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We are given an array of size N. The array has all 0’s initially. The task is to count the no. of 1’s in the array after N moves. Each Nth move has a rule associated. The rules are −

1st Move − Change element at positions 1, 2, 3, 4…………..

2nd Move − Change element at positions 2, 4, 6, 8…………..

3rd Move − Change element at positions 3, 6, 9, 12…………..

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

Let’s understand with examples.

**Input**

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

**Output**

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

**Explanation** − Array after following moves −

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 following moves −

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.

We take an integer array Arr[] initialized with 0’s and integer N.

Function Onecount takes Arr[] and it’s size N as input and returns no. of ones in the final array after N moves.

The for loop starts from 1 till the end of the array.

Each i represents the ith move.

Nested for loop starts from 0th index till end of array.

For each ith move, if the index j is multiple of i (j%i==0), replace the 0 with 1 at that position.

This process continues for each i till the end of array.

**Note**− Index starts from i=1,j=1 but array indexes are from 0 to N-1. For this reason arr[j1] is converted each time.Finally traverse the whole array again and count no. of 1’s in it and store in count.

- Return count as desired result.

#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'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; }

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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