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 {

    public static void main(String[] args) {
        TestCase testCase = readFromInput();

        final long result = process(testCase);

        output(result);
    }

    private static TestCase readFromInput() {
        Scanner sc = new Scanner(System.in);

        long K = sc.nextLong();
        long A = sc.nextLong();
        long B = sc.nextLong();

        sc.close();
        return new TestCase(K, A, B);
    }

    public static long process(TestCase testCase) {
        long K = testCase.K;
        long A = testCase.A;
        long B = testCase.B;

        boolean isPossibleToFullExchange = K >= (A - 1 + 2);
        final long diff = B - A;
        boolean isBeneficialToExchange = diff > 1;
        if (isPossibleToFullExchange && isBeneficialToExchange) {
            final long numRemainingOps = K - (A - 1);
            return A + numRemainingOps / 2 * diff + numRemainingOps % 2;
        } else {
            return 1L + K;
        }
    }

    private static void output(long result) {
        System.out.println(result);
    }

    public static class TestCase {
        final long K;
        final long A;
        final long B;

        public TestCase(long k, long a, long b) {
            K = k;
            A = a;
            B = b;
        }
    }
}