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]<<" ";
}
}