Question

我为Yahtzee编写了代码。我的代码询问用户是否要玩Yahtzee或顺子。然后滚动直到达到Yahtzee或直线。

它运行良好，但是需要花费很长时间和很多回合，因此我希望我的代码更有效并能够重新发布。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，所以我对编程的知识有限。我知道如何使用循环，切换，if和else语句。任何建议或简单的解决方案都将有所帮助。

我对如何重新启动计算机有一些想法，但是它们似乎没有作用。如果您能够弄清楚如何重新启动计算机，请向我提供对代码的修改，并指导我如何做。

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...只是开始。

您大概知道从哪里开始，知道基本结构，这就是您所需要的。

不要一口气做所有的事情，只是走一小步，测试它是否有效，然后再做一个。

您将犯错，但事情将无法正常进行，这是完全正常的现象。学习编码有点像学习马拉松走路，跌跌撞撞，跌落在地上，但是只需要振作起来然后再次向前迈进即可。每次尝试时，您都可以跑得更远，进度会令人振奋。

编辑：响应Q的编辑，具体询问如何重新滚动。

因此要掷出第二个骰子，您需要执行以下代码：

die2 = rand() % 6 + 1;

要重新滚动它，请使用同一行。

最终，您将想要将代码转换为使用六个骰子的数组。这将使重复更少的代码更加动态和智能。但这是第二个版本，只需一些小步骤，先完成一些工作，然后再进行其他工作，然后对其进行调整以使其工作得更好，等等。

需要在Yahtzee中进行计算机重新注册

1 个答案: