A. Transformation: from A to B

time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
 

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.

Examples
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：

有两种变换方式：

（1）x*10+1

（2）x*2

问a能否经过两种变换转换成b

 

如果b是奇数，则必是由（1）方式转换而来，若为偶数则必是（2）转换而来。

 

附AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 int c[100010];
 5 
 6 int main(){
 7     int a,b;
 8     cin>>a>>b;
 9     int ans=1;
10     c[0]=b;
11     int flag=1;
12     while(a<b){
13         if((b-1)%10==0){
14             b=(b-1)/10;
15             c[ans++]=b;
16         }
17         else if(b%2==0){
18             b/=2;
19             c[ans++]=b;
20         }
21         else{
22             flag=0;
23             break;
24         }
25     }
26     if(flag==0){
27         cout<<"NO"<<endl;
28     }
29     else if(a==b){
30         cout<<"YES"<<endl;
31         cout<<ans<<endl;
32         for(int i=ans-1;i>=0;i--){
33             cout<<c[i]<<" ";
34         }
35     }
36     else{
37         cout<<"NO"<<endl;
38     }
39     return 0;
40 }

 

posted @ 2016-11-05 19:46  Kiven#5197  阅读(275)  评论(0)  编辑  收藏  举报