Sum of All Odd Length Subarrays — C++

// Time:  O(n)
// Space: O(1)

class Solution {
public:
    int sumOddLengthSubarrays(vector<int>& arr) {
        // begin\pos |0   i       (n-1)
        // ----------------------------
        // 0          --101....      ↑
        // 1           -010....    i-0+1
        // i            101....      ↓
        //              ← (n-1-i+1) →
        // 
        // for each number x with its position i, as the graph depicted above,
        // (begin, pos) pair represents a subarray arr[begin:pos+1] containing x, marked 1 if odd length else 0,
        // so the total number of 0 and 1 are exactly the total number of subarrays with x, which is (i-0+1)*((len(arr)-1)-i+1),
        // because the number of 1 is always equal to or one more than the number of 0, (always begins with 1010... and alternatively flips)
        // so there are ceil(count/2) odd length subarrays with x
        int result = 0;
        for (int i = 0; i < size(arr); ++i) {
            result += arr[i] * ceil_divide((i - 0 + 1) * ((size(arr) - 1) - i + 1), 2);
        }
        return result;
    }

private:
    int ceil_divide(int a, int b) {
        return (a + (b - 1)) / b;
    }
};