c++ - 循环 rand()，在下一个 rand() 上不要使用前一个？

Question

我每个循环都需要两个随机数，但不能使用前一个循环的随机数。我迷路了，我已经搜索过，但不知道该怎么做。请帮忙!我把我的代码放在下面。因此，我具体需要的是生成存储在 n1 和 n2 中的两个随机数。然后，在下一个循环中，不要使用那些以前的数字。但是，在没有连续使用过两次之后使用它们是可以的。

#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;

int main()
{
    //declare variables
    int numAttempts = 3;
    int response;
    char playAgain;
    int points;
    bool previousAnswer;
    srand(time(0));

    //Welcome user to the program and explain the rules
    cout << "\nWelcome! This is the Multiplication Practice Game!" << endl
    << "A multiplication problem will be presented which you must solve." << endl << endl
    << "THESE ARE THE RULES:" << endl
    << "*You start with 3 lives." << endl
    << "*Each correct answer earns you 5 points!" << endl
    << "*An incorrect answer results in 1 life lost." << endl
    << "*If you're incorrect but within 5 of the answer:" << endl
    << "\t-you are granted another attempt" << endl
    << "\t-you earn 3 points if correct" << endl
    << "\t-you lose a life if incorrect" << endl
    << "*Once you lose all of your lives, it's game over!" << endl << endl
    << "Good luck, let's begin..." << endl << endl;

    //Do while numAttempts is not equal to 0
    do{
            //Random numbers for n1 and n2
            int n1 = rand() % 13;
            int n2 = rand() % 13;

            //Present the problem and prompt for response
            cout << "Answer the problem: ";
            cout << n1 << "*"<< n2 << ": ";
            cin >> response;

            //If response is correct, congratulate
            if(response == n1*n2)
            {
                cout << "CORRECT, great job. Keep going! \n\n";
                points += 5;
                previousAnswer = true;
            }

            //If response is not correct and lives are not equal to 0
            if((response != (n1*n2)) && (numAttempts != 0))
            {
                //If response is not within 5 of the correct answer, no second chance and subtract 1 from numAttempts
                if((response > (n1*n2)+5) || (response < (n1*n2)-5) || (previousAnswer != true))
                {
                    cout << "That answer is incorrect." << endl;
                    numAttempts -= 1;
                    previousAnswer = false;
                    cout << "You have " << numAttempts << " lives remaining" << endl << endl;
                }

                //If response is within 5 of correct answer and previousAnswer is true, offer second attempt
                if(response <= ((n1*n2)+5) && (response >= (n1*n2)-5) && (previousAnswer == true))
                {
                    cout << "So close, try once more: ";
                    cin >> response;
                    if(response == n1 * n2)
                    {
                        cout << "CORRECT, great job. Keep going! \n\n";
                        points +=3;
                        previousAnswer = true;
                    }

                    //If answer is still incorrect, subtract 1 from numAttempts
                    else{
                    cout << "Sorry, that answer is still incorrect" << endl;
                    numAttempts -= 1;
                    previousAnswer = false;
                    cout << "You have " << numAttempts << " lives remaining" << endl << endl;
                    }
                }
            }


            //If user runs out of lives, notify and ask if they would like to play again
            if (0 == numAttempts)
            {
                cout << "You're all out of lives!" << endl
                << "Your total score is " << points << ", great job!" << endl
                << "Would you like to play again? Y/N: ";
                cin >> playAgain;
                if('y' == tolower(playAgain))
                {
                    cout << "\nGreat! Let's try again! Good luck!" << endl;
                    numAttempts += 3;
                    cout << "Let's begin..." << endl << endl;
                }else{
                cout << "\nThanks for playing, see you next time!" << endl;
                }
            }
      }while(numAttempts != 0); //ends loop if attempts are equal to 0


    return 0;
}

Answer 1

执行随机选择而不重复的一种简单方法是保留一个随机顺序的数字列表，范围内的每个数字对应一个条目。然后你只需取出列表前面的那个，使用它，然后将它移到后面的位置。

在你的情况下，你会将它移到至少一个远离前面的地方(在删除它之后)，这样前面的下一个数字就不会与旧的前面相同。

例子:

初始列表:

6、5、10、1、0、11、8、12、4、2、3、7、9

取第一个数字并从列表中删除。

现在的列表是:

5、10、1、0、11、8、12、4、2、3、7、9

在非前面的随机位置插入数字。

现在的列表是:

5、10、1、6、0、11、8、12、4、2、3、7、9