ABC-402-D - Line Crossing（圆上之弦相交之几何）

思路讲解

赛时过的，记录一下。

首先，注意到，$(u+v)\mod N$ 值相等的线平行。全部组合中，减去平行的组合，就是答案。

	FOR(i,1,M){
		ll u,v;
		cin>>u>>v;
		// G.pb(u,v);
		Sum.pb((u+v)%N);
	}
	ll ans=M*(M-1),lans=0;
	sort(all(Sum));
	FOR(i,0,SZ(Sum)-1){
		ll num=upper_bound(all(Sum),Sum[i])-lower_bound(all(Sum),Sum[i])-1;
		ans-=num;
// #ifdef LOCAL
	// debug(Sum[i]);
    // debug(num);
// #endif
	}

AC代码

https://atcoder.jp/contests/abc402/submissions/66034983

源代码

// Problem: D - Line Crossing
// Contest: AtCoder - Tokio Marine & Nichido Fire Insurance Programming Contest 2025  (AtCoder Beginner Contest 402)
// URL: https://atcoder.jp/contests/abc402/tasks/abc402_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
    do { \
        std::cerr << #variable << ":"; \
        for (const auto& elem : variable) { \
            std::cerr << elem << " "; \
        } \
        std::cerr << "\n"; \
    } while (0)
#define uniVec(var) \
    do { \
        sort(var.begin(),var.end());\
        var.resize(unique(var.begin(),var.end())-var.begin());\
    } while (0)
#define debugN(var,N) \
	do{ \
		std::cerr<<#var<<":"; \
		FOR(i,1,N){ \
			std::cerr<<var[i]<<" "; \
		} \
		std::cerr<<"\n"; \
	}while(0)
#define debugMap(variable) \
    do { \
        std::cerr << #variable << ":\n"; \
        for (const auto& pair : variable) { \
            std::cerr << "  " << pair.first << " => " << pair.second << "\n"; \
        } \
        std::cerr << std::endl; \
    } while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr3;
typedef array<ll,2> arr2;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,Q,K,T,A[MAXN];
/*

*/
inline void solve(){
	cin>>N>>M;
	vector<arr2> G;
	vector<ll> Sum;
	FOR(i,1,M){
		ll u,v;
		cin>>u>>v;
		// G.pb(u,v);
		Sum.pb((u+v)%N);
	}
	ll ans=M*(M-1),lans=0;
	sort(all(Sum));
	FOR(i,0,SZ(Sum)-1){
		ll num=upper_bound(all(Sum),Sum[i])-lower_bound(all(Sum),Sum[i])-1;
		ans-=num;
#ifdef LOCAL
	debug(Sum[i]);
    debug(num);
#endif
	}
#ifdef LOCAL
    debug(ans);
#endif
	ans/=2;
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	// cin>>T;
	// while(T--){
		// solve();
	// }
	solve();
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc402/submissions/66034983
*/

心路历程（WA，TLE，MLE……）