c - 如何在循环中为数组设置特定值？ - C

Question

我正在学习初级编程类(class)，我们的任务是编写一个 10 人开始的游戏程序。你从第一个人开始，数到 3 个人，然后丢弃那个人。你继续这样做，直到只剩下一个人。

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n[10] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, i = 0;
    int * ptrN = &n[i];

    for( i = 0; n[i]; n[i+3]){

        if (n[i+3] = 1){ 
            *ptrN = 0; 
        }
        else if (n[i] = 0)
            continue;
        else
            printf("Wrong");
    }
    printf("The value of n0 %d\tn1 %d\tn2 %d\tn3 %d\tn4 %d\tn5 %d\tn6 %d\tn7 %d\tn8 %d\tn9 %d", n[0], n[1], n[2], n[3], n[4], n[5], n[6], n[7], n[8], n[9]);
    return 0;
}

我现在离我的答案还很远，但我已经遇到了问题。当我运行上面的代码时，它只会更改 n[0] 的值，然后退出程序。将不胜感激任何指导。谢谢

Answer 1

正如评论中提到的:

1. 使用==进行比较，=用于赋值。
2. 您不需要使用指针。您可以使用n[i]
3. 您需要增加i。
4. 您需要跳过 0。

正如您所提到的，当只剩下一个 1 时，您也需要停止。

解决这个问题的方法不止一种，但你可以通过一个计数器来计算有多少个 1，并使用一个 while 循环来实现这一点。当计数器减至 1 时结束:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n[10] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, count = 10;

    /* `i` needs to start at -1 so that we start counting:
     *     "0, 1, 2. n[2] is out!"
     *  instead of counting:
     *     "1, 2, 3. n[3] is out!"
     */
    int i = -1;

    while (count > 1) {

        /* We need to count three of those persons that are not out.
         * That means we have to search for the next person that is not
         * out (by skipping over `0`) and repeat that three times.
         *
         * If we just increase `i` by 3 ( `(i + 3) % 10` ) then we are
         * not checking how many of those persons between `n[i]` and
         * `n[(i + 3) % 10]` needed to be skipped over.
         *
         * So repeat three times:
         */
        for (int j = 0; j < 3; j++) {

            /* This will search for the next person that is not out by 
             * increasing `i` by one until `n[i]` is not `0`:
             */
            do {
                i = (i + 1) % 10;
            } while (n[i] == 0); // Check next person if this one is out.

        } // I forgot to close this bracket in my first version of the answer

        n[i] = 0;
        count--;
    }

    printf("The value of n0 %d\tn1 %d\tn2 %d\tn3 %d\tn4 %d\tn5 %d\tn6 %d\tn7 %d\tn8 %d\tn9 %d", n[0], n[1], n[2], n[3], n[4], n[5], n[6], n[7], n[8], n[9]);
    return 0;
}