Educational Codeforces Round 16 E. Generate a String-优快云博客

本文介绍了一种通过动态规划解决的问题：如何用最短时间生成包含n个字符'a'的文件。具体操作包括插入或删除字符'a'以及复制粘贴整个文件内容。

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

这题的大意是一个人要输出n个‘a’，他打出或删除一个‘a’需要x秒钟，复制粘贴要y秒钟，问你要输入n个字符‘a’最少要多长时间。

这题其实是一个不难的DP，分奇偶进行DP就行，一个转态只和前面的某两个状态有关。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

long long dp[10000010];
int main(void)
{
    int n,i,j,x,y;
    while(scanf("%d%d%d",&n,&x,&y)==3)
    {
        dp[0] = 0;
        for(i=1;i<=n;i++)
        {
            if(i % 2 == 1)
                dp[i] = min(dp[i-1]+x,dp[(i+1)/2]+x+y);
            else
                dp[i] = min(dp[i-1]+x,dp[i/2]+y);
        }
        printf("%I64d\n",dp[n]);
    }
}