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: Accepted)

import java.util.Scanner;

public class Main {
	static Scanner sc = new Scanner(System.in);
	static long K = Long.parseLong(sc.next());
	static long A = Long.parseLong(sc.next());
	static long B = Long.parseLong(sc.next());
	public static void main(String[] args) {
		if (B-A <= 2 || K-A <= 0) {
			System.out.println(K+1);
			return;
		}
//		long trade = (K+1)/(A+2);
		// 一度Aまでビスケットを増やしてしまえば何回でも叩ける
		long trade = (K+1 - A) / 2;
//		System.out.println(trade);
//		System.out.println(B-A-2);
		long ans = trade * (B-A-2) + K + 1;
		System.out.println(ans);
	}
}