Problem:
Score : 400 points

Problem Statement

Snuke has one biscuit and zero Japanese yen (the currency) in his pocket.
He will perform the following operations exactly K times in total, in the order he likes:

Hit his pocket, which magically increases the number of biscuits by one.

Exchange A biscuits to 1 yen.

Exchange 1 yen to B biscuits.

Find the maximum possible number of biscuits in Snuke's pocket after K operations.

Constraints

1 \leq K,A,B \leq 10^9

K,A and B are integers.

Input

Input is given from Standard Input in the following format:

K A B

Output

Print the maximum possible number of biscuits in Snuke's pocket after K operations.

Sample Input 1

4 2 6

Sample Output 1

7

The number of biscuits in Snuke's pocket after K operations is maximized as follows:

Hit his pocket. Now he has 2 biscuits and 0 yen.

Exchange 2 biscuits to 1 yen. his pocket. Now he has 0 biscuits and 1 yen.

Hit his pocket. Now he has 1 biscuits and 1 yen.

Exchange 1 yen to 6 biscuits. his pocket. Now he has 7 biscuits and 0 yen.

Sample Input 2

7 3 4

Sample Output 2

8

Sample Input 3

314159265 35897932 384626433

Sample Output 3

48518828981938099

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	static long biscuits = 1;
	static long money = 0;
	static long K;
	static long A;
	static long B;

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		K = scanner.nextInt() - 1;
		A = scanner.nextInt();
		B = scanner.nextInt();

		long A1 = A + 1;
		for (int i = 0; i < K; i++) {
			if (i == K && money == 0) {
				knock();
			} else if (B > A1) {
				if (money == 1) {
					moneyToBiscuit();
				} else if (biscuits >= A) {
					biscuitToMoney();
				} else {
					knock();
				}
			} else {
				knock();
			}
		}

		System.out.println(biscuits);
	}

	private static void knock() {
		biscuits++;
	}

	private static void biscuitToMoney() {
		biscuits -= A;
		money++;
	}

	private static void moneyToBiscuit() {
		money--;
		biscuits += B;
	}
}