c - 需要在 Yahtzee 中重新启动计算机

Question

我为 Yahtzee 编写了代码。我的代码询问用户是否想玩 Yahtzee 还是顺子。然后它会滚动，直到获得快艇或直道。

它运行良好，但需要很长时间并且需要太多轮次，所以我希望我的代码更加高效并且能够自行重新滚动。因此，如果滚动的数字是 1 2 2 3 2，我希望计算机重新滚动 1 和 3，直到它们变成 2。

我刚刚开始学习如何用 C++ 编程，所以我对编程主题的了解有限。我知道如何使用循环、switch、if 和 else 语句。任何建议或简单的解决方案都会有所帮助。

我有一些关于如何让计算机重新滚动的想法，但它们似乎不起作用。如果您能够弄清楚如何让计算机重新滚动，请向我提供对代码的修改，并指导我如何做到这一点。

int die3 = 0;
int die4 = 0;
int die5 = 0;
int roundCounter = 0;
bool yahtzeeNotFound = true;
bool yahtzeeGame = true;
char game[4] = "";

//ask user if they want to play for Yahtzee or Straight
printf("Lets play Yahtzee!\n Do you want to play for a Yahtzee or a Straight?  (Y/S)");
scanf("%s", game);

//if user wants to play Yahtzee, program rolls for same numbers from all the dice.
if (tolower(game[0]) == 'y')
{
    yahtzeeGame = true;

    //this is included so that new numbers are rolled after each round.
    srand(time(0));

    //if the condition of an unfound Yahtzee is met, the program proceeds to enter the loop.
    while (yahtzeeNotFound)
    {
        roundCounter++;

        //the dice are all rolled at same time.
        die1 = rand() % 6 + 1;
        die2 = rand() % 6 + 1;
        die3 = rand() % 6 + 1;
        die4 = rand() % 6 + 1;
        die5 = rand() % 6 + 1;

        //the numbers rolled are printed for user to see.
        //the counter counts each round, so user knows how many rounds it take for a Yahtzee.
        printf("ROUND # %d\n\n", roundCounter);
        printf("   Die 1 = %d\n", die1);
        printf("   Die 2 = %d\n", die2);
        printf("   Die 3 = %d\n", die3);
        printf("   Die 4 = %d\n", die4);
        printf("   Die 5 = %d\n\n\n\n", die5);

        //when all the dice have rolled the same number, a Yahtzee is achieved.
        if ((die1 == die2) && (die1 == die3) && (die1 == die4) && (die1 == die5))

        {
            printf(" Congratulations! You finally reached Yahtzee after %d rounds!\n", roundCounter);
            //when the Yahtzee is achieved, the program exits the loop.
            break;
        }
    }
}

else
{
//if the user does not play for a Yahtzee, they must play for a straight.
yahtzeeGame = false;

//this ensures that new number are rolled after each round.
srand(time(0));

//the program enters the loop.
while (yahtzeeNotFound)
{
    //the counter is declared.
    roundCounter++;

    //the 5 dice are rolled at the same time.
    die1 = rand() % 6 + 1;
    die2 = rand() % 6 + 1;
    die3 = rand() % 6 + 1;
    die4 = rand() % 6 + 1;
    die5 = rand() % 6 + 1;

    //the numbers are printed out with the round number for the user to see.
    printf("ROUND # %d\n\n", roundCounter);
    printf("   Die 1 = %d\n", die1);
    printf("   Die 2 = %d\n", die2);
    printf("   Die 3 = %d\n", die3);
    printf("   Die 4 = %d\n", die4);
    printf("   Die 5 = %d\n\n\n\n", die5);

    //if the numbers rolled have unique values, the user has a Straight.
    if ((die1 != die2) && (die1 != die3) && (die1 != die4) && (die1 != die5) && (die2 != die3)
            && (die2 != die4) && (die2 != die5) && (die3 != die4) && (die3 != die5) && (die4 != die5))
    {
        // The user is told how many rounds it took for them to get a Straight.
        printf(" Congratulations! You finally reached a Straight after %d rounds!\n", roundCounter);

        //when the user has a Straight, the program exits the loop.
        break;
    }
}
}
return 0;
}

Answer 1

我帮助教人们如何编码，这是一个常见问题。

答案是......刚刚开始。

您大致知道从哪里开始，知道基本结构，这就是您所需要的。

不要试图一次性完成所有工作，只需迈出一小步，测试它是否有效，然后再采取另一步。

你会犯错误，事情不会顺利进行，这是完全正常的和预期的。学习编码有点像通过马拉松学习走路，你会绊倒，你会摔倒，但你只需要爬起来，然后再次向前迈进。每次你尝试，你就能跑得更远一点，进步是令人振奋的。

编辑:回应Q的编辑，特别询问如何重新滚动。

因此，要掷第二个骰子，您需要使用以下代码行:

die2 = rand() % 6 + 1;

要重新滚动它，请使用同一条线。

最终您将希望将代码转换为对六个骰子使用数组。这将允许更动态、更智能的代码，并且重复次数更少。但这是第二个版本，小步骤，让一些东西工作，然后让其他东西工作，然后调整它们以更好地工作等等。