CF 295div2 B dp（记忆化搜索）

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http://codeforces.com/contest/520/problem/B

Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number n.

Bob wants to get number m on the display. What minimum number of clicks he has to make in order to achieve this result?

Input

The first and the only line of the input contains two distinct integers n and m (1 ≤ n, m ≤ 104), separated by a space .

Output

Print a single number — the minimum number of times one needs to push the button required to get the number m out of number n.

        Sample test(s)
      
          input
        
4 6

          output
        
2

          input
        
10 1

          output
        
9

Note

In the first example you need to push the blue button once, and then push the red button once.

In the second example, doubling the number is unnecessary, so we need to push the blue button nine times.

/**
CF 295div2 B dp（记忆化搜索）
题目大意：给定一个n，每次操作可以减1，可以乘2，问最少几次操作到m
解题思路：若n>=m，答案为n-m；否则dp[x]=min(dp[x-1],dp[2*x]),采用记忆化搜索来写
*/
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;

int n,m,dp[20005],vis[20005];

int dfs(int x)
{
    if(x<=0)
        return 9999999;
    if(x>=m)
    {
        dp[x]=x-m;
        ///printf("%d,%d\n",x,dp[x]);
        return dp[x];
    }
    if(vis[x])
    {
        ///printf("%d,%d\n",x,dp[x]);
        return dp[x];
    }
    vis[x]=1;
    dp[x]=min(dfs(x-1),dfs(2*x))+1;
   /// printf("%d,%d\n",x,dp[x]);
    return dp[x];
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(dp,0x3f3f3f3f,sizeof(dp));
        memset(vis,0,sizeof(vis));
        printf("%d\n",dfs(n));
    }
    return 0;
}