0 
#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