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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"fmt"
	"math"
)

func main() {
	var k, a, b int
	fmt.Scan(&k, &a, &b)

	var bis int
	if b%a != 0 {
		if k-(a-2)%3 == 0 {
			fmt.Println("A")
			bis = ((k - a + 2) / 3 * b)
		} else {
			fmt.Println("B")
			bis = (k-a+2)/3*b + ((k - a + 2) % 3)
		}
	} else {
		if (k-a+2)%2 == 0 {
			fmt.Println("C")
			bis = (k-a+2)/3*b + 1
		} else {
			fmt.Println("D")
			//bis = ((k-a+2)/3 + 1) * b
			bis = a * int(math.Pow(float64(3), float64((k-a+2)/3+1)))
		}
	}
	fmt.Println(bis)
}