Educational Codeforces Round 16 E 题 Generate a String

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本文探讨了如何通过动态规划解决字符串生成问题，特别是在限定时间和操作条件下，找到生成特定数量字符'a'所需的最短时间。

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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

题意：输入一个数n，x，y n表示生成字符的个数，每增加一个字符a需要x时间，是复制前面的串需要y时间。

求生成n个字符a至少需要多长时间。

我们用dp[i]表示生成i个字符需要时间当i是偶数时，dp[i]=min(dp[i-1]+x,dp[i/2]+y)(增加或复制); 当i为奇数时，dp[i]=min(dp[i-1]+x,dp[(i+1)/2]+x+y)(增加或复制和删除)。

#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<stdlib.h>
#include<stack>
#include<vector>
#include<string.h>
#include<map>
#define INF 0x3f3f3f3f3f
using namespace std;
long long dp[10000009];

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


}