Question

my code shows all 100 "lights" with values of 1 and 0, what I need to do (and not managing to..) is to show only the lights which has value of 0 (so it's these lights [and these numbers are what needs to be shown] 1, 4, 9, 16, 25, 36, 49, 64, 81, 100). Thanks for the help.

The current code:

#include <stdio.h>
#include <conio.h>
#define lights 100
int main()
{
    int arr[lights] = { 0 }, i, j;
    printf("Lights that are off:\n");
    for (i = 0; i <= lights; i++)
        arr[i] = 1;
    for (i = 1; i <= lights; i++)
    {
        for (j = i; j <= lights; j += i)
        {
            if (arr[j] == 0)
                arr[j] = 1;
            else if (arr[j] == 1)
                arr[j] = 0;
        }
    }
    for (i = 1; i <= lights; i++)
        printf("%d ", arr[i]);
    return 0;
    _getch;
}

Answer 1

所以看起来你正试图解决灯泡拼图。这里有一个很好的解决方案：http://puzzles.nigelcoldwell.co.uk/six.htm

我已修复您的代码以使其正常工作，您必须小心索引。

#define NUM_OF_LIGHTS             (100)

void main(void)
{
  int bulbs[NUM_OF_LIGHTS] = { 0 };
  int i, j;

  for (i = 1; i < (NUM_OF_LIGHTS + 1); i++)
  {
    for (j = i; j < (NUM_OF_LIGHTS + 1); j += i)
    {
      /* Toggle the indexed bulb.
       * Note: adjust the array as it runs from 0 to 99, not 1 to 100.
       */
      bulbs[j - 1] ^= 1;
    }
  }
  /* Print out the result. */
  printf("Lights that are off:\n");
  for (i = 0; i < NUM_OF_LIGHTS; i++)
  {
    /* Only showing the bulb index for lights that are switched off. */
    if (bulbs[i] == 0)
    {
      printf("%d ", i + 1);
    }
  }
}

Need to show index number for lamps that have value of 0

1 个答案: