Codeforces Round 867 (Div. 3) G2. Magic Triples (Hard Version)

题意是多少对互不相同的下标i,j,k满足对应数等比

很自然的想到了枚举最大数，然后找到因子就做完了

1e9以内因子最多1e3，这个不难发现，知道怎么通过质因数分解得到因子个数就知道怎么算了。

毫无优化，全是暴力。居然过了

可以通过手写哈希去掉map的log，而且不难

我用了一个写的时候想到的哈希写法试了试，不过似乎常数很大？并没有优化多少

map写法

#include<bits/stdc++.h>
using namespace std;
const int P=998244353,N=4e4+10,inf=1e9+10;
using ll=long long;
bool st[N];
int pri[N],cnt;
void init()
{for(int i=2;i<N;++i){if(!st[i])pri[++cnt]=i;for(int j=1;pri[j]*i<N;++j){st[pri[j]*i]=1;if(i%pri[j]==0)break;}}
}
vector<array<int,2>>vec;
void func(int n)
{vec.clear();for(int i=1;i<=cnt;++i){if(n==1)return ;if(n%pri[i]==0){vec.push_back({pri[i],0});while(n%pri[i]==0)++vec.back()[1],n/=pri[i];}}if(n>1)vec.push_back({n,1});
}
vector<int>e;
void dfs(int now,int x)
{if(now==vec.size()){e.push_back(x);return ;}dfs(now+1,x);for(int i=1;i<=vec[now][1];++i){x=x*vec[now][0];dfs(now+1,x);}
}
ll C(int n)
{if(n<3)return 0;return 1ll*n*(n-1)*(n-2);
}
void solve()
{int n;scanf("%d",&n);vector<int>a(n);for(auto &v:a)scanf("%d",&v);map<int,int>mp;for(auto &v:a)mp[v]++;sort(a.begin(),a.end());a.erase(unique(a.begin(),a.end()),a.end());ll rs=0;for(auto &v:a){func(v);e.clear();dfs(0,1);//for(auto t:e)cout<<t<<" \n"[t==e.back()];for(auto &t:e){if(t==1)rs+=C(mp[v]);else{int x=v/t;if(x%t)continue;int y=x/t;rs+=1ll*mp[x]*mp[y]*mp[v];}}}printf("%lld\n",rs);
}
int main()
{init();int T=1;scanf("%d",&T);while(T--)solve();
}

取模哈希试了下

只从2500ms卡到了1900ms

#include<bits/stdc++.h>
using namespace std;
const int P=33331,N=33333+10,inf=1e9+10;
using ll=long long;
bool st[N];
int pri[N],cnt;
vector<array<int,2>>hsh[P+2];
void init()
{for(int i=2;i<N;++i){if(!st[i])pri[++cnt]=i;for(int j=1;pri[j]*i<N;++j){st[pri[j]*i]=1;if(i%pri[j]==0)break;}}
}
vector<array<int,2>>vec;
void func(int n)
{vec.clear();for(int i=1;i<=cnt;++i){if(n==1)return ;if(n%pri[i]==0){vec.push_back({pri[i],0});while(n%pri[i]==0)++vec.back()[1],n/=pri[i];}}if(n>1)vec.push_back({n,1});
}
vector<int>e;
void dfs(int now,int x)
{if(now==vec.size()){e.push_back(x);return ;}dfs(now+1,x);for(int i=1;i<=vec[now][1];++i){x=x*vec[now][0];dfs(now+1,x);}
}
ll C(int n)
{if(n<3)return 0;return 1ll*n*(n-1)*(n-2);
}
int find(int x)
{array<int,2>z={x,0};auto t=lower_bound(hsh[x%P].begin(),hsh[x%P].end(),z)-hsh[x%P].begin();if(t==hsh[x%P].size())return 0;if(hsh[x%P][t][0]==x)return hsh[x%P][t][1];return 0;
}
void solve()
{int n;scanf("%d",&n);vector<int>a(n);for(auto &v:a)scanf("%d",&v);map<int,int>mp;for(auto &v:a)mp[v]++;sort(a.begin(),a.end());a.erase(unique(a.begin(),a.end()),a.end());ll rs=0;for(auto v:a)hsh[v%P].push_back({v,mp[v]});set<int>S;for(auto v:a){if(S.find(v%P)!=S.end())continue;S.insert(v%P);sort(hsh[v%P].begin(),hsh[v%P].end());}for(auto &v:a){func(v);e.clear();dfs(0,1);//for(auto t:e)cout<<t<<" \n"[t==e.back()];for(auto &t:e){if(t==1)rs+=C(find(v));else{int x=v/t;if(x%t)continue;int y=x/t;rs+=1ll*find(x)*find(y)*find(v);}}}printf("%lld\n",rs);for(auto v:a)hsh[v%P].clear();
}
int main()
{init();//cout<<pri[cnt]<<endl;int T=1;scanf("%d",&T);while(T--)solve();
}

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