이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
#include <chrono>
#include <random>
using namespace std;
#define F first
#define S second
#define pb push_back
#define vll vector<ll>
#define pll pair<ll, ll>
typedef long long ll;
struct P{
ll x, y;
bool operator< (const P &tar) const{
if(x!=tar.x){
return x<tar.x;
}
return y<tar.y;
}
};
void solve(){
ll n;
cin>>n;
vector<P> p(n);
for(auto &it:p){
cin>>it.x>>it.y;
}
if(n==1){
cout<<"-1\n";
return;
}
sort(p.begin(), p.end());
set<ll> st;
for(auto &it:p){
st.insert(it.x);
}
if(st.size()<=1){
cout<<"-1\n";
return;
}
ll gcd1=(*prev(st.end()))-(*st.begin());
ll pre=(*st.begin());
for(auto &it:st){
if(it==*st.begin()) continue;
gcd1=__gcd(gcd1, it-pre);
pre=it;
}
st.clear();
for(auto &it:p){
st.insert(it.y);
}
if(st.size()<=1){
cout<<"-1\n";
return;
}
ll gcd2=(*prev(st.end()))-(*st.begin());
pre=(*st.begin());
for(auto &it:st){
if(it==*st.begin()) continue;
gcd2=__gcd(gcd2, it-pre);
pre=it;
}
bool type1=1;
pll comp={(p[1].x-p[0].x)/gcd1, (p[1].y-p[0].y)/gcd2};
for(ll i=2;i<n;i++){
ll f=(p[i].x-p[0].x)/gcd1;
ll s=(p[i].y-p[0].y)/gcd2;
if(comp.F!=f || comp.S!=s){
type1=0;
break;
}
}
if(type1){
cout<<"-1\n";
return;
}
bool type2=1;
for(ll i=1;i<n;i++){
ll f=(p[i].x-p[0].x)/gcd1;
ll s=(p[i].y-p[0].y)/gcd2;
if((abs(f+s))%2!=0){
type2=0;
break;
}
}
if(type2){
cout<<gcd1*gcd2*2<<'\n';
return;
}
cout<<gcd1*gcd2<<'\n';
return;
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
solve();
return 0;
}
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