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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::stdin;

fn main() {
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    let vs: Vec<u64> = s.trim().split_whitespace().map(|token| token.parse().ok().unwrap()).collect();
    let k = vs[0];
    let a = vs[1];
    let b = vs[2];

    if b <= a + 2 {
        println!("{}", k + 1);
        return;
    }

    let mut n:u64 = 1;
    let mut yen:u64 = 0;
    // シミュレーション方針
    for i in (0..k).rev() {
        // println!("{}", i);
        if i <= yen {
            if yen > 0 {
                n += b;
                yen -= 1;
                // println!("  yen to n n:{}, yen:{}", n, yen);
            } else {
                n += 1;
                // println!("  n++ n:{}, yen:{}", n, yen);
            }
        } else if n >= a {
            n -= a;
            yen += 1;
            // println!("  n to yen n:{}, yen:{}", n, yen);
        } else if yen > 0 {
            n += b;
            yen -= 1;
            // println!("  yen to n n:{}, yen:{}", n, yen);
        } else {
            n += 1;
            // println!("  n++ n:{}, yen:{}", n, yen);
        }
    }
    println!("{}", n);
}