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 PHP (5.6.30) (judge verdict: Wrong Answer)

<?php

$input = trim(fgets(STDIN));

list($k, $a, $b) = explode(" ", $input);

$biscuits = 1;
$money = 0;

if ($a >= $b) {
    // 交換するより叩いたほうが儲かる
    $biscuits += $k;
    echo $biscuits;
    exit;
}

if ( $a + 1 + 1 - 1 > $k ) {
    // 交換枚数に達しないので叩くだけで終わる
    $biscuits += $k;
    echo $biscuits;
    exit;
}

// 交換枚数まで叩いて消費
$k -= ($a - 1);
$biscuits = $a;

// 残り回数で何回交換できるか
$change_num = floor($k / 2);
$amari = $k % 2;

$biscuits += (($b - $a) * $change_num);
$biscuits += $amari;

echo $biscuits;