CodeForces - 520B Two Buttons bfs

最小操作数转换

最新推荐文章于 2021-12-08 20:14:28 发布

mmk27

最新推荐文章于 2021-12-08 20:14:28 发布

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本文链接：https://blog.youkuaiyun.com/mmk27_word/article/details/85254787

本文介绍了一个算法问题，即如何通过最少的操作（乘2或减1）将一个正整数n转换为另一个正整数m。使用了BFS算法进行解决，并提供了一段C++代码示例，展示了如何有效地实现这一过程。

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.

Examples

Input

4 6

Output

Input

10 1

Output

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.

题目链接：点击查看

题意：可以 *2 和 -1 问至少多少次操作将n转化成m

题解：bfs一遍即可，注意数组扩大一倍

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
int vis[20010];
int n,m;
int main()
{
	cin>>n>>m;
	memset(vis,INF,sizeof(vis));
	queue<int> q;
	q.push(n);
	vis[n]=0;
	while(!q.empty())
	{
		int now=q.front();q.pop();
		if(now==m) break;
		if(now*2<20000&&vis[now]+1<vis[now*2])
		{
			vis[now*2]=vis[now]+1;
			q.push(now*2);
		}
		if(now-1>0&&vis[now]+1<vis[now-1])
		{
			vis[now-1]=vis[now]+1;
			q.push(now-1);
		}
	}
	cout<<vis[m]<<endl;
	return 0;
}