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>
#include <stdlib.h>

int main(void)
{
  unsigned long K, A, B, ANS;
  scanf("%ld %ld %ld", &K, &A, &B);
  printf("K(%ld) A(%ld) B(%ld)\n", K, A, B);

  // 売れない
  if (K < ((A - 1) + 1)) {
    ANS = 1 + K;
  // 売れる
  } else {
    // 最初に叩いた残り回数が偶数の場合
    if ((K - (A - 1)) % 2 == 0) {
      // 売買で儲かる
      if (A + 2 < B) {
        ANS = 1 + (A - 1) + (B - A) * (K - (A - 1)) / 2;
      // 売買で儲からない
      } else {
        ANS = 1 + K;
      }
    // 最初に叩いた残り回数が奇数の場合
    } else {
      // 売買で儲かる
      if (A + 2 < B) {
        ANS = (A - 1) + (B - A) * (K - (A - 1)) / 2;
      // 売買で儲からない
      } else {
        ANS = 1 + K;
      }
    }
  }
  printf("%ld\n", ANS);
  return 0;
}