Bytelandian Gold Coins | CodeChef Solution

Hello coders, today we are going to solve Bytelandian Gold Coins CodeChef Solution whose Problem Code is COINS.

Task

In Byteland they have a very strange monetary system.

Each Bytelandian gold coin has an integer number written on it. A coin n can be exchanged in a bank into three coins: n/2, n/3 and n/4. But these numbers are all rounded down (the banks have to make a profit).

You can also sell Bytelandian coins for American dollars. The exchange rate is 1:1. But you can not buy Bytelandian coins.

You have one gold coin. What is the maximum amount of American dollars you can get for it?

Input Format

The input will contain several test cases (not more than 10). Each testcase is a single line with a number n, 0 <= n <= 1 000 000 000. It is the number written on your coin.

Output Format

For each test case output a single line, containing the maximum amount of American dollars you can make.

Example

Sample Input

12
2

Sample Output

13
2

Explanation

You can change 12 into 6, 4 and 3, and then change these into 6+4+3=13. If you try changing the coin 2 into 3 smaller coins, you will get 1, 0 and 0, and later you can get no more than 1 out of them. It is better just to change the 2 coin directly into 2.

Solution – Bytelandian Gold Coins

C++

#include<bits/stdc++.h>
#define ll long long
using namespace std;

unordered_map<ll, ll> mp;

ll solve(ll n) {
    if(n<2) {
        return n;
    }

    if(mp.count(n)) {
        return mp[n];
    }
    
    mp[n] = max(n, solve(n/2)+solve(n/3)+solve(n/4));

    return mp[n];
}

int main() {
    ll n;
    while(cin>>n)
        cout<<solve(n)<<'\n';
}

Python

# cook your dish her
dp={0:0,1:1,2:2,3:3,4:4}
    
while(True):
    try:
        n=int(input())
        def solve(n,dp):
            if(n in dp):
                return dp[n] 
            else:
                out=max(n,solve(n//2,dp)+solve(n//3,dp)+solve(n//4,dp))
                dp[n]=out
                return out
                
        (solve(n,dp))
        print(dp[n])
    except:
        break
    

Java

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
     public static void main(String[] args) {
        final HashMap<Long,Long> map = new HashMap();
        Scanner sc = new Scanner(System.in);

        while (sc.hasNextLong()) {
            long N = sc.nextLong();
            long max = find_max(N, map);
            System.out.println(max);
        }
    }

    static long find_max (long n, Map<Long,Long> map)
    {
        if(n ==0|| n==1) return n;
        if(map.containsKey(n)) return map.get(n);
        long max = Math.max(n, find_max(n/2,map) + find_max(n/3,map) + find_max(n/4,map));
        if(n<10000) map.put(n,max);
        return max;
    }
    
}

Disclaimer: The above Problem (Bytelandian Gold Coins) is generated by CodeChef but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.