結果

問題 No.2308 [Cherry 5th Tune B] もしかして、真?
ユーザー k1suxu
提出日時 2023-05-19 23:30:47
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 242 ms / 2,000 ms
コード長 5,760 bytes
コンパイル時間 3,347 ms
コンパイル使用メモリ 257,276 KB
実行使用メモリ 25,480 KB
最終ジャッジ日時 2024-12-20 02:05:46
合計ジャッジ時間 12,475 ms
ジャッジサーバーID
(参考情報)
judge2 / judge3
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
other AC * 39
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 #
プレゼンテーションモードにする

// #pragma GCC target("avx")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
#define FOR(n) for(int i = 0; i < (int)n; i++)
#define repi(i,a,b) for(int i = (int)a; i < (int)b; i++)
#define all(x) x.begin(),x.end()
//#define mp make_pair
#define vi vector<int>
#define vvi vector<vi>
#define vvvi vector<vvi>
#define vvvvi vector<vvvi>
#define pii pair<int,int>
#define vpii vector<pair<int,int>>
template<typename T>
void chmax(T &a, const T &b) {a = (a > b? a : b);}
template<typename T>
void chmin(T &a, const T &b) {a = (a < b? a : b);}
using ll = long long;
using ld = long double;
using ull = unsigned long long;
const ll INF = numeric_limits<long long>::max() / 2;
const ld pi = 3.1415926535897932384626433832795028;
const ll mod = 998244353;
int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
#define int long long
namespace internal {
// @param n `0 <= n`
// @return minimum non-negative `x` s.t. `n <= 2**x`
int ceil_pow2(int n) {
int x = 0;
while ((1U << x) < (unsigned int)(n)) x++;
return x;
}
// @param n `1 <= n`
// @return minimum non-negative `x` s.t. `(n & (1 << x)) != 0`
constexpr int bsf_constexpr(unsigned int n) {
int x = 0;
while (!(n & (1 << x))) x++;
return x;
}
// @param n `1 <= n`
// @return minimum non-negative `x` s.t. `(n & (1 << x)) != 0`
int bsf(unsigned int n) {
#ifdef _MSC_VER
unsigned long index;
_BitScanForward(&index, n);
return index;
#else
return __builtin_ctz(n);
#endif
}
} // namespace internal
template <class S, S (*op)(S, S), S (*e)()> struct segtree {
public:
segtree() : segtree(0) {}
explicit segtree(int n) : segtree(std::vector<S>(n, e())) {}
explicit segtree(const std::vector<S>& v) : _n((int)v.size()) {
log = internal::ceil_pow2(_n);
size = 1 << log;
d = std::vector<S>(2 * size, e());
for (int i = 0; i < _n; i++) d[size + i] = v[i];
for (int i = size - 1; i >= 1; i--) {
update(i);
}
}
void set(int p, S x) {
assert(0 <= p && p < _n);
p += size;
d[p] = x;
for (int i = 1; i <= log; i++) update(p >> i);
}
S get(int p) const {
assert(0 <= p && p < _n);
return d[p + size];
}
S prod(int l, int r) const {
assert(0 <= l && l <= r && r <= _n);
S sml = e(), smr = e();
l += size;
r += size;
while (l < r) {
if (l & 1) sml = op(sml, d[l++]);
if (r & 1) smr = op(d[--r], smr);
l >>= 1;
r >>= 1;
}
return op(sml, smr);
}
S all_prod() const { return d[1]; }
template <bool (*f)(S)> int max_right(int l) const {
return max_right(l, [](S x) { return f(x); });
}
template <class F> int max_right(int l, F f) const {
assert(0 <= l && l <= _n);
assert(f(e()));
if (l == _n) return _n;
l += size;
S sm = e();
do {
while (l % 2 == 0) l >>= 1;
if (!f(op(sm, d[l]))) {
while (l < size) {
l = (2 * l);
if (f(op(sm, d[l]))) {
sm = op(sm, d[l]);
l++;
}
}
return l - size;
}
sm = op(sm, d[l]);
l++;
} while ((l & -l) != l);
return _n;
}
template <bool (*f)(S)> int min_left(int r) const {
return min_left(r, [](S x) { return f(x); });
}
template <class F> int min_left(int r, F f) const {
assert(0 <= r && r <= _n);
assert(f(e()));
if (r == 0) return 0;
r += size;
S sm = e();
do {
r--;
while (r > 1 && (r % 2)) r >>= 1;
if (!f(op(d[r], sm))) {
while (r < size) {
r = (2 * r + 1);
if (f(op(d[r], sm))) {
sm = op(d[r], sm);
r--;
}
}
return r + 1 - size;
}
sm = op(d[r], sm);
} while ((r & -r) != r);
return 0;
}
private:
int _n, size, log;
std::vector<S> d;
void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
};
using S = int;
S op(S x, S y) {
return x+y;
}
S e() {
return 0;
}
void solve() {
int n;
cin >> n;
vector<string> x(n);
FOR(n) cin >> x[i];
vector<string> y(n-1);
FOR(n-1) cin >> y[i];
segtree<S, op, e> xid(vi(n, 1)), yid(vi(n-1, 1));
int s;
auto f = [&](int x) -> bool {
return x < s;
};
FOR(n-1) {
cin >> s;
int xpos = xid.max_right<decltype(f)>(0, f);
int ypos = yid.max_right<decltype(f)>(0, f);
string fr = x[xpos];
string op = y[ypos];
xid.set(xpos, 0);
yid.set(ypos, 0);
int x_back = xid.max_right<decltype(f)>(0, f);
string bc = x[x_back];
xid.set(x_back, 0);
int P = (fr == "True");
int Q = (bc == "True");
int ans;
if(op == "and") ans = P&Q;
else if(op == "or") ans = P|Q;
else if(op == "xor") ans = P^Q;
else ans = (P?Q:1);
xid.set(xpos, 1);
x[xpos] = (ans? "True" : "False");
}
cout << x[0] << endl;
//erase x
//get ith element
}
signed main() {
cin.tie(nullptr);
ios::sync_with_stdio(false);
int t;
cin >> t;
while(t--) solve();
return 0;
}
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