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+// A Divide and Conquer based program for maximum subarray sum problem 
+#include <stdio.h> 
+#include <limits.h> 
+  
+// A utility funtion to find maximum of two integers 
+int max(int a, int b) { return (a > b)? a : b; } 
+  
+// A utility funtion to find maximum of three integers 
+int max(int a, int b, int c) { return max(max(a, b), c); } 
+  
+// Find the maximum possible sum in arr[] auch that arr[m] is part of it 
+int maxCrossingSum(int arr[], int l, int m, int h) 
+{ 
+    // Include elements on left of mid. 
+    int sum = 0; 
+    int left_sum = INT_MIN; 
+    for (int i = m; i >= l; i--) 
+    { 
+        sum = sum + arr[i]; 
+        if (sum > left_sum) 
+          left_sum = sum; 
+    } 
+  
+    // Include elements on right of mid 
+    sum = 0; 
+    int right_sum = INT_MIN; 
+    for (int i = m+1; i <= h; i++) 
+    { 
+        sum = sum + arr[i]; 
+        if (sum > right_sum) 
+          right_sum = sum; 
+    } 
+  
+    // Return sum of elements on left and right of mid 
+    // returning only left_sum + right_sum will fail for [-2, 1] 
+    return max(left_sum + right_sum, left_sum, right_sum); 
+} 
+  
+// Returns sum of maxium sum subarray in aa[l..h] 
+int maxSubArraySum(int arr[], int l, int h) 
+{ 
+   // Base Case: Only one element 
+   if (l == h) 
+     return arr[l]; 
+  
+   // Find middle point 
+   int m = (l + h)/2; 
+  
+   /* Return maximum of following three possible cases 
+      a) Maximum subarray sum in left half 
+      b) Maximum subarray sum in right half 
+      c) Maximum subarray sum such that the subarray crosses the midpoint */
+   return max(maxSubArraySum(arr, l, m), 
+              maxSubArraySum(arr, m+1, h), 
+              maxCrossingSum(arr, l, m, h)); 
+} 
+  
+/*Driver program to test maxSubArraySum*/
+int main() 
+{ 
+   int arr[] = {2, 3, 4, 5, 7}; 
+   int n = sizeof(arr)/sizeof(arr[0]); 
+   int max_sum = maxSubArraySum(arr, 0, n-1); 
+   printf("Maximum contiguous sum is %dn", max_sum); 
+   getchar(); 
+   return 0; 
+}