CodeForces 727A Transformation: from A to B （DFS）

Vasily has a number a, which he wants to turn into a number b. For this purpose, he can do two types of operations:

multiply the current number by 2 (that is, replace the number x by 2·x);
append the digit 1 to the right of current number (that is, replace the number x by 10·x + 1).
You need to help Vasily to transform the number a into the number b using only the operations described above, or find that it is impossible.

Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform a into b.

Input
The first line contains two positive integers a and b (1 ≤ a < b ≤ 109) — the number which Vasily has and the number he wants to have.

Output
If there is no way to get b from a, print "NO" (without quotes).

Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer k — the length of the transformation sequence. On the third line print the sequence of transformations x1, x2, ..., xk, where:

x1 should be equal to a,
xk should be equal to b,
xi should be obtained from xi - 1 using any of two described operations (1 < i ≤ k).
If there are multiple answers, print any of them.

Example
Input
2 162
Output
YES
5
2 4 8 81 162
Input
4 42
Output
NO
Input
100 40021
Output
YES
5
100 200 2001 4002 40021

题意：

给两个数a,b。对a每次操作，只能在末尾添1或者乘2。问最终能否得到b，输出步数和变化的中间数。

题解：

DFS水水就好了啊。

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=1e7+5;
int a[maxn];
bool flag=false;
int ans,n,k;
void dfs(int x,int step)
{
	if(x==k)
	{
		flag=true;
		ans=step;
		return ;
	}
	if(x>k)
		return ;
	a[step]=x*10+1;
	if(x<=2e8)//防止乘10后爆int
		dfs(x*10+1,step+1);
	if(flag)
		return ;
	a[step]=x*2;
	dfs(x*2,step+1);
}
int main()
{
	while(cin>>n>>k)
	{
		flag=false;
		memset(a,0,sizeof(a));
		dfs(n,1);
		if(flag)
		{
			cout<<"YES\n"<<ans<<endl;
			a[0]=n;
			for(int i=0;i<ans;i++)
				cout<<a[i]<<' ';
			cout<<endl;
		}
		else
		{
			cout<<"NO"<<endl;
		}
	}
	return 0;
}