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[7007];
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=0;k<kase;k++){
        int n;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
        }
        vector<set<long long>> v(7000);

        for(int i=n-1;i>=0;i--)
        {
            for(int j=i+1;j<n;j++)
                v[i].insert(a[j]);
        }

        long long ans=0;
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
            {

                if(a[i]*a[j]<=200000&&v[j].count(a[i]*a[j]))
                    ans++;
                else if(a[j]!=0&&(a[i]/float(a[j])==a[i]/a[j])&&v[j].count(a[i]/a[j]))
                    ans++;
                else if(a[i]!=0&&(a[j]/float(a[i])==a[j]/a[i])&&v[j].count(a[j]/a[i]))
                    ans++;
            }
        cout<<"Case #"<<k+1<<": "<<ans<<endl;
    }
    return 0;
}