AtCoder Regular Contest 124 C - LCM of GCDs (记忆化搜索)

• 题意:有两个容器\(x\)和\(y\),\(n\)对数\(a[i]\)和\(b[i]\),每次选一对数将\(a[i]\)或者\(b[i]\)放入容器\(x\)或\(y\)中,全部放完后将\(x\)和\(y\)中所有数求gcd,然后得到的两个数求lcm,问能得到的最大lcm是多少.

• 题解:这题的\(n\)给的很小,但是直接dfs肯定是不行的,因为每次都是求gcd,所以可考虑每层搜索的时候记忆化一下,如果搜到某一层的时候容器内的两个数已经出现过了,那么直接return即可.

• 代码:

  #include <bits/stdc++.h>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
  ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}
  
  int n;
  ll a[N],b[N];
  vector<map<PLL,ll>> dp(N);
  
  ll dfs(int u,ll xa,ll xb){
      if(dp[u-1][make_pair(xa,xb)]) return dp[u-1][make_pair(xa,xb)];
      if(u==n+1) return dp[u-1][make_pair(xa,xb)]=lcm(1ll*xa,1ll*xb);
      ll res1=dfs(u+1,gcd(xa,a[u]),gcd(xb,b[u]));
      ll res2=dfs(u+1,gcd(xa,b[u]),gcd(xb,a[u]));
      return dp[u-1][make_pair(xa,xb)]=max(res1,res2);
  }
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      cin>>n;
      rep(i,1,n) cin>>a[i]>>b[i];
  
      cout<<dfs(2,a[1],b[1])<<'\n';
  
      return 0;
  }