結果
問題 | No.2990 Interval XOR |
ユーザー |
|
提出日時 | 2024-12-16 20:23:43 |
言語 | C++17(gcc12) (gcc 12.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 387 ms / 2,000 ms |
コード長 | 14,431 bytes |
コンパイル時間 | 5,170 ms |
コンパイル使用メモリ | 277,272 KB |
実行使用メモリ | 14,576 KB |
最終ジャッジ日時 | 2024-12-16 20:23:58 |
合計ジャッジ時間 | 13,360 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 37 |
ソースコード
#ifndef HIDDEN_IN_VS // 折りたたみ用// 警告の抑制#define _CRT_SECURE_NO_WARNINGS// ライブラリの読み込み#include <bits/stdc++.h>using namespace std;// 型名の短縮using ll = long long; using ull = unsigned long long; // -2^63 ~ 2^63 = 9e18(int は -2^31 ~ 2^31 = 2e9)using pii = pair<int, int>; using pll = pair<ll, ll>; using pil = pair<int, ll>; using pli = pair<ll, int>;using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>; using vvvvi = vector<vvvi>;using vl = vector<ll>; using vvl = vector<vl>; using vvvl = vector<vvl>; using vvvvl = vector<vvvl>;using vb = vector<bool>; using vvb = vector<vb>; using vvvb = vector<vvb>;using vc = vector<char>; using vvc = vector<vc>; using vvvc = vector<vvc>;using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;template <class T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;using Graph = vvi;// 定数の定義const double PI = acos(-1);int DX[4] = { 1, 0, -1, 0 }; // 4 近傍(下,右,上,左)int DY[4] = { 0, 1, 0, -1 };int INF = 1001001001; ll INFL = 4004004003094073385LL; // (int)INFL = INF, (int)(-INFL) = -INF;// 入出力高速化struct fast_io { fast_io() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(18); } } fastIOtmp;// 汎用マクロの定義#define all(a) (a).begin(), (a).end()#define sz(x) ((int)(x).size())#define lbpos(a, x) (int)distance((a).begin(), std::lower_bound(all(a), (x)))#define ubpos(a, x) (int)distance((a).begin(), std::upper_bound(all(a), (x)))#define Yes(b) {cout << ((b) ? "Yes\n" : "No\n");}#define rep(i, n) for(int i = 0, i##_len = int(n); i < i##_len; ++i) // 0 から n-1 まで昇順#define repi(i, s, t) for(int i = int(s), i##_end = int(t); i <= i##_end; ++i) // s から t まで昇順#define repir(i, s, t) for(int i = int(s), i##_end = int(t); i >= i##_end; --i) // s から t まで降順#define repe(v, a) for(const auto& v : (a)) // a の全要素(変更不可能)#define repea(v, a) for(auto& v : (a)) // a の全要素(変更可能)#define repb(set, d) for(int set = 0, set##_ub = 1 << int(d); set < set##_ub; ++set) // d ビット全探索(昇順)#define repis(i, set) for(int i = lsb(set), bset##i = set; i < 32; bset##i -= 1 << i, i = lsb(bset##i)) // set の全要素(昇順)#define repp(a) sort(all(a)); for(bool a##_perm = true; a##_perm; a##_perm = next_permutation(all(a))) // a の順列全て(昇順)#define uniq(a) {sort(all(a)); (a).erase(unique(all(a)), (a).end());} // 重複除去#define EXIT(a) {cout << (a) << endl; exit(0);} // 強制終了#define inQ(x, y, u, l, d, r) ((u) <= (x) && (l) <= (y) && (x) < (d) && (y) < (r)) // 半開矩形内判定// 汎用関数の定義template <class T> inline ll powi(T n, int k) { ll v = 1; rep(i, k) v *= n; return v; }template <class T> inline bool chmax(T& M, const T& x) { if (M < x) { M = x; return true; } return false; } // 最大値を更新(更新されたら trueを返す)template <class T> inline bool chmin(T& m, const T& x) { if (m > x) { m = x; return true; } return false; } // 最小値を更新(更新されたら trueを返す)template <class T> inline T getb(T set, int i) { return (set >> i) & T(1); }template <class T> inline T smod(T n, T m) { n %= m; if (n < 0) n += m; return n; } // 非負mod// 演算子オーバーロードtemplate <class T, class U> inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }template <class T> inline istream& operator>>(istream& is, vector<T>& v) { repea(x, v) is >> x; return is; }template <class T> inline vector<T>& operator--(vector<T>& v) { repea(x, v) --x; return v; }template <class T> inline vector<T>& operator++(vector<T>& v) { repea(x, v) ++x; return v; }#endif // 折りたたみ用#if __has_include(<atcoder/all>)#include <atcoder/all>using namespace atcoder;#ifdef _MSC_VER#include "localACL.hpp"#endifusing mint = modint998244353;//using mint = static_modint<1000000007>;//using mint = modint; // mint::set_mod(m);string mint_to_frac(mint x, int v_max = 31595) {repi(dnm, 1, v_max) {int num = (x * dnm).val();if (num == 0) {return "0";}if (num <= v_max) {if (dnm == 1) return to_string(num);return to_string(num) + "/" + to_string(dnm);}if (mint::mod() - num <= v_max) {if (dnm == 1) return "-" + to_string(mint::mod() - num);return "-" + to_string(mint::mod() - num) + "/" + to_string(dnm);}}return to_string(x.val());}namespace atcoder {inline istream& operator>>(istream& is, mint& x) { ll x_; is >> x_; x = x_; return is; }#ifdef _MSC_VERinline ostream& operator<<(ostream& os, const mint& x) { os << mint_to_frac(x); return os; }#elseinline ostream& operator<<(ostream& os, const mint& x) { os << x.val(); return os; }#endif}using vm = vector<mint>; using vvm = vector<vm>; using vvvm = vector<vvm>; using vvvvm = vector<vvvm>; using pim = pair<int, mint>;#endif#ifdef _MSC_VER // 手元環境(Visual Studio)#include "local.hpp"#else // 提出用(gcc)inline int popcount(int n) { return __builtin_popcount(n); }inline int popcount(ll n) { return __builtin_popcountll(n); }inline int lsb(int n) { return n != 0 ? __builtin_ctz(n) : 32; }inline int lsb(ll n) { return n != 0 ? __builtin_ctzll(n) : 64; }inline int msb(int n) { return n != 0 ? (31 - __builtin_clz(n)) : -1; }inline int msb(ll n) { return n != 0 ? (63 - __builtin_clzll(n)) : -1; }#define dump(...)#define dumpel(...)#define dump_list(v)#define dump_mat(v)#define input_from_file(f)#define output_to_file(f)#define Assert(b) { if (!(b)) { vc MLE(1<<30); EXIT(MLE.back()); } } // RE の代わりに MLE を出す#endif//【アダマール変換】: O(2^n n)/** a[0..2^n) を* A[set] = Σset2 (-1)^popcount(set ∩ set2) a[set2]* なる A[0..2^n) に上書きする.*/template <class T>void hadamard(vector<T>& a) {// verify : https://judge.yosupo.jp/problem/bitwise_xor_convolution// 具体例:// A[0] = a[0] + a[1] + a[2] + a[3] + a[4] + a[5] + a[6] + a[7] + ...// A[1] = a[0] - a[1] + a[2] - a[3] + a[4] - a[5] + a[6] - a[7] + ...// A[2] = a[0] + a[1] - a[2] - a[3] + a[4] + a[5] - a[6] - a[7] + ...// A[3] = a[0] - a[1] - a[2] + a[3] + a[4] - a[5] - a[6] + a[7] + ...// A[4] = a[0] + a[1] + a[2] + a[3] - a[4] - a[5] - a[6] - a[7] + ...// A[5] = a[0] - a[1] + a[2] - a[3] - a[4] + a[5] - a[6] + a[7] + ...// A[6] = a[0] + a[1] - a[2] - a[3] - a[4] - a[5] + a[6] + a[7] + ...// A[7] = a[0] - a[1] - a[2] + a[3] - a[4] + a[5] + a[6] - a[7] + ...int n = msb(sz(a));rep(i, n) repb(set, n) {if (!(set & (1 << i))) {T x = a[set];T y = a[set | (1 << i)];a[set] = x + y;a[set + (1 << i)] = x - y;}}}//【逆アダマール変換(mint)】: O(2^n n + log(mod))/** A[0..2^n) を* A[set] = Σset2 (-1)^popcount(set ∩ set2) a[set2]* なる a[0..2^n) に上書きする.** 制約:mint の法は 2 の倍数でない** 利用:【アダマール変換】*/void hadamard_inv(vm& A) {// verify : https://atcoder.jp/contests/abc265/tasks/abc265_hhadamard(A);// まとめて商をとらないと log(mod) 倍遅くなる.mint inv = mint(sz(A)).inv();rep(i, sz(A)) A[i] *= inv;}vm TLE(int n, int m, vi l, vi r) {vm f(1LL << n, 1);rep(j, m) {dump("--- j:", j, "---");dump("w:", r[j] - l[j]);vm g(1LL << n);repi(i, l[j], r[j] - 1) g[i]++;dump(g);hadamard(g);dump(g);rep(i, 1 << n) f[i] *= g[i];}dump("f:"); dump(f);hadamard_inv(f);return f;}void zikken() {int n = 4;int N = 1 << n;auto dump2 = [&](vm a) {rep(b, n) {rep(i, N) if (lsb(i) == b) cout << right << setw(2) << a[i] << " ";cout << "|\n"[b == n - 1];}};repi(k, 0, N) {// dump("--- k:", k, "---");vm g(N);rep(i, k) g[i] = 1;// dump2(g);hadamard(g);dump2(g);}exit(0);}/*r\i 1 3 5 7 9 11 13 15 | 2 6 10 14 | 4 12 | 80 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 01 1 1 1 1 1 1 1 1 | 1 1 1 1 | 1 1 | 12 0 0 0 0 0 0 0 0 | 2 2 2 2 | 2 2 | 23 1 -1 1 -1 1 -1 1 -1 | 1 1 1 1 | 3 3 | 34 0 0 0 0 0 0 0 0 | 0 0 0 0 | 4 4 | 45 1 1 -1 -1 1 1 -1 -1 | 1 -1 1 -1 | 3 3 | 56 0 0 0 0 0 0 0 0 | 2 -2 2 -2 | 2 2 | 67 1 -1 -1 1 1 -1 -1 1 | 1 -1 1 -1 | 1 1 | 78 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 89 1 1 1 1 -1 -1 -1 -1 | 1 1 -1 -1 | 1 -1 | 710 0 0 0 0 0 0 0 0 | 2 2 -2 -2 | 2 -2 | 611 1 -1 1 -1 -1 1 -1 1 | 1 1 -1 -1 | 3 -3 | 512 0 0 0 0 0 0 0 0 | 0 0 0 0 | 4 -4 | 413 1 1 -1 -1 -1 -1 1 1 | 1 -1 -1 1 | 3 -3 | 314 0 0 0 0 0 0 0 0 | 2 -2 -2 2 | 2 -2 | 215 1 -1 -1 1 -1 1 1 -1 | 1 -1 -1 1 | 1 -1 | 116 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 0A_r[i] = Σj∈[0..r) (-1)^popcount(i ∩ j)*/void zikken2() {int n = 4;int N = 1 << n;cout << "r\\i";rep(s, n) {rep(i, N) {if (lsb(i) == s) {cout << right << setw(2) << i << " ";}}cout << "|\n"[s == n - 1];}repi(r, 0, N) {cout << right << setw(2) << r << " ";rep(s, n) {rep(i, N) {if (lsb(i) == s) {int t = ((i >> s) - 1) / 2;int q = r / (1 << (s + 1));int m = r % (1 << (s + 1));int val = (popcount(t & q) & 1 ? -1 : 1) * min(m, (1 << (s + 1)) - m);cout << right << setw(2) << val << " ";}}cout << "|\n"[s == n - 1];}}exit(0);}/*r\i 1 3 5 7 9 11 13 15 | 2 6 10 14 | 4 12 | 80 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 01 1 1 1 1 1 1 1 1 | 1 1 1 1 | 1 1 | 12 0 0 0 0 0 0 0 0 | 2 2 2 2 | 2 2 | 23 1 -1 1 -1 1 -1 1 -1 | 1 1 1 1 | 3 3 | 34 0 0 0 0 0 0 0 0 | 0 0 0 0 | 4 4 | 45 1 1 -1 -1 1 1 -1 -1 | 1 -1 1 -1 | 3 3 | 56 0 0 0 0 0 0 0 0 | 2 -2 2 -2 | 2 2 | 67 1 -1 -1 1 1 -1 -1 1 | 1 -1 1 -1 | 1 1 | 78 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 89 1 1 1 1 -1 -1 -1 -1 | 1 1 -1 -1 | 1 -1 | 710 0 0 0 0 0 0 0 0 | 2 2 -2 -2 | 2 -2 | 611 1 -1 1 -1 -1 1 -1 1 | 1 1 -1 -1 | 3 -3 | 512 0 0 0 0 0 0 0 0 | 0 0 0 0 | 4 -4 | 413 1 1 -1 -1 -1 -1 1 1 | 1 -1 -1 1 | 3 -3 | 314 0 0 0 0 0 0 0 0 | 2 -2 -2 2 | 2 -2 | 215 1 -1 -1 1 -1 1 1 -1 | 1 -1 -1 1 | 1 -1 | 116 0 0 0 0 0 0 0 0 | 0 0 0 0 | 0 0 | 0A_r[i] = Σj∈[0..r) (-1)^popcount(i ∩ j)i = (2t+1) 2^s, r = q 2^(s+1) + m のときA_r[i] = (-1)^popcount(t ∩ q) min(m, 2^(s+1)-m)*/vm TLE2(int n, int m, vi l, vi r) {int N = 1 << n;vm res(N, 1); vm sub(N, 1);rep(j, m) res[0] *= r[j] - l[j];rep(s, n) {int Ns = N >> (s + 1);vi ql(m), A(m), qr(m), B(m);rep(j, m) {qr[j] = r[j] / (1 << (s + 1));ql[j] = l[j] / (1 << (s + 1));int mr = r[j] % (1 << (s + 1));A[j] = min(mr, (1 << (s + 1)) - mr);int ml = l[j] % (1 << (s + 1));B[j] = min(ml, (1 << (s + 1)) - ml);}rep(t, Ns) {int i = (2 * t + 1) << s;rep(j, m) {int sgnA = popcount(t & qr[j]) & 1 ? -1 : 1;int sgnB = popcount(t & ql[j]) & 1 ? -1 : 1;res[i] *= sgnA * A[j] - sgnB * B[j];sub[i] *= sgnB;}}}dump(res); dump(sub);hadamard_inv(res);return res;}/*f[i] = Π(A_r[i] - A_l[i])i = (2t+1) 2^s, r = q 2^(s+1) + m のときf[i] = Π((-1)^popcount(t ∩ qr) min(mr, 2^(s+1)-mr) - (-1)^popcount(t ∩ ql) min(ml, 2^(s+1)-ml))*/vm TLE3(int n, int m, vi l, vi r) {int N = 1 << n;vm res(N, 1);rep(j, m) res[0] *= r[j] - l[j];rep(s, n) {int Ns = N >> (s + 1);vi cnt(n - s);rep(j, m) {int q = r[j] >> (s + 1);repis(b, q) cnt[b]++;}rep(t, Ns) {int i = (2 * t + 1) << s;int sum = 0;repis(b, t) sum += cnt[b];res[i] *= sum & 1 ? -1 : 1;}vi q(m), A(m), B(m);rep(j, m) {q[j] = (r[j] ^ l[j]) / (1 << (s + 1));int mr = r[j] % (1 << (s + 1));A[j] = min(mr, (1 << (s + 1)) - mr);int ml = l[j] % (1 << (s + 1));B[j] = min(ml, (1 << (s + 1)) - ml);}rep(t, Ns) {int i = (2 * t + 1) << s;rep(j, m) {int sgn = popcount(t & q[j]) & 1 ? -1 : 1;res[i] *= A[j] - sgn * B[j];}}}dump(res);hadamard_inv(res);return res;}/*f[i] = Π(A_r[i] - A_l[i])i = (2t+1) 2^s, r = q 2^(s+1) + m のときf[i] = Π(-1)^popcount(t ∩ qr) Π(min(mr, 2^(s+1)-mr) - (-1)^popcount(t ∩ (qr XOR ql)) min(ml, 2^(s+1)-ml))= Π(-1)^popcount(t ∩ qr) Π_j (A_j - (-1)^popcount(t ∩ q_j) B_j)*///【アダマール変換】O(2^n n)(の改変)/** a[0..2^n) を* A[set] = Σset2 (-1)^popcount(set ∩ set2) a[set2]* なる A[0..2^n) に上書きする.*/void hadamard2(vector<pair<mint, mint>>& a) {int n = msb(sz(a));rep(i, n) repb(set, n) {if (!(set & (1 << i))) {auto [xu, xv] = a[set];auto [yu, yv] = a[set | (1 << i)];// (log xu - log xv) + (log yu - log yv)// = log (xu yu) - log (xv yv)a[set] = { xu * yu, xv * yv };// (log xu - log xv) - (log yu - log yv)// = log (xu yv) - log (xv yu)a[set + (1 << i)] = { xu * yv, xv * yu };}}}vm solve(int n, int m, vi l, vi r) {int N = 1 << n;vm res(N, 1);rep(j, m) res[0] *= r[j] - l[j];rep(s, n) {int Ns = N >> (s + 1);vi cnt(n - s);rep(j, m) {int q = r[j] >> (s + 1);repis(b, q) cnt[b]++;}rep(t, Ns) {int i = (2 * t + 1) << s;int sum = 0;repis(b, t) sum += cnt[b];res[i] *= sum & 1 ? -1 : 1;}vector<pair<mint, mint>> CD(Ns, { 1, 1 });rep(j, m) {int q = (r[j] ^ l[j]) / (1 << (s + 1));int mr = r[j] % (1 << (s + 1));int A = min(mr, (1 << (s + 1)) - mr);int ml = l[j] % (1 << (s + 1));int B = min(ml, (1 << (s + 1)) - ml);CD[q % Ns].first *= A - B;CD[q % Ns].second *= A + B;}hadamard2(CD);rep(t, Ns) {int i = (2 * t + 1) << s;res[i] *= CD[t].first;}}dump(res);hadamard_inv(res);return res;}int main() {// input_from_file("input.txt");// output_to_file("output.txt");// zikken2();int n, m;cin >> n >> m;vi l(m), r(m);rep(j, m) cin >> l[j] >> r[j];++r;dump(TLE(n, m, l, r)); dump("-----");dump(TLE2(n, m, l, r)); dump("-----");dump(TLE3(n, m, l, r)); dump("-----");auto res = solve(n, m, l, r);repb(set, n) cout << res[set] << "\n";}