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.*;
import java.io.*;

import static java.lang.System.in;

public class Main {
    public static void main(String[] args)throws IOException{
        Scanner sc = new Scanner(System.in);
        int K =sc.nextInt(),A=sc.nextInt(),B=sc.nextInt();
        long ans = help(K,A,B);
        System.out.println(ans);
    }
    static long help(long K, long A, long B){
        if(B-A<=2) return K+1;
        if(K==1) return 2;
        if(1+K-2<A) return 1+K; // after K-2 operations still can't reach A biscuits
        long ans = A;
        K -= A-1;
        ans += K/2*(B-A)+K%2;
        return ans;
    }
}