CF1661B Getting Zero

Getting Zero

题面翻译

给定 $v$，可以将 $v$ 如下操作：

变成 $(v + 1) \% 32768$ 或者 $2\times v\%32768$，

求最少经过几次操作能将 $v$ 变回 $0$。

题目描述

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

• either set $v = (v + 1) \bmod 32768$
• or set $v = (2 \cdot v) \bmod 32768$ .

You are given $n$ integers $a_1, a_2, \dots, a_n$ . What is the minimum number of operations you need to make each $a_i$ equal to $0$ ?

输入格式

The first line contains the single integer $n$ ( $1 \le n \le 32768$ ) — the number of integers.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ( $0 \le a_i < 32768$ ).

输出格式

Print $n$ integers. The $i$ -th integer should be equal to the minimum number of operations required to make $a_i$ equal to $0$ .

样例 #1

样例输入 #1

4
19 32764 10240 49

样例输出 #1

14 4 4 15

提示

Let's consider each $a_i$ :

• $a_1 = 19$ . You can, firstly, increase it by one to get $20$ and then multiply it by two $13$ times. You'll get $0$ in $1 + 13 = 14$ steps.
• $a_2 = 32764$ . You can increase it by one $4$ times: $32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$ .
• $a_3 = 10240$ . You can multiply it by two $4$ times: $10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$ .
• $a_4 = 49$ . You can multiply it by two $15$ times.

思路

32768实际上是2的15次方。看到最少次数的时候，可以想到bfs宽搜到最短路的性质，而两种操作方式对应的就是扩展方法，依次，我们就能求出来扩展的最少次数。

代码

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int dist[65537];
int bfs(int u){
    memset(dist,-1,sizeof dist);
    queue<int> q;
    q.push(u);
    dist[u]=0;
    while(q.size()){
        int t=q.front();
        q.pop();
        if(!t) return dist[0];
        int f1=(t+1)%32768,f2=(t<<1)%32768;
        if(!~dist[f1]&&dist[t]+1<=15){
            dist[f1]=dist[t]+1;
            q.push(f1);
        }
        if(!~dist[f2]&&dist[t]+1<=15){
            dist[f2]=dist[t]+1;
            q.push(f2);
        }
    }
}
int main(){
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        int x;
        cin>>x;
        bfs(x);
        cout<<dist[0]<<" ";
    }

}