Generate a String CodeForces - 710E（dp）

zscoder wants to generate an input file for some programming competition problem.

His input is a string consisting of n letters ‘a’. He is too lazy to write a generator so he will manually generate the input in a text editor.

Initially, the text editor is empty. It takes him x seconds to insert or delete a letter ‘a’ from the text file and y seconds to copy the contents of the entire text file, and duplicate it.

zscoder wants to find the minimum amount of time needed for him to create the input file of exactly n letters ‘a’. Help him to determine the amount of time needed to generate the input.

Input

The only line contains three integers n, x and y (1 ≤ n ≤ 107, 1 ≤ x, y ≤ 109) — the number of letters ‘a’ in the input file and the parameters from the problem statement.

Output

Print the only integer t — the minimum amount of time needed to generate the input file.

Examples

Input

8 1 1

Output

4

Input

8 1 10

Output

8

思路：這個題目要分奇偶讨論。

如果n是偶數，有兩種轉換形式

①n-1->n

②n/2->n(n/2直接複制到n，花費y）

如果n是奇數，有三種轉換形式

①n-1->n

②n/2->n（n的一半複制到n-1花費y，然後還少一個，采用增加一個的形式，花費x）

③n/2+1->n（n/2+1複制到n+2花費y,多了一個，采用減少一個的形式,花費x）

狀态轉移方程：

if(i&1) dp[i]=min(dp[i-1]+x,min(dp[i/2]+x+y,dp[i/2+1]+x+y));
else dp[i]=min(dp[i-1]+x,dp[i/2]+y);
           

代碼如下：

#include<bits/stdc++.h>
#define ll long long
using namespace std;

const int maxx=1e7+100;
ll n,x,y;
ll dp[maxx];

int main()
{
	scanf("%lld%lld%lld",&n,&x,&y);
	memset(dp,-1,sizeof(dp));
	dp[0]=0;
	dp[1]=x;
	for(int i=2;i<=n;i++)
	{
		if(i&1) dp[i]=min(dp[i-1]+x,min(dp[i/2]+x+y,dp[i/2+1]+x+y));
		else dp[i]=min(dp[i-1]+x,dp[i/2]+y);
	}
	cout<<dp[n]<<endl;
	return 0;
}
           

努力加油a啊，(o)/~