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>

#define FOR(i, l, r) for(ll i = l; i < r; i++)
#define rep(i, N) FOR(i, 0, N)

#define MOD 1000000007
#define INF 1000000000

using ll = long long int;
using namespace std;

typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;

int main() {
    ll K, A, B; cin >> K >> A >> B;
    ll ans = 1;
    if(A+2 >= B || K < A+1) ans += K;
    else{
      ans = A + (((K-A+1)/2)*(B-A)) + (K-A+1)%2;
    }

    cout << ans << "\n";
    return 0;
}