digit[1] = 0; digit[1] < alphabet.length(); digit[1]++)
for (digit[0] = 1; digit[0] < alphabet.length(); digit[0]++) {
attempt++;
if (attempt % 2500000 == 0) std::cout << ".";
test = alphabet[digit[0]];
for (int i = 1; i < passwordLength; i++)
if (alphabet[digit[i]] != '-')test += alphabet[digit[i]];
if (pass.compare(test) == 0) {
end_t = clock(); std::cout << endl << endl << endl << ">\n>> CRACKED THE PASSWORD! >>\n>" << endl << endl << "The password : " << pass;
std::cout << "The time duration passed : " << (double)(end_t - start_t) / 1000 << " seconds" << endl << endl;
system("pause");
}
}
alphabetSet++;
}
}
void cases4(int alphabetSet, string pass, int passwordLength) {
while (alphabetSet < 8) {
alphabets(alphabetSet);
for (digit[3] = 0; digit[3] < alphabet.length(); digit[3]++)
for (digit[2] = 0; digit[2] < alphabet.length(); digit[2]++)
for (digit[1] = 0; digit[1] < alphabet.length(); digit[1]++)
for (digit[0] = 1; digit[0] < alphabet.length(); digit[0]++) {
attempt++;
if (attempt % 2500000 == 0) std::cout << ".";
test = alphabet[digit[0]];
for (int i = 1; i < passwordLength; i++)
if (alphabet[digit[i]] != '-')test += alphabet[digit[i]];
if (pass.compare(test) == 0) {
end_t = clock(); std::cout << endl << endl << endl << ">\n>> CRACKED THE PASSWORD! >>\n>" << endl << endl << "The password : " << pass;
std::cout << "The time duration passed : " << (double)(end_t - start_t) / 1000 << " seconds" << endl << endl;
system("pause");
}
}
alphabetSet++;
}
}
void cases3(int alphabetSet, string pass, int passwordLength) {
while (alphabetSet < 8) {
alphabets(alphabetSet);
for (digit[2] = 0; digit[2] < alphabet.length(); digit[2]++)
for (digit[1] = 0; digit[1] < alphabet.length(); digit[1]++)
for (digit[0] = 1; digit[0] < alphabet.length(); digit[0]++) {
attempt++;
if (attempt % 2500000 == 0) std::cout << ".";
test = alphabet[digit[0]];
for (int i = 1; i < passwordLength; i++)
if (alphabet[digit[i]] != '-')test += alphabet[digit[i]];
if (pass.compare(test) == 0) {
end_t = clock(); std::cout << endl << endl << endl << ">\n>> CRACKED THE PASSWORD! >>\n>" << endl << endl << "The password : " << pass;
std::cout << "The time duration passed : " << (double)(end_t - start_t) / 1000 << " seconds" << endl << endl;
system("pause");
}
}
alphabetSet++;
}
}
void cases2(int alphabetSet, string pass,int passwordLength) {
while (alphabetSet < 6) {
alphabets(alphabetSet);
for (digit[1] = 0; digit[1] < alphabet.length(); digit[1]++)
for (digit[0] = 1; digit[0] < alphabet.length(); digit[0]++) {
attempt++;
if (attempt % 2500000 == 0) std::cout << ".";
test = alphabet[digit[0]];
for (int i = 1; i < passwordLength; i++)
if (alphabet[digit[i]] != '-')test += alphabet[digit[i]];
if (pass.compare(test) == 0) {
end_t = clock(); std::cout << endl << endl << endl << ">\n>> CRACKED THE PASSWORD! >>\n>" << endl << endl << "The password : " << pass;
std::cout << "The time duration passed : " << (double)(end_t - start_t) / 1000 << " seconds" << endl << endl;
system("pause");
}
}
alphabetSet++;
}
}
void cases1(int alphabetSet, string pass, int passwordLength) {
while (alphabetSet < 4) {
alphabets(alphabetSet);
for (digit[1] = 0; digit[1] < alphabet.length(); digit[1]++)
for (digit[0] = 1; digit[0] < alphabet.length(); digit[0]++) {
attempt++;
if (attempt % 2500000 == 0) std::cout << ".";
test = alphabet[digit[0]];
for (int i = 1; i < passwordLength; i++)
if (alphabet[digit[i]] != '-')test += alphabet[digit[i]];
if (pass.compare(test) == 0) {
end_t = clock(); std::cout << endl << endl << endl << ">\n>> CRACKED THE PASSWORD! >>\n>" << endl << endl << "The password : " << pass;
std::cout << "The time duration passed : " << (double)(end_t - start_t) / 1000 << " seconds" << endl << endl;
system("pause");
}
}
alphabetSet++;
}
}
void crackPassword(string pass) {
start_t = clock();
while (1) {
#pragma omp parallel num_threads(7)
#pragma omp parallel
{
#pragma omp single
{
#pragma omp task
cases1(alphabetSet, pass, 1);
#pragma omp task
cases2(alphabetSet, pass, 2);
#pragma omp task
cases3(alphabetSet, pass, 3);
#pragma omp task
cases4(alphabetSet, pass, 4);
#pragma omp task
cases7(alphabetSet, pass, 7);
#pragma omp task
cases5(alphabetSet, pass, 5);
#pragma omp task
cases6(alphabetSet, pass, 6);
}
}
}
}
我现在使用的是g ++编译器，因为可视化studip不支持openMP 3.0，现在我编译时就这样做了
g++ Hello.cpp -o Hello.exe -fopenmp -lpthread
在我打开exe并输入密码后，它完全崩溃了，我还注意到，当我输入类似1的非常短的密码时，我看到似乎是无限循环或只是线程正在执行此操作，这也值得提到我使用的是Windows，而不是Linux。
答案 0 :(得分：1)
即使是对任何事物的成功暴力破解通常也涉及某种领域知识或渗透。要通过考试，请从教师桌的抽屉中获取密码。
如果没有介绍中的建议，则可以使用迭代器构建密码生成器。这样可以轻松地与标准函数（例如for循环等）集成。取消引用的迭代器随后将返回密码。牢记领域知识，可以使它在使用次数较少的密码之前生成使用最多的密码-但是对于直截了当的攻击，只需逐步解决所有可能的字符组合。 operator++将使迭代器进入下一个密码。在我的示例中，我使用从<space>到~（包括两端）的所有ASCII字符。
生成器还将需要边界，开始和停止密码，以便能够生成从aaa到bbb的所有密码。这样就可以并行运行许多生成器-为它们提供唯一的生成密码范围。
还需要一个分区程序来划分工作，以便您可以为所有生成器分配其唯一的密码范围。它也应该有一个开始和停止密码，但是要设置更大的范围，例如，从no password到~~~~~~~（7个波浪号，这是我生成器中7个字符密码的最后一个）。然后，它将使用硬件知识（可以同时运行多少个线程是合理的），并使用一个生成器启动那么多线程，每个生成器将提供要测试的整个密码范围的唯一部分。
要使其更进一步，可以构建一个分区程序，并使用不限数量的密码进行测试。它会为本地分区程序集群提供服务，并为它们提供新的范围，以在它们用尽时进行测试。我没有在示例中包括它。相反，我制作了本地分区程序，以便它可以动态创建新范围。
我建议您研究execution policies和find_if来运行分区程序（可以运行，但是...）。但是，在此示例中，我实现了一些与执行策略类似的方法，但是具有以预定顺序创建密码生成器的方式，并且还具有在找到答案后停止循环的方式。另外，还没有多少编译器支持执行策略。但是，此示例将在C ++ 17模式下与Visual Studio，g ++和clang ++一起使用。
在Intel Core i9-7920X（12核，2线程/核）上收集的一些令人沮丧的统计数据：
在不到3.9分钟的时间内即可找到任何0-6个字符的密码
在不到7.3小时的时间内找到了0-7个字符的密码
我没有为8个字符的密码运行它：-）