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 Python (3.4.3) (judge verdict: Wrong Answer)

k, a, b = map(int, input().split())

if a < b - 2:
    # k が a+2より大きい場合
    if k > (a + 1):
        # まずa+2 -1 回叩いて b 枚を得る 最初に１枚あることを考慮する
        k2 = k - (a + 1)
        res = b
        # 残りの k2回を a枚を1円に，1円をB枚に交換の2回の操作をできるだけ繰り返す
        # 2回の操作で (b-a)円増える
        # この操作は　k2//2 + 1 回行える
        res += (k2//2) * (b - a) + (k2 // 2)

    # kがa+2より小さい場合は全部叩くしかない
    else:
        res = k + 1
else:
    # a枚をb枚に交換しても増えなければ全部叩いた方が良い
    res = k + 1

print(res)