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#include <stdio.h>
#include <time.h>
int main(void)
{
    int games = 0;
    int stayWins = 0;
    int switchWins = 0;
    int chosenDoor;
    int remainingDoor;
    int revealedDoor;
    int winningDoor;
    int option;

    srand (time(NULL));

    do
    {
    chosenDoor = rand() % 3 + 1;
    winningDoor = rand() % 3 + 1;
        do
        {
            revealedDoor = rand() % 3 + 1;
        } while (revealedDoor == chosenDoor || revealedDoor == winningDoor);

        do
        {
            remainingDoor = rand() % 3+1;
        } while (remainingDoor == chosenDoor || remainingDoor == revealedDoor);

        option = rand() % 2 + 1;
        if (option == 1)
        {
            if (chosenDoor == winningDoor)
            {
                stayWins++;
            }
        }
        if (option == 2)
        {
            chosenDoor = remainingDoor;
            if (chosenDoor == winningDoor)
            {
                    switchWins++;
            }
        }
        games++;
    } while (games < 10000);

printf("Out of 10,000 games, the contestant won %d times by staying with his/her original choice and won %d times by switching his/her choice.",stayWins,switchWins);

    return 0;
}

晚安,这是一个完整的蒙蒂霍尔问题代码,它打印了 10,000 场比赛的结果。代码将为用户选择所选的门。我该如何更改它,以便程序允许用户自己选择?同样,我该如何修改它,以便程序不会为程序随机化“1”或“2”的值,而是允许我选择切换?我的进步......而不是这个:

chosenDoor = rand() % 3 + 1;

使用这个,只有可接受的输入是 1、2 或 3 :

printf("Choice:");
scanf("%d",&chosenDoor);

这是正确的轨道吗?我知道此时用户需要在程序“完成”之前输入他的选择 10,000 次,那么有没有办法可以将第一选择应用于其他 9,999 次试验?

4

1 回答 1

1

有没有办法可以将第一选择应用于其他 9,999 项试验?

移动

printf("Choice: ");
scanf("%d", &chosenDoor);

循环外的部分代码。

于 2012-10-28T21:56:30.493 回答