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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>
#include <string.h>
#include <math.h>

int main (int argc, char const *argv[])
{
    int64_t K, A, B;
    int64_t i, ans = 0;

    scanf("%"PRId64"%"PRId64"%"PRId64, &K, &A, &B);

    if (A >= (B - 1)) {
        printf("%"PRId64"\n", 1 + K);
        return 0;
    }

    i = (1 + K) / A;
    for (;;) {
        if ((1 + K - (i * 2)) / (A * i) > 0) break;
        i--;
    }

    ans = 1 + (B - A) * i + K - (i * 2);

    printf("%"PRId64"\n", ans);
}