kickstart-G

感觉自己很蠢，large数据只能交一次，忘记这回事了

A题 O(n^2)解法，用vector<set> 缓存j后面的数据，减少一重循环

 #include <string>
 #include <vector>
 #include<iostream>
 #include<cstdio>
 #include<queue>
 #include<cmath>
 #include <set>
 #include<algorithm>
 #include<cstring>
 
 using namespace std;
 int a[];
 int main(){
     //freopen("/Users/zjg/CLionProjects/ac/A-large-attempt0.in","r",stdin);
     //freopen("/Users/zjg/CLionProjects/ac/A-large-attempt0.out","w",stdout);
     int kase;
     cin>>kase;
     for(int k=;k<kase;k++){
         int n;
         cin>>n;
         for(int i=;i<n;i++)
         {
             cin>>a[i];
         }
         vector<set<long long>> v();
 
         for(int i=n-;i>=;i--)
         {
             for(int j=i+;j<n;j++)
                 v[i].insert(a[j]);
         }
 
         long long ans=;
         for(int i=;i<n;i++)
             for(int j=i+;j<n;j++)
             {
 
                 if(a[i]*a[j]<=&&v[j].count(a[i]*a[j]))
                     ans++;
                 else if(a[j]!=&&(a[i]/float(a[j])==a[i]/a[j])&&v[j].count(a[i]/a[j]))
                     ans++;
                 else if(a[i]!=&&(a[j]/float(a[i])==a[j]/a[i])&&v[j].count(a[j]/a[i]))
                     ans++;
             }
         cout<<"Case #"<<k+<<": "<<ans<<endl;
     }
     return ;
 }