wa&&tle:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[200005],ans=0,idex;
map<ll,pair<ll,ll> >mp;
map<ll,ll>mmp;
int main()
{
int n;
bool flag1=0;
cin>>n;
for(int i=0; i<n; i++)
{
cin>>a[i];
mmp[abs(a[i])]=a[i];
}
for(int i=0; i<n; i++)
for(int j=i+1; j<n; j++)
{
if(mp.count(abs(a[i]-a[j])))
mp[abs(a[i]-a[j])].first= a[i];
mp[abs(a[i]-a[j])].second= a[j];
}
ll temp=1;
for(int k=0; k<30; k++)
{
if(mmp.count(temp))
idex=mmp[temp];
if(mp.count(temp))
{
ans=temp;
for(int i=0; i<n; i++)
{
if(abs(mp[temp].first-a[i])==temp&&a[i]!=mp[temp].second)
{
ans=a[i];
flag1=1;
break;
}
if(abs(mp[temp].second-a[i])==temp&&a[i]!=mp[temp].first)
{
ans=a[i];
flag1=1;
break;
}
}
}
if(flag1)
break;
temp = temp << 1;
}
if(flag1)
cout<<3<<endl<<mp[temp].first<<" "<<mp[temp].second<<" "<<ans<<endl;
else if(ans!=0)
cout<<2<<endl<<mp[ans].first<<" "<<mp[ans].second<<endl;
else
cout<<1<<endl<<idex<<endl;
return 0;
}
//THE most important is translate the view;
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
set<int >T;
int a[maxn];
int main()
{
int n,x1,x2,ans=0;
cin>>n;
for(int i=0; i<n; i++)
{
cin>>a[i];
T.insert(a[i]);
}
for(int i=0; i<n; i++)
for(int k=0; k<31; k++)
{
if(T.count(a[i]+(1<<k))&&T.count(a[i]-(1<<k)))
{
cout<<3<<endl<<a[i]<<" "<<a[i]+(1<<k)<<" "<<a[i]-(1<<k)<<endl;
return 0;
}
else if(T.count(a[i]+(1<<k))&&ans==0)
{
x1=a[i];
x2=a[i]+(1<<k);
ans=2;
}
}
if(ans==2)
cout<<ans<<endl<<x1<<" "<<x2<<endl;
else
cout<<1<<endl<<a[0];
return 0;
}
用一个数的和加上
去找存不存在,