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：這個題目要分奇偶討論。
如果n是偶數，有兩種轉換形式
①n-1->n
②n/2->n(n/2直接復制到n，花費y）
如果n是奇數，有三種轉換形式
①n-1->n
②n/2->n（n的一半復制到n-1花費y，然后還少一個，采用增加一個的形式，花費x）
③n/2+1->n（n/2+1復制到n+2花費y,多了一個，采用減少一個的形式,花費x）
狀態轉移方程：

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)/~

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