The fee within the \(h\) o’clock (from exactly \(h\) o’clock to exactly \((h+1)\) o’clock) can be computed as follows:

• \(X\) if \(L \leq h \leq R-1\)
• \(Y\) otherwise

For the integer \(h\), run a loop within the range \(A \leq h \leq B-1\), find the fee within \(h\) o’clock as described above, and sum them up.

Sample code (C++)

#include <iostream>
using std::cin;
using std::cout;
using std::cerr;
using std::endl;

int main (void) {
	int x, y, l, r, a, b;
	cin >> x >> y >> l >> r >> a >> b;

	int sum = 0;
	for (int i = a; i <= b-1; i++) {
		// cost of i-th hour
		if (l <= i && i <= r-1) {
			sum += x;
		} else {
			sum += y;
		}
	}

	cout << sum << "\n";

	return 0;
}