結果

問題 No.3208 Parse AND OR Affection
ユーザー 遭難者
提出日時 2025-07-18 16:07:56
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 94 ms / 5,000 ms
コード長 4,193 bytes
コンパイル時間 5,451 ms
コンパイル使用メモリ 335,240 KB
実行使用メモリ 28,040 KB
最終ジャッジ日時 2025-07-18 16:08:05
合計ジャッジ時間 8,451 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 20
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;
using std::cerr;
using std::cin;
using std::cout;
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using mint = atcoder::modint998244353;
istream &operator>>(istream &is, mint &a) {
	long long t;
	is >> t;
	a = t;
	return is;
}
ostream &operator<<(ostream &os, mint a) { return os << a.val(); }
#endif
typedef long double ld;
#define long long long
#define uint unsigned int
#define ull unsigned long
#define overload3(a, b, c, name, ...) name
#define rep3(i, a, b) for (int i = (a); i < (b); i++)
#define rep2(i, n) rep3(i, 0, n)
#define rep1(n) rep2(__i, n)
#define rep(...) overload3(__VA_ARGS__, rep3, rep2, rep1)(__VA_ARGS__)
#define per3(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define per2(i, n) per3(i, 0, n)
#define per1(n) per2(__i, n)
#define per(...) overload3(__VA_ARGS__, per3, per2, per1)(__VA_ARGS__)
#define all(a) a.begin(), a.end()
#define UNIQUE(a)                                                              \
	sort(all(a));                                                              \
	a.erase(unique(all(a)), a.end())
#define sz(a) (int)a.size()
#define vec vector
#ifndef DEBUG
#define cerr                                                                   \
	if (0)                                                                     \
	cerr
// #undef assert
// #define assert(...) void(0)
#undef endl
#define endl '\n'
#endif
template <typename T> ostream &operator<<(ostream &os, vector<T> a) {
	const int n = a.size();
	rep(i, n) {
		os << a[i];
		if (i + 1 != n)
			os << " ";
	}
	return os;
}
template <typename T, size_t n>
ostream &operator<<(ostream &os, array<T, n> a) {
	rep(i, n) {
		os << a[i];
		if (i + 1 != n)
			os << " ";
	}
	return os;
}
template <typename T> istream &operator>>(istream &is, vector<T> &a) {
	for (T &i : a)
		is >> i;
	return is;
}
template <typename T, typename S> bool chmin(T &x, S y) {
	if ((T)y < x) {
		x = (T)y;
		return true;
	}
	return false;
}
template <typename T, typename S> bool chmax(T &x, S y) {
	if (x < (T)y) {
		x = (T)y;
		return true;
	}
	return false;
}
template <typename T> void operator++(vector<T> &a) {
	for (T &i : a)
		++i;
}
template <typename T> void operator--(vector<T> &a) {
	for (T &i : a)
		--i;
}
template <typename T> void operator++(vector<T> &a, int) {
	for (T &i : a)
		i++;
}
template <typename T> void operator--(vector<T> &a, int) {
	for (T &i : a)
		i--;
}
struct S {
	long c0, c1;
	long x[4] = {};
	int f, v;
	long ans;
};
ostream &operator<<(ostream &os, S a) {
	return os << "c : " << a.c0 << " " << a.c1 << '\n'
			  << "x : " << a.x[0] << " " << a.x[1] << " " << a.x[2] << " "
			  << a.x[3] << '\n'
			  << "f v : " << a.f << " " << a.v << '\n'
			  << "ans : " << a.ans;
}
S e() {
	S res = S();
	res.v = -1;
	return res;
}
int ap[4][4] = {{0, 0, 3, 3}, {0, 1, 2, 3}, {0, 2, 1, 3}, {0, 3, 0, 3}};
int fx[4][2] = {{0, 0}, {0, 1}, {1, 0}, {1, 1}};
S op(S l, S r) {
	if (r.v == -1)
		return l;
	if (l.v == -1)
		return r;
	long c0 = r.c0;
	long c1 = r.c1;
	(r.f == 0 || r.f == 1 ? c0 : c1) += l.c0;
	(r.f == 0 || r.f == 2 ? c0 : c1) += l.c1;
	long x[4] = {};
	rep(k, 4) x[k] = l.x[k];
	rep(k, 4) x[ap[l.f][k]] += r.x[k];
	int f = ap[l.f][r.f];
	int v = fx[r.f][l.v];
	long ans =
		l.ans + r.ans + l.c0 * (r.x[2] + r.x[3]) + l.c1 * (r.x[1] + r.x[3]);
	S res = {c0, c1, {x[0], x[1], x[2], x[3]}, f, v, ans};
	return res;
}

void solve() {
	int n, que;
	string s;
	cin >> n >> que >> s;
	s = "+" + s;
	n = (n + 1) / 2;
	vec<S> a(n);
	rep(i, n) {
		if (s[2 * i] == '+') {
			a[i].f = s[2 * i + 1] == 'T' ? 3 : 1;
		} else if (s[2 * i] == '*') {
			a[i].f = s[2 * i + 1] == 'T' ? 1 : 0;
		} else {
			a[i].f = s[2 * i + 1] == 'T' ? 2 : 1;
		}
		a[i].x[a[i].f]++;
		if (s[2 * i + 1] == 'T') {
			a[i].ans++;
			a[i].c1++;
			a[i].v = 1;
		} else
			a[i].c0++;
	}
	atcoder::segtree<S, op, e> seg(a);
	while (que--) {
		int l, r;
		cin >> l >> r;
		l--;
		l = (l + 1) / 2, r = (r + 1) / 2;
		S res = seg.prod(l, r);
		cout << res.ans << endl;
	}
}
int main() {
	// srand((unsigned)time(NULL));
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(20);
	int t = 1;
	// cin >> t;
	while (t--)
		solve();
}
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