Codeforces Round #479 (Div. 3) D. Divide by three, multiply by two (DFS)

• 题意:给你一个长度为\(n\)的序列\(a\).对它重新排列,使得\(a_{i+1}=a_{i}/3\)或\(a_{i+1}=2*a_{i}\).输出重新排列后的序列.

• 题解:经典DFS,遍历这个序列,对每个数dfs,每次dfs使\(len+1\),当\(len=n\)时,说明这个序列已经满足条件了,输出并且标记一下,防止还有另外的情况导致多输出.

• 代码:

  #include <iostream>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<long,long> PLL;
  
  int n;
  bool st[N];
  int len;
  ll ans[N];
  ll a[N];
  bool ok;
  void dfs(int x,int len){
      if(len==n && !ok){
          ok=1;
          for(int i=1;i<=n;++i) printf("%lld ",ans[i]);
          return;
      }
      st[x]=true;
       for(int i=1;i<=n;++i){
           if(!st[i] && (a[i]*3==ans[len] || ans[len]*2==a[i])){
               ans[len+1]=a[i];
               dfs(i,len+1);
           }
       }
       st[x]=false;
  }
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);
      cin>>n;
       for(int i=1;i<=n;++i)  cin>>a[i];
  
       for(int i=1;i<=n;++i){
           ans[1]=a[i];
           memset(st,0,sizeof(st));
           dfs(i,1);
       }
  
      return 0;
  }