Reach John to top - c++

Question

Problem Statement
John wants to climb up a stair of n steps. He can climb 1 or 2 steps at each move. John wants the number of moves to be a multiple of an integer m.
What is the minimal number of steps making him climb to the top of the stairs that satisfies his condition?
Input
The single line contains two space separated integers n, m (0 < n ≤ 10000, 1 < m ≤ 10).
Output
Print a single integer — the minimal steps being a multiple of m. If there is no way he can climb satisfying condition print  - 1 instead.
Sample test(s)
Input
10 2
Output
6
Input
3 5
Output
-1
Notes:
For the first sample, John could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.
My code:
I have been trying to solve this problem ,so i thought of using nearest power of 2 less than the given number but got wrong answer
#include<stdio.h>
#include<math.h>
using namespace std;
int main(){
int n,m;
scanf("%d %d",&n,&m);
int x= pow (2,floor (log2(n)) );
int rem = n-x;
int ans = ((x/2)+rem);
if ( ans % m == 0 )
printf (" %d \n ",ans);
else
printf("-1\n");
return 0;
}

Answer 1

I personally don't see how the powers of two are useful at all.
Let's write a different algorithm in pseudocode first:
N = number of steps
M = desired multiple
# Excluding any idea of the multiple restraint, what is the maximum and minimum
# number of steps that John could take?
If number of steps is even:
minimum = N / 2
maximum = N
If number of steps is odd:
minimum = N / 2 + 1
maximum = N
# Maybe the minimum number of steps is perfect?
If minimum is a multiple of M:
Print minimum
# If it isn't, then we need to increase the number of steps up to a multiple of M.
# We then need to make sure that it didn't surpass the maximum number of steps.
Otherwise:
goal = minimum - (minimum % M) + M
if goal <= maximum:
Print goal
Otherwise:
Print -1
We can then convert this to code:
#include <cstdio>
int main(){
int n,m;
scanf("%d %d", &n, &m);
const int minimum = (n / 2) + (n % 2);
const int maximum = n;
if (minimum % m == 0) {
printf("%d\n", minimum);
return 0;
}
const int guess = minimum - (minimum % m) + m;
if (guess <= maximum) {
printf("%d\n", guess);
return 0;
}
printf("%d\n", -1);
return 0;
}
The key tool that I'm using here is that I know that John can scale the stairs in any combination of steps between (and including) [minimum, maximum]. How can we determine this?
We know the minimum number of steps is using 2 at a time as much as possible.
We know the maximum number of steps is using 1 at a time.
We know that if the current number of steps is not the maximum, then we can replace one of the steps (that must be using 2 at a time) and replace it with taking each step one at a time. That would increase the total number of steps by 1.