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>
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <complex>
#include <tuple>

#define FAST ios_base::sync_with_stdio(false); cin.tie(0)
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
// #define repr(i, a, b) for(int i=int(a); i>=int(b); i--)
#define reps(i,n) for(int i=1;i<=int(n);++i)
#define repr(i,x) for(int i=((int)(x)-1);i>=0;i--)
#define reprs(i,x) for(int i=((int)(x));i>0;i--)
#define iceil(x,y) (x+y-1)/y
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin(); it!=(c).end(); ++it)
#define all(x) (x).begin(),(x).end()
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() )
#define LF "\n"

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef pair<int, PII> PP;

ll K,A,B;

int main(){
	FAST;

	cin >> K >> A >> B;

	if (B-A<=2 || K+1<=A) {
		cout << K+1 <<LF;
	}
	else {
		K -= A-1;
		ll ans = A + K/2 * (B-A) + K % 2;
		cout << ans << LF;
	}

	return 0;
}