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: Wrong Answer)

#include<stdio.h>

main(){
  int K,A,B,i;
  int count=0,sum=1,a=0;
  scanf("%d %d %d",&K,&A,&B);
  if(A>=B)printf("%d\n",K+1);
  else if((B-A)>=3){
  	for(i=1;i<A;++i){
    	++count;
    }
    sum+=(B-A)/2;
    a+=2;
    while((B-A)>2*A){for(i=0;i<A;++i)++count;
    sum+=(K-count)*(B-A)/2;
                    a+=2;}
    sum+=K-count-a;
                  
    if((K-count)%2!=0)sum+=1;
    printf("%d\n",sum);
  }
  else printf("%d\n",K+1);
  return 0;
}