In C language, count the number of 1's in the array after N moves

Given an array of size N. The array is initially all zeros. The task is to count. The number of 1's in the array after N moves. Each Nth step has an associated rule. The rule is -

• The first move-change the element at position 1, 2, 3, 4………….

• th The second move - change the elements at positions 2, 4, 6, 8... ……..

• Count the number of 1’s in the last array.

We understand through examples.

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Input

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

Output

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

Explanation - Array after subsequent movement-

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.

Input

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

Output

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

Explanation - Array after subsequent movement-

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.

The method used in the following program is as follows

We use an integer array Arr[] initialized with 0 and an integer N.

• Function Onecount takes an Arr[] and its size N as input and returns no. The number in the final array after N moves.

• The for loop starts from 1 and goes to the end of the array.

• Each i represents step i.

• Nested for loop starts at index 0 and goes to the end of the array.

• For each i-th move, if index j is a multiple of i (j%i==0), replace 0 at that position with 1.

• Continue this process for each i until the end of the array.

• Note
- Indexing starts at i=1,j=1, but array indexing goes from 0 to N-1. So arr[j1] will be converted every time.

• Finally traverse the entire array again, counting no. It contains 1 and is stored in the count.

• Returns the count of the desired results.

Example
• Live Demonstration

#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;
}

Output

If we run the above code, it will generate the following output-

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

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