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.*;
public class Main {
	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int K = sc.nextInt();
		long A = sc.nextLong();
		long B = sc.nextLong();
		/*long dp[] = new long[1000000000];
		dp[0] = 1;
		dp[1] = 2;
		for(int i = 2; i <= K; i++) {
			long temp = (long)(dp[i - 2] / A) * B - ((long)(dp[i - 2] / A) * A);
			dp[i] = Math.max(dp[i - 1] + 1, dp[i - 2] + temp);
			//System.out.println(dp[i]);
		}*/
		
		/*
		long dp[] = new long[3];
		dp[0] = 1;
		dp[1] = 2;
		long temp = (long)(dp[0] / A) * B - ((long)(dp[0] / A) * A);
		dp[2] = Math.max(dp[1] + 1, dp[0] + temp);
		for(int i = 2; i <= K; i++) {
			if(i > 2) {
				dp[0] = dp[1];
				dp[1] = dp[2];
			}
			temp = (long)(dp[0] / A) * B - ((long)(dp[0] / A) * A);
			dp[2] = Math.max(dp[1] + 1, dp[0] + temp);
		}*/
		
		long ans = 1;
		for(int i = 0; i < K; i++) {
				if(ans < A || i == K - 1 || B - A < 2) {
					ans++;
					continue;
				} else {
					//long temp = ans / A;
					ans = ans - A + B;
					//System.out.println(ans);
					i++;
				}
		}
		
		System.out.println(ans);
	}
}