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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <math.h>

int main(int argc, char *argv[])
{
    int k, a, b;
    int diff;
    long long biscuit;
    long long money;
    int i;

    scanf("%d %d %d", &k, &a, &b);

    diff = b - a; //売買による増分

    biscuit = 1;
    money = 0;

    for (i = 0; i < k; i++) {
        if (diff > 2) { //売買による増分のほうが多い
            if (money > 0) {
                // 買う
                money--;
                biscuit += b;
                continue;
            } else if (biscuit >= a && k - i >= 2) {
                // 売る
                biscuit -= a;
                money++;
                continue;
            }
        }
        biscuit++;
    }

    printf("%lld\n", biscuit);

    return 0;
}