제출 #1226398

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1226398		kaiboy	Bowling (BOI15_bow)	C++20	100 / 100	651 ms	2412 KiB

bow

#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

const  int   N = 10;
const  int   S = 300;
const  int   K = 13;
const char *AA = "0123456789x-/";

char aa[N * 2 + 2]; int ss[N];
long long dp[S + 1][K][K][2], dq[S + 1][K][K][2];
vector<tuple<int, int, int, int>> qu;

bool check(char a, char b) {
	if (a == 'x')
		return b == '-';
	if (!isdigit(a))
		return false;
	if (b == '/')
		return true;
	if (!isdigit(b))
		return false;
	return a - '0' + b - '0' < 10;
}

int geta(char a) {
	return a == 'x' ? 10 : a - '0';
}

int getb(char a, char b) {
	return b == 'x' ? 10 : b == '/' ? 10 - (a - '0') : b == '-' ? 0 : b - '0';
}

int calc(char a, char b) {
	return a == 'x' || b == '/' ? 10 : a - '0' + b - '0';
}

int count(char a, char b) {
	return a == 'x' ? 2 : b == '/' ? 1 : 0;
}

bool check(char a, char b, char c) {
	if (a == 'x' && b == 'x' && c == 'x')
		return true;
	if (a == 'x' && b == 'x')
		return isdigit(c);
	if (a == 'x' && c == '/')
		return isdigit(b);
	if (a == 'x')
		return isdigit(b) && isdigit(c) && b - '0' + c - '0' < 10;
	if (b == '/' && c == 'x')
		return isdigit(a);
	if (b == '/')
		return isdigit(a) && isdigit(c);
	return isdigit(a) && isdigit(b) && c == '-' && a - '0' + b - '0' < 10;
}

int main() {
	ios_base::sync_with_stdio(false), cin.tie(NULL);
	int t; cin >> t;
	while (t--) {
		int n; cin >> n >> aa;
		for (int i = 0; i < n; i++)
			cin >> ss[i];
		for (int s = 0; s <= S; s++)
			for (int ha = 0; ha < K; ha++)
				for (int hb = 0; hb < K; hb++)
					for (int f = 0; f <= 1; f++)
						dp[s][ha][hb][f] = 0;
		dp[0][0][0][0] = 1;
		for (int i = 0; i + 1 < n; i++) {
			qu.clear();
			for (int s = 0; s <= S; s++)
				for (int ha = 0; ha < K; ha++)
					for (int hb = 0; hb < K; hb++)
						for (int f = 0; f <= 1; f++)
							if (dp[s][ha][hb][f])
								qu.push_back({ s, ha, hb, f });
			for (int s = 0; s <= S; s++)
				for (int ha = 0; ha < K; ha++)
					for (int hb = 0; hb < K; hb++)
						for (int f = 0; f <= 1; f++)
							dq[s][ha][hb][f] = 0;
			char c_ = aa[i * 2], d_ = aa[i * 2 + 1];
			for (int hc = 0; hc < K; hc++) {
				char c = AA[hc];
				if (c_ == '?' || c == c_)
					for (int hd = 0; hd < K; hd++) {
						char d = AA[hd];
						if ((d_ == '?' || d == d_) && check(c, d))
							for (auto &z : qu) {
								int s = get<0>(z), ha = get<1>(z), hb = get<2>(z), f = get<3>(z);
								long long x = dp[s][ha][hb][f];
								if (x) {
									char a = AA[ha], b = AA[hb];
									int s_ = s + (f ? geta(c) : 0);
									if (i < 2 || ss[i - 2] == -1 || s_ == ss[i - 2]) {
										int t = s_ + calc(a, b), g = 0, k = count(a, b);
										if (k)
											t += geta(c);
										if (k >= 2)
											if (c == 'x')
												g = 1;
											else
												t += getb(c, d);
										dq[t][hc][hd][g] += x;
									}
								}
							}
					}
			}
			for (int s = 0; s <= S; s++)
				for (int ha = 0; ha < K; ha++)
					for (int hb = 0; hb < K; hb++)
						for (int f = 0; f <= 1; f++)
							dp[s][ha][hb][f] = dq[s][ha][hb][f];
		}
		qu.clear();
		for (int s = 0; s <= S; s++)
			for (int ha = 0; ha < K; ha++)
				for (int hb = 0; hb < K; hb++)
					for (int f = 0; f <= 1; f++)
						if (dp[s][ha][hb][f])
							qu.push_back({ s, ha, hb, f });
		long long ans = 0;
		char c_ = aa[n * 2 - 2], d_ = aa[n * 2 - 1], e_ = aa[n * 2];
		for (int hc = 0; hc < K; hc++) {
			char c = AA[hc];
			if (c_ == '?' || c == c_)
				for (int hd = 0; hd < K; hd++) {
					char d = AA[hd];
					if (d_ == '?' || d == d_)
						for (int he = 0; he < K; he++) {
							char e = AA[he];
							if ((e_ == '?' || e == e_) && check(c, d, e))
								for (auto &z : qu) {
									int s = get<0>(z), ha = get<1>(z), hb = get<2>(z), f = get<3>(z);
									long long x = dp[s][ha][hb][f];
									if (x) {
										char a = AA[ha], b = AA[hb];
										int s_ = s + (f ? geta(c) : 0);
										if (n < 3 || ss[n - 3] == -1 || ss[n - 3] == s_) {
											int t = s_ + calc(a, b), k = count(a, b);
											if (k)
												t += geta(c);
											if (k >= 2)
												t += getb(c, d);
											if (ss[n - 2] == -1 || ss[n - 2] == t) {
												t += geta(c) + getb(c, d) + getb(d, e);
												if (ss[n - 1] == -1 || ss[n - 1] == t)
													ans += x;
											}
										}
									}
								}
						}
				}
		}
		cout << ans << '\n';
	}
	return 0;
}

• Subtask 1: 16 / 16
• Subtask 2: 17 / 17
• Subtask 3: 26 / 26
• Subtask 4: 23 / 23
• Subtask 5: 18 / 18