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;

class Main{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		long K=sc.nextLong();
		long A=sc.nextLong();
		long B=sc.nextLong();

		if(A>=B) {
			pl((K+1));		//ビスケットを交換しないでそのままため続けたほうが良い
			//最初に1枚持っている
			System.exit(0);
		}
		else if(A<B) {	//交換したほうがお得なとき
			long res=K-A;
			long ans=(B-A)*((res)/2)+A;
			if(res%2==0) {
				pl(ans);
				System.exit(0);
			}
			else {
				ans=(B-A)*(1+(res)/2)+A;
				pl(ans);
				System.exit(0);
			}
		}
	}
	public static void pl(Object o) {
		System.out.println(o);
	}
	public static void pl() {
		System.out.println();
	}
	public static void p(Object o) {
		System.out.print(o);
	}
}