9.A - 作业题(建议使用BFS)

Suppose you have an integer v�. In one operation, you can:

• either set v=(v+1)mod32768�=(�+1)mod32768
• or set v=(2⋅v)mod32768�=(2⋅�)mod32768.

You are given n� integers a1,a2,…,an�1,�2,…,��. What is the minimum number of operations you need to make each ai�� equal to 00?

Input

The first line contains the single integer n� (1≤n≤327681≤�≤32768) — the number of integers.

The second line contains n� integers a1,a2,…,an�1,�2,…,�� (0≤ai<327680≤��<32768).

Output

Print n� integers. The i�-th integer should be equal to the minimum number of operations required to make ai�� equal to 00.

Sample 1

Inputcopy	Outputcopy
4 19 32764 10240 49	14 4 4 15

Note

Let's consider each ai��:

• a1=19�1=19. You can, firstly, increase it by one to get 2020 and then multiply it by two 1313 times. You'll get 00 in 1+13=141+13=14 steps.
• a2=32764�2=32764. You can increase it by one 44 times: 32764→32765→32766→32767→032764→32765→32766→32767→0.
• a3=10240�3=10240. You can multiply it by two 44 times: 10240→20480→8192→16384→010240→20480→8192→16384→0.
• a4=49�4=49. You can multiply it by two 1515 times.

• #include<bits/stdc++.h>
  typedef long long ll;
  #define endl '\n'
  using namespace std;
  int t;
  int v,ans;
  int vis[33000];
  struct nmb
  {
  	int x,temp;
  };  nmb a; nmb b; nmb c;
  queue<nmb> q;
  int main()
  {
  	ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);
  	cin>>t;
  	while(t--)
  	{
  		cin>>v;
  		a.x=v,a.temp=0;
  		vis[v]=1;
  		q.push(a);
  		while(!q.empty())
  		{
  			a=q.front();
  			q.pop();
  			if(a.x%32768==0)
  			{
  				ans=a.temp;
  				break;
  			}
  			if(vis[(a.x+1)%32768]==0)
  			{
  				b.x=(a.x+1)%32768;
  				b.temp=a.temp+1;
  				vis[(a.x+1)%32768]=1;
  				q.push(b);
  			}
  			if(vis[(a.x*2)%32768]==0)
  			{
  				b.x=(a.x*2)%32768;
  				b.temp=a.temp+1;
  				vis[(a.x*2)%32768]=1;
  				q.push(b);
  			}
  		}
  		cout<<ans<<" ";
  		memset(vis,0,sizeof(vis));
  		while(!q.empty()) q.pop();
  	}
  }