What's wrong in my approach in a problem of game of death in a circle?

#include<bits/stdc++.h>
#include<vector>
#include<algorithm>
using namespace std;
void safe(int n,int k,int a[])
{
    vector<int>s(a+0,a+n);
    vector<int>b(a+0,a+n);
    while(s.size()>1)
    {
        int r;
        r=s.size();
       int count=1;
        while(count<=r)
        {
            b[count%r]=0;
            count=count+k;
        }
    b.erase(remove(b.begin(),b.end(),0),b.end());
    s=b;    
    }
    
    for(int x:s)
    {
        cout<<x<<"\n";
    }
}


int main()
{
    int t;
    cin>>t;
    for(int i=0;i<t;i++)
    {
        int n,k;
        cin>>n>>k;
        int *a=new int[n];
        for(int j=1;j<=n;j++)
        {
            a[j-1]=j;
        }
        if(n>2)
        {
        safe(n,k,a);
        }
        else if(n==2)
        {
     cout<<a[1]<<"\n";
        }
        else
        cout<<a[0]<<"\n";
    }
}

Input:

Output:

1
3
2
19  --> it should be 36

If you don't know this problem Description of question: There are n people standing in a circle (numbered clockwise 1 to n) waiting to be executed. The counting begins at point 1 in the circle and proceeds around the circle in a fixed direction (clockwise). In each step, a certain number of people are skipped and the next person is executed. The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom. Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle. The task is to choose the place in the initial circle so that you are the last one remaining and so survive. Consider if n = 5 and k = 2, then the safe position is 3. Firstly, the person at position 2 is killed, then person at position 4 is killed, then person at position 1 is killed. Finally, the person at position 5 is killed. So the person at position 3 survives.

Solution

Here is a possible solution, hopefully the comments make it clear. There's no need to allocate an int[] and copy it into the vector, you can init the vector directly with iota. Likewise you can remove elements as you go, rather than zeroing and removing as a later step.

#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>

using namespace std;
void safe(int n, int k) {
  vector<int> s(n);
  // Init the vector 1..n
  iota(s.begin(), s.end(), 1);
  int i = 0;
  // While more than one survive...
  while (s.size() > 1) {
    // Skip forward (k-1)
    i = (i + k - 1) % s.size();
    // Kill the next one
    s.erase(s.begin() + i);
  }
  for (int x : s) {
    cout << x << "\n";
  }
}

int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n, k;
    cin >> n >> k;
    if (n > 2) {
      safe(n, k);
    } else if (n == 2) {
      cout << 2 << "\n";
    } else
      cout << 1 << "\n";
  }
}