循环 rand()，在下一个 rand() 上不要使用前一个？

Question

我每次循环需要两个随机数，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。不过连续两次没用过就可以用了

#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持随机顺序的数字列表，您范围内的每个数字一个条目。然后你就拿列表前面的那个，使用它，把它移到后面的位置。

在你的情况下，你会将它移到至少一个远离前面的位置（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。

示例：

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。

列表现在是：

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。

列表现在是：

5、10、1、6、0、11、8、12、4、2、3、7、9

Answer 2

您可以在 rand()%13 之前使用 randomize(); 然后程序每次都会生成随机数并且不会重复..

循环 rand()，在下一个 rand() 上不要使用前一个？

loop rand(), on next rand() don't use previous?

c++

random

ctime