Optimal Strategy for a game

GameTheoryProblem.cpp

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+// C++ program to find out maximum value from a 
+// given sequence of coins 
+#include <bits/stdc++.h> 
+using namespace std; 
+
+// Returns optimal value possible that a player can 
+// collect from an array of coins of size n. Note 
+// than n must be even 
+int optimalStrategyOfGame(int* arr, int n) 
+{ 
+	// Create a table to store solutions of subproblems 
+	int table[n][n]; 
+
+	// Fill table using above recursive formula. Note 
+	// that the table is filled in diagonal fashion (similar 
+	// to http:// goo.gl/PQqoS), from diagonal elements to 
+	// table[0][n-1] which is the result. 
+	for (int gap = 0; gap < n; ++gap) { 
+		for (int i = 0, j = gap; j < n; ++i, ++j) { 
+
+			// Here x is value of F(i+2, j), y is F(i+1, j-1) and 
+			// z is F(i, j-2) in above recursive formula 
+			int x = ((i + 2) <= j) ? table[i + 2][j] : 0; 
+			int y = ((i + 1) <= (j - 1)) ? table[i + 1][j - 1] : 0; 
+			int z = (i <= (j - 2)) ? table[i][j - 2] : 0; 
+
+			table[i][j] = max(arr[i] + min(x, y), arr[j] + min(y, z)); 
+		} 
+	} 
+
+	return table[0][n - 1]; 
+} 
+
+// Driver program to test above function 
+int main() 
+{ 
+	int arr1[] = { 8, 15, 3, 7 }; 
+	int n = sizeof(arr1) / sizeof(arr1[0]); 
+	printf("%d\n", optimalStrategyOfGame(arr1, n)); 
+
+	int arr2[] = { 2, 2, 2, 2 }; 
+	n = sizeof(arr2) / sizeof(arr2[0]); 
+	printf("%d\n", optimalStrategyOfGame(arr2, n)); 
+
+	int arr3[] = { 20, 30, 2, 2, 2, 10 }; 
+	n = sizeof(arr3) / sizeof(arr3[0]); 
+	printf("%d\n", optimalStrategyOfGame(arr3, n)); 
+
+	return 0; 
+}