CF1028F. Make Symmetrical

题目描述

Solution

结论1：两个点$(x_1,y_1),(x_2,y_2)$关于$(0,0),(x_3,y_3)$对称的必要条件为$(x_1,y_1)$和$(x_2,y_2)$在同一个以$(0,0)$为圆心的圆上，即$x_1^2+y_1^2=x_2^2+y_2^2$。

结论2：$x^2+y^2=N(n\le 2\ast 112904^2)$的解数$M$小于等于$144$。

因此我们每次添加或删除点时，只会影响到在同一个圆上的点，而圆上最多有$144$个整点，暴力枚举这些圆上的点分别更新对称轴的值即可。

时间复杂度$O(Mq)$。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
int num=0,cnt=0;
map<ll,int> Map;
map<PR,int> Ans;
set<PR> Set[MAXN];
int gcd(int x,int y) { return y==0?x:gcd(y,x%y); }
void add(int x,int y,int c)
{int g=gcd(x,y);x/=g,y/=g;Ans[MP(x,y)]+=c;
}
int query(int x,int y)
{int g=gcd(x,y);x/=g,y/=g;return Ans[MP(x,y)];
}
int main()
{int Case=read();while (Case--){ll opt=read(),x=read(),y=read();if (opt==1){ll z=x*x+y*y;if (!Map.count(z)) Map[z]=++cnt;z=Map[z];	for (auto V:Set[z]) add(V.fi+x,V.se+y,2);Set[z].insert(MP(x,y)),add(x,y,1),num++;}else if (opt==2){ll z=x*x+y*y;if (!Map.count(z)) Map[z]=++cnt;z=Map[z];Set[z].erase(Set[z].find(MP(x,y))),add(x,y,-1),num--;for (auto V:Set[z]) add(V.fi+x,V.se+y,-2);}else printf("%d\n",num-query(x,y));}return 0;
}