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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
    public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		PrintWriter ou = new PrintWriter(System.out);
		int k = Integer.parseInt(sc.next());
		int a = Integer.parseInt(sc.next());
		int b = Integer.parseInt(sc.next());
		sc.close();
		if(a >= b) ou.println(k + 1);
		else{
			if(k < a) ou.println(k + 1);
			else{
				long yono = a;
				long s = k + 1;
				k -= a - 1;
				if(k % 2 == 1) yono++;
				k /= 2;
				yono += (b - a) * (long)k;
				ou.println(Math.max(yono , s));
			}
		}
		ou.flush();
	}
}