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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace みんプロ
{
    class C
    {
        static ulong[] kab;
        static void Main(string[] args)
        {
            kab = Console.ReadLine().Split().Select(ulong.Parse).ToArray();
            ulong ans = 0;
            if (kab[1] >= kab[2] - 1 || kab[0] <= kab[1]) { ans = kab[0] + 1; }
            else
            {
                if ((kab[0] - kab[1] - 1) % 2 == 0)
                {
                    ans = kab[2];
                    ans += (kab[2] - kab[1]) * ((kab[0] - kab[1] - 1) / 2);
                    /*for (int i = 0; i < kab[0] - kab[1] - 1; i+=2)
                    {
                        ans += kab[2] - kab[1];
                    }*/
                }
                else
                {
                    ans = kab[2];
                    ans += (kab[2] - kab[1]) * ((kab[0] - kab[1] - 2) / 2);
                    ans++;
                    /*for (int i = 0; i < kab[0] - kab[1] - 3; i+=2)
                    {
                        ans += kab[2] - kab[1];
                    }
                    ans++;*/
                }
            }
            Console.WriteLine(ans);
        }
    }
}