제출 #255043

#제출 시각아이디문제언어결과실행 시간메모리
255043vioalbertCards (LMIO19_korteles)C++14
100 / 100
147 ms20312 KiB
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 456976+5, MAX1 = 26*26+5, MAX2 = 26*26*26+5, MAX3 = 26*26*26*26+5;
int n, ans;
int val[N][4];
int cnt[4][MAX3][4];

inline int parse(char a, char b) {
	return (int)(a - 'A') * 26 + (int)(b - 'A');
}

inline int parse(char a, char b, char c) {
	return (int)(a - 'A') * 26 * 26 + (int)(b - 'A') * 26 
	+ (int)(c - 'A');
} 
inline int parse(char a, char b, char c, char d) {
	return (int)(a - 'A') * 26 * 26 * 26 + (int)(b - 'A') * 26 * 26 
	+ (int)(c - 'A') * 26 + (int)(d - 'A');
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	ans = 0;
	cin >> n;
	for(int i = 0; i < n; i++) {
		char c[4];
		for(int j = 0; j < 4; j++)
			cin >> c[j];
		swap(c[2], c[3]);
		val[i][0] = parse(c[0], c[1]);
		val[i][1] = parse(c[1], c[2]);
		val[i][2] = parse(c[3], c[2]);
		val[i][3] = parse(c[0], c[3]);

		//ans
		//type 0
		for(int j = 0; j < 4; j++)
			ans += cnt[0][val[i][j]][j^2];
		//type 1
		for(int j = 0; j < 4; j++) {
			char lst = 'a';
			bool ok = 1;
			for(int k = 0; k < 4 && ok; k++) {
				if(k == j || k == (j^2)) continue;
				if(lst == 'a') lst = c[k];
				else if(lst != c[k]) ok = 0;
			}
			if(ok) {
				ans -= cnt[1][parse(c[j^2], lst, c[j^2])][j^2];
			}
		}
		//type 2
		ans -= cnt[2][parse(c[3], c[2], c[0], c[1])][0];
		ans -= cnt[2][parse(c[0], c[3], c[1], c[2])][1];
		//type 3
		if(c[0] == c[2] && c[1] == c[3]) {
			ans += 3 * cnt[3][parse(c[1], c[0], c[3], c[2])][0];
		}

		//cnt
		//type 0
		for(int j = 0; j < 4; j++)
			cnt[0][val[i][j]][j]++;
		//type 1
		for(int j = 0; j < 4; j++) {
			char lst = 'a';
			bool ok = 1;
			for(int k = 0; k < 4 && ok; k++) {
				if(k == j || k == (j^2)) continue;
				if(lst == 'a') lst = c[k];
				else if(lst != c[k]) ok = 0;
			}
			if(ok) {
				cnt[1][parse(lst, c[j^2], lst)][j]++;
			}
		}
		//type 2
		cnt[2][parse(c[0], c[1], c[3], c[2])][0]++;
		cnt[2][parse(c[1], c[2], c[0], c[3])][1]++;
		//type 3
		if(c[0] == c[2] && c[1] == c[3]) {
			cnt[3][parse(c[0], c[1], c[2], c[3])][0]++;
		}
	}

	cout << ans << '\n';

	return 0;
}
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