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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#107999#2838. 2D Geometryl1924365846TL 0ms0kbC++172.9kb2023-05-23 13:31:222023-05-23 13:31:26

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-05-23 13:31:26]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:0kb
  • [2023-05-23 13:31:22]
  • 提交

answer

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#include <unordered_map>
#include <unordered_set>
#define ll long long
#define int ll
#define ms(a,b) memset((a),(b),sizeof(a))
#define mc(a,b) memcpy((a),(b),sizeof(a))
#define PI acos(-1)
#define ssin(x) sin(x*PI/180.00)
#define ccos(x) cos(x*PI/180.00)
#define ttan(X) tan(x*PI/180.00)
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, n, a) for (int i = n - 1; i >= a; i--)
#define IOS ios::sync_with_stdio(false)
#define si(x) scanf("%d",&x)
#define sii(x, y) scanf("%d %d",&x, &y)
#define siii(x, y, z) scanf("%d %d %d",&x, &y, &z)
#define siiii(x, y, z, w) scanf("%d %d %d %d",&x, &y, &z, &w)
#define sl(x) scanf("%lld",&x)
#define sll(x, y) scanf("%lld %lld",&x, &y)
#define slll(x, y, z) scanf("%lld %lld %lld",&x, &y, &z)
#define sllll(x, y, z, w) scanf("%lld %lld %lld %lld",&x, &y, &z, &w)
#define pr(x) printf("%d",x)
#define pl(x) printf("%lld",x)
#define sd(x) scanf("%lf",&x)
#define sdd(x, y) scanf("%lf %lf",&x, &y)
#define sddd(x, y, z) scanf("%lf %lf %lf",&x, &y, &z)
#define sdddd(x, y, z, w) scanf("%lf %lf %lf %lf",&x, &y, &z, &w)
#define fi first
#define se second
#define pb push_back
#define mk make_mair

using namespace std;

typedef unsigned long long ull;

typedef pair<int, int> PII;

const int mod = 1e9 + 7;

const int P = 131;

const int N = 2e5 + 10, M = 3e6 + 10;

const int inf = 0x3f3f3f3f;

const ll lnf = 9e18;

const double eps = 1e-6;

int n;
struct F {
	int x, y;
} f[N];
map<int, int> ma;
map<int, int> mb;
map<PII, int> mp;

int check (int a, int b) {
	int x = f[a].x, y = f[a].y, xx = f[b].x, yy = f[b].y;
	int cnt = 2;

	for (int i = 0; i < n; i ++) {
		if (i != a && i != b) {
			int xxx = f[i].x, yyy = f[i].y;

			if ( (xxx - xx) * (y - yy) == (yyy - yy) * (x - xx) ) {
				cnt++;
			}
		}
	}

	return cnt;
}

void solve() {
	srand ( (unsigned) time (NULL) );

	while (~scanf ("%lld", &n) ) {
		ma.clear(), mb.clear();
		mp.clear();
		int kk = 0;

		for (int i = 0; i < n; i ++) {
			int x, y;
			sll (x, y);
			f[i] = {x, y};
//			ma[x]++;
//			mb[y]++;
//			kk = max (kk, max (ma[x], mb[y]) );
		}

		for (int i = 0; i < 500; i ++) {
			int x = rand() % n, y = rand() % n;

			while (x == y || mp[ {x, y}] == 1)
				y = rand() % n;
			mp[ {x, y}] = 1;

			kk = max (kk, check (x, y) );
		}

		if (kk <= (n - kk) * 2) {
			printf ("%lld\n", n % 3);
		}
		else {
			printf ("%lld\n", kk - 2 * (n - kk) );
		}
	}
}

signed main() {
	//	IOS;
	//	cin.tie (0);
	//	cout.tie (0);
	int T (1);
	//	sl (T);

	while ( T-- ) {
		solve();
	}

	return 0;
}
/*
4
0 0
0 1
0 2
1 1
3
0 0
0 1
1 0
6
0 0
0 1
0 2
0 3
1 1
1 2
*/

详细

Test #1:

score: 0
Time Limit Exceeded

input:

3
0 0
0 1
0 2
3
0 0
0 1
1 0
6
0 0
0 1
0 2
0 3
1 1
1 2

output:


result: