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.io.*;
import java.util.*;

public class Main {
    static MyReader in = new MyReader();

    public static void main(String[] args) {
        int k = in.i() + 1;
        int a = in.i();
        long b = in.i() - a;
        int c = k - a;
        long ans = Math.max(k, a + c / 2 * b + c % 2);
        System.out.println(ans);
    }
}

class MyReader extends BufferedReader {
    char[] cbuf = new char[1024];
    int head = 0;
    int tail = 0;

    MyReader() {
        super(new InputStreamReader(System.in));
    }

    char next() {
        if (head == tail) {
            try {
                tail = super.read(cbuf, 0, cbuf.length);
            } catch (IOException e) {
                e.printStackTrace();
            }
            head = 0;
        }
        return cbuf[head++];
    }

    int i() {
        int k = next() - '0';
        boolean minus = k == -3;
        int n = minus ? 0 : k;
        while (0 <= (k = next() - '0') && k <= 9) n = 10 * n + k;
        return minus ? -n : n;
    }

    int[] ii(final int N) {
        int[] a = new int[N];
        for (int j = 0; j < a.length; j++) a[j] = i();
        return a;
    }

    long l() {
        int k = next() - '0';
        boolean minus = k == -3;
        long n = minus ? 0 : k;
        while (0 <= (k = next() - '0') && k <= 9) n = 10 * n + k;
        return minus ? -n : n;
    }

    char[] s(final int N) {
        char[] s = new char[N];
        for (int i = 0; i < N; i++) {
            s[i] = next();
        }
        next();
        return s;
    }

    public int read(char[] cbuf) {
        int i;
        char c;
        for (i = 0; (c = next()) != ' ' && c != '\n'; i++) cbuf[i] = c;
        return i;
    }

    public String readLine() {
        try {
            return super.readLine();
        } catch (IOException e) {
            return null;
        }
    }
}