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 <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <math.h>
int compare(const void *a,const void *b){
    if(*(int*)a>*(int*)b)return 1;
    return -1;
}
int main()
{   long long  a,b,k;
    scanf("%lld%lld%lld",&k,&a,&b);
    if(k<=a){
    printf("%lld",k+1);
    return 0;
    }
    if(a+2>=b){
        printf("%lld",k+1);
        return 0;
    }
    long long ans;
    if((k-a+1)%2==0){
        ans=(b-a)*(k-a+1)/2+a;
    }else{
        ans=(b-a)*(k-a)/2+1+a;
    }
    printf("%lld",ans);
    
    return 0;
}