Algorithm
Problem link- https://www.spoj.com/problems/COINS/
COINS - Bytelandian gold coins
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
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
For each test case output a single line, containing the maximum amount of American dollars you can make.
Example
Input: 12 2 Output: 13 2
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.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <map>
using namespace std;
long long int get_coins(int N, map<int, long long int> &memoArray);
int main()
{
map<int, long long int> M;
int N;
while (scanf("%lld", &N) != EOF)
{
cout << get_coins(N, M) << endl;
}
return 0;
}
long long int get_coins(int N, map<int, long long int> &memoArray)
{
if (memoArray[N] != 0)
{
return memoArray[N];
}
else
{
if (N >= 0 && N <= 11)
{
memoArray[N] = N;
return N;
}
else
{
return memoArray[N] = get_coins(N / 2, memoArray) + get_coins(N / 3, memoArray) + get_coins(N / 4, memoArray);
}
}
}Input
2
Output
2
#2 Code Example with Java Programming
Code -
Java Programming
//This problem uses a combination of
//topdown and bottom up approach in Dynamic programming
import java.util.Scanner;
class ByteLandian
{
static long[] arr;
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
arr=new long[100000+1];
//calculates the value of maximum dollar for coin upto 100000
// for coin greater than 100000 it uses recursion
for(int i=1;i<=100000;i++)
{
arr[i]=Math.max((arr[i/2]+arr[i/3]+arr[i/4]),i);
}
while(sc.hasNext())
{
int n=sc.nextInt();
System.out.println(max_dollar(n));
}
}
static long max_dollar(int n)
{
if(n<=100000)
return arr[n];
else
return(max_dollar(n/2)+max_dollar(n/3)+max_dollar(n/4));
}
}Input
2
Output
2
Demonstration
SPOJ Solution-Bytelandian gold coins-Solution in C, C++, Java, Python