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: Runtime Error)

#include <stdio.h>

int main()
{
	long long int k, a, b;
	scanf("%lld%lld%lld", &k, &a, &b);
	
	int i, j;
	long long int pre[k+1];
	for (i = 0; i < k+1; i++) {
		pre[k] = 0;
	}
	if (a >= b) {
		printf("%lld\n", k + 1);
	} else {
		pre[0] = 1;
		pre[1] = 2;
		pre[2] = 3;
		for (i = 3; i < k+1; i++) {
			if ((pre[i-2] >= a) && ((pre[i-2] + 2) < (pre[i-2] - a + b))) {
				pre[i] = pre[i-2] - a + b;
			} else {
				pre[i] = pre[i-1] + 1;
			}
		}
		printf("%lld\n", pre[i-1]);
	}
	return 0;
}