Question

#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;
}

晚上的人，这是一个完整的Monty Hall问题代码，用于打印10,000个游戏的结果。代码将为用户选择所选的门。如何更改它以便程序允许用户自己选择？同样，我如何修改它以便程序不会随机化程序的“1”或“2”值，而是允许我选择切换？我的进步...... 而不是：

chosenDoor = rand() % 3 + 1;

使用此选项，只有可接受的输入为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，那么有没有办法将第一选择应用于其他9,999次试验？

Answer 1

我有没有办法将第一选择应用于其他9,999次试验？

移动

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

循环外的部分代码。

Monty Hall Game：如何让用户选择门代替程序自动随机选择？

1 个答案: