Maximum Interval Sum

Given a array of natural numbers, we have to find the maximum interval sum.
5    5   -1    5    5
Now we have these intervals (5 + 5), ( 5 + 5 + (-1)), (5 + 5 + (-1) + 5), (5 + 5 + (-1) + 5 + 5). 
Which interval has maximum sum?
(5 + 5) = 10
( 5 + 5 + (-1)) = 9
(5 + 5 + (-1) + 5) = 14
(5 + 5 + (-1) + 5 + 5) = 19
So, the last interval has maximum sum. These type of problems are called Maximum Interval Sum.
C++ implementation:

#include <iostream>
using namespace std;
const int INF = (-(1<<30));  // negative infinity
int MIS( ) {
    sum = 0; // initially sum is 0
    Max = -INF; // Maximum is less than 0
    for( i=0; i<N; i++ ) {
       sum += A[i]; // "A[]" array of intejars
        if( sum > Max ) Max = sum;
        if( sum < 0 ) sum = 0;  // sum below 0 is not useful, right?
    }
    return Max;
}