C 언어에서는 N번 이동한 후 배열에 있는 1의 개수를 셉니다.

Arr[]={ 0,0,0,0 } N=4

Move 1: { 1,1,1,1 }
Move 2: { 1,0,1,0 }
Move 3: { 1,0,0,3 }
Move 4: { 1,0,0,1 }
Number of ones in the final array is 2.

Arr[]={ 0,0,0,0,0,0} N=6

Move 1: { 1,1,1,1,1,1,1 }
Move 2: { 1,0,1,0,1,0,1 }
Move 3: { 1,0,0,1,0,0,1 }
Move 4: { 1,0,0,0,1,0,0 }
Move 5: { 1,0,0,0,0,1,0 }
Move 4: { 1,0,0,0,0,0,1 }
Number of ones in the final array is 2.

#include <stdio.h>
int Onecount(int arr[], int N){
   for (int i = 1; i <= N; i++) {
      for (int j = i; j <= N; j++) {
         // If j is divisible by i
         if (j % i == 0) {
            if (arr[j - 1] == 0)
               arr[j - 1] = 1; // Convert 0 to 1
            else
               arr[j - 1] = 0; // Convert 1 to 0
         }
      }
   }
   int count = 0;
   for (int i = 0; i < N; i++)
      if (arr[i] == 1)
         count++; // count number of 1&#39;s
   return count;
}
int main(){
   int size = 6;
   int Arr[6] = { 0 };
   printf("Number of 1s in the array after N moves: %d", Onecount(Arr, size));
return 0;
}

Number of 1s in the array after N moves: 2