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<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<map>
#include<iomanip>
#define rep(index,num) for(int index=0;index<num;index++)
#define rep1(index,num) for(int index=1;index<=num;index++)
#define scan(argument) cin>>argument
#define prin(argument) cout<<argument<<endl
#define kaigyo cout<<endl
#define eps 1e-7
#define mp(a1,a2) make_pair(a1,a2)
typedef long long ll;
using namespace std;
typedef pair<ll,ll> pll;
typedef pair<int,int> pint;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<pint> vpint;
typedef vector<pll> vpll;
ll INFl=1e+18+1;
int INF=1e+9+1;
int main(){
	ll K,A,B;
	scan(K);scan(A);scan(B);
	ll ans=K+1;
	if(K>=A-1&&B>A){
		if((K-A+1)%2==0) ans=max(ans,(K-A+1)/2*(B-A)+A);
		else ans=max(ans,(K-A+1)/2*(B-A)+A+1);
	}
	prin(ans);
	return 0;
}