AMCAT: INACTIVE AND ACTIVE Puzzle solution in C

Question :There are 8 peoples which are standing in a row. They can have two states on a  respective day: Inactive(0) or Active(1). We were given an array of 8 elements which shows the states of these people today and we have to calculate the state of these people after the given no. of days.

Assumption 1: State of person will be Inactive on the next day if both the adjacent persons are having the Active state or Inactive state today.

Assumption 2: State of person will be Active on the next day if one adjacent person is having Active state and other adjacent person is having Inactive state or vice versa.

Assumption 3: Person on the extreme left and extreme right have only one adjacent person, so we can imagine, the other adjacent person is Inactive.

Ex- 1

Input 1: [0,1,1,0,0,1,0,1], 1

Output 1: [1,1,1,1,1,0,0,0]

Ex-2

Input 2: [0,1,0,1,1,0,1,1], 2

Output 2: [0,1,1,1,1,0,1,1]

 /* After day 1, states will be like this [1,0,0,1,1,0,1,1] 

#include <iostream>
using namespace std; 
void equal(int *arr1, int *arr2, int n) 
{
 for(int i=0;i<n;i++) 
    { arr1[i]=arr2[i]; } 
} 
void changestate(int *arr, int n, int days) 
 { int *tmp=(int*)malloc(n*sizeof(int)); 
 for(int i=0;i<days;i++) 
  { for(int j=0;j<n;j++) 
 {  if(j==0) 
   { tmp[j]=arr[j+1]; } 
  else if(j>0&&j<n-1)
  { 
  if(arr[j-1]==arr[j+1]) 
   tmp[j]=0; 
  else tmp[j]=1; 
  }
  else 
  { tmp[j]=arr[j-1]; }
 }
  equal(arr,tmp,n); //Copy the tmp array in arr 
 }
} 
int main() 
{ int *state, n,days; 
 cout<<"Enter the no. of states: ";
 cin>>n; 
 cout<<"\nEnter the "<<n<<" states: ";
 state = (int*)malloc(n*sizeof(int));
 for(int i=0;i<n;i++) { cin>>state[i]; } 
 cout<<"\nEnter the no. of days: ";
 cin>>days; cout<<"\n\nOutput: "; 
 changestate(state,n,days);
 for(int i=0;i<n;i++) 
 { cout<<state[i]<<" "; } 
return 0;
 }