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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> mypair;
ll max(ll a,ll b){
	if(a>b)return a;
	else return b;
}
int main(){
	ll k,a,b;
	cin>>k>>a>>b;
	if(k<a+1|| a>=b)cout<<k+1<<endl;
	else if(b-a<=2) cout<<k+1<<endl;
	else{
		ll rest  = k;
		rest = k-(a-1);
		ll ans = a;
		ll t = max(0, (rest/2) );
			ans += (b-a)*t;
		
		/*
		while(t!=0){
			ans+= (b-a)*t;
			t /= a/(b-a);

		}*/

		cout<<ans+rest%2<<endl;
	}

	return 0;
}