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 "pch.h"
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

#define llong long long
#define fr(i,l,r) for(i=(l);i<(r);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))
#define INF 1000000000000//10^12

int main(void)
{
	//変数の宣言
	llong int k,a,b;
	llong int num[5];
	llong int ans;
	llong int i;
	llong int flag;
	llong int mod;
	
	//データの読み込み
	
	scanf("%lld %lld %lld",&k,&a,&b);

//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	if(b-a<=2){
		ans=1+k;
	}else{
		k=k-a+1;
		mod=k%2;
		k=k-mod;
		ans=a+(b-a)*k/2+mod;
	}
	
//	printf("計算部分終了\n");
	
	//出力
	printf("%lld",ans);
	
//	printf("結果の出力終了\n");
	
	return 0;
}