HTML & JS
/*  A Dynamic Programming (Memoization-based) approach to 
 calculate Grundy Number of a Game
 Game Description-
 Just like a one-pile version of Nim, the game starts with
 a pile of n stones, and the player to move may take any 
 positive number of stones.
 The last player to move wins. Which player wins the game? */
#include<bits/stdc++.h>
using namespace std;

// A Function to calculate Mex of all the values in that set
// This function remains same 
int calculateMex(unordered_set<int> Set)
{
	int Mex = 0;
	
	while (Set.find (Mex) != Set.end())
		Mex++;
	
	return (Mex);
}

// A function to Compute Grundy Number of 'n'
// Only this function varies according to the game
int calculateGrundy(int n, int Grundy[])
{
	if (n == 0)
		return (0);
	
	if (Grundy[n] != -1)
		return (Grundy[n]);
	
	unordered_set<int> Set; // A Hash Table
	
	for (int i=0; i<=n-1; i++)
			Set.insert(calculateGrundy(i, Grundy));
	
	// Store the result
	Grundy[n] = calculateMex (Set);
	
	return(Grundy[n]);
}

// Driver program to test above functions
int main()
{
	int n = 10;	
	
	// An array to cache the sub-problems so that
	// re-computation of same sub-problems is avoided
	int Grundy[n+1];
	memset (Grundy, -1, sizeof(Grundy));
	
	printf ("%d", calculateGrundy(n, Grundy));
	
	return (0);
}




Generated URL:

Output:

10
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