Let amount = 101 and coins = [1 18 28]. To get to 101, we can have s coins that have a denomination of 20 and 1 coin that has a denomination of 1. That is, (285) + (1 1)= 101. So, with 6 coins, we can get to the amount of 101. We can also get to 101 with 18 coins of denomination 10 and 1 coin of denomination 1. (1818) (11) 181. But, this will need 11 coins in total. So, the answer is 6 because we are looking for the minimum number of coins.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int amount = 101;
    int coins [] = {1, 18, 28};
    int min = 0;
   
    while (amount > 0){
        while (amount >= coins[2]){
            amount -= coins[2];
            min++;
        }
        while (amount >= coins[1]){
            amount -= coins[1];
            min++;
        }
        while (amount >= coins[0]){
            amount -= coins[0];
            min++;
        }
    }
    cout<<min;
    return 0;
}