QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#581057#9380. MatchshiqiaqiayaAC ✓67ms4200kbC++1714.2kb2024-09-22 05:57:402024-09-22 05:57:40

Judging History

你现在查看的是最新测评结果

  • [2024-09-22 05:57:40]
  • 评测
  • 测评结果:AC
  • 用时:67ms
  • 内存:4200kb
  • [2024-09-22 05:57:40]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
#define int long long
template <class T, class ... A>
void debug(const T & t, const A & ... a) { cerr << "[" << t, ((cerr << ", " << a), ...), cerr << "]\n"; }
mt19937_64 rd(time(0));

constexpr int mod = 998244353, g[2] = {332748118, 3};
constexpr int MOD[] = {998244353, 1004535809, 754974721, 469762049}, G[][2] = {{332748118, 3}, {334845270, 3}, {617706590, 11}, {156587350, 3}};

auto binpow = [](int a, int b, int mod, int res = 1) { // 模数为质数,求逆元,可以对次幂 mod (mod - 1) 加速
    for (a %= mod; b; b >>= 1, (a *= a) %= mod) {
        if (b & 1) {
            (res *= a) %= mod;
        }
    }
    return res;
};

auto Cipolla = [](int n, int mod) -> set<int> { // size() 为 0 无解,为 1 相同解,为 2 两个解
    if (!(n %= mod) || mod == 2) return {mod == 2};
    if (binpow(n, mod - 1 >> 1, mod) == mod - 1) return {};
    while (true) {
        int a = rd() % mod, w = (a * a - n + mod) % mod;
        if (binpow(w, mod - 1 >> 1, mod) == mod - 1) {
            return [&](array<int, 2> a, int b, array<int, 2> res = {1, 0}) {
                auto mul = [&](auto & a, auto & b) {
                    a = {(a[0] * b[0] + a[1] * b[1] % mod * w) % mod, (a[0] * b[1] + a[1] * b[0]) % mod};
                };
                for ( ; b; b >>= 1, mul(a, a)) {
                    if (b & 1) mul(res, a);
                }
                return res[0] << 1 == mod ? set{res[0]} : set{res[0], mod - res[0]};
            } ({a, 1}, mod + 1 >> 1);
        }
    }
};

template <int mod = mod, const int *g = g>
struct Poly : vector<int> {
    int deg;
    static vector Inv;
    static array<vector, 2> W;

    Poly & Resize(int sz, int x = 0) {   // !!改度必须用这个改,不能手动改,别用错 Resize()
        resize(sz, x), resize((deg = sz - 1) > 0 ? (1ll << 1 + __lg(deg)) : 1, x);
        for (int i = Inv.size(); i <= size(); i++) {
            Inv.emplace_back((mod - mod / i) * Inv[mod % i] % mod);
        }
        for (int bit = W[0].size(); 1ll << bit < size(); bit++) {
            for (auto op : {0, 1}) {
                W[op].emplace_back(::binpow(g[op], mod >> bit + 1, mod));
            }
        }
        return *this;
    } Poly(int sz = 1, int x = 0) {
        Resize(sz, x);
    } Poly(const Poly & f, int sz = 0, int x = 0) : vector(f), deg(f.deg) {
        if (sz) Resize(sz, x);  // 拷贝,有要改 sz 才改
    } Poly(vector::iterator l, vector::iterator r, int sz = 0, int x = 0) : vector(l, r) {
        Resize(sz ? sz : max<int>(1, size()), x);   // 要注意传入的 size 为 0 的情况
    } Poly(initializer_list<int> f) : vector(f) {
        Resize(max<int>(1, size()));
    }

    void NTT(int op) {    // 1 为 DFT, 0 为 IDFT
        vector<int> Rev(size());
        for (int i = 0; i < size(); i++) {
            Rev[i] = ((i & 1) * size() | Rev[i >> 1]) >> 1;
            if (i < Rev[i]) ::swap(at(i), at(Rev[i]));
        }
        for (int h = 1, t = 0; h < size(); h <<= 1, t++) {
            for (int j = 0; j < size(); j += h << 1) {
                for (int k = j, tw = 1; k < j + h; k++, (tw *= W[op][t]) %= mod) {
                    (at(k + h) = at(k) - tw * at(k + h) % mod + mod) %= mod;
                    (at(k) += at(k) - at(k + h) + mod) %= mod;
                }
            }
        }
        for (int i = 0; !op && i < size(); i++) {
            (at(i) *= Inv[size()]) %= mod;
        }
    } Poly & operator *= (const Poly & t) {
        Resize(deg + t.deg + 1);
        Poly b(t, deg + 1);
        NTT(1), b.NTT(1);
        for (int i = 0; i < size(); i++) {
            (at(i) *= b[i]) %= mod;
        }
        return NTT(0), *this;
    } Poly operator * (const Poly & t) { return Poly(*this) *= t; }
    Poly & operator *= (const int & k) {
        for (int i = 0; i <= deg; i++) {
            (at(i) *= k) %= mod;
        }
        return *this;
    } Poly operator * (const int & k) { return Poly(*this) *= k; }

    Poly & operator += (const Poly & t) {
        Resize(max(deg, t.deg) + 1);
        for (int i = 0; i <= t.deg; i++) {
            (at(i) += t[i]) %= mod;
        }
        return *this;
    } Poly operator + (const Poly & t) { return Poly(*this) += t; }
    Poly & operator -= (const Poly & t) {
        Resize(max(deg, t.deg) + 1);
        for (int i = 0; i <= t.deg; i++) {
            (at(i) += mod - t[i]) %= mod;
        }
        return *this;
    } Poly operator - (const Poly & t) { return Poly(*this) -= t; }

    Poly & operator /= (const Poly & t) {
        return *this = DivMod(t)[0];
    } Poly operator / (const Poly & t) { return DivMod(t)[0]; }
    Poly & operator %= (const Poly & t) {
        return *this = DivMod(t)[1];
    } Poly operator % (const Poly & t) { return DivMod(t)[1]; }

    Poly & operator >>= (const int & k) { // 消灭低次项
        if (deg < k) return *this = Poly();
        copy(begin() + k, begin() + deg + 1, begin());
        return Resize(deg - k + 1);
    } Poly operator >> (const int & k) { return Poly(*this) >>= k; }
    Poly & operator <<= (const int & k) { // 增加低次项
        insert(begin(), k, 0);
        return Resize(deg + k + 1);
    } Poly operator << (const int & k) { return Poly(*this) <<= k; }

    Poly inv() {  // 存在当且仅当 at(0) != 0 ,使用时用临时变量开至模 x^sz 大小
        Poly f(1, binpow(at(0), mod - 2, mod));  // f 为逆
        for (int bit = 2; bit >> 1 <= deg; bit <<= 1) {   // h 为原函数
            Poly h(begin(), begin() + bit, bit << 1);   // 要开四倍空间
            f.Resize(h.size()); // 俩个函数 (bit >> 1) 和 一个为 bit 相乘
            f.NTT(1), h.NTT(1);
            for (int i = 0; i < h.size(); i++) {
                (f[i] *= 2 - f[i] * h[i] % mod + mod) %= mod;
            }
            f.NTT(0), f.Resize(bit);
        }
        return f.Resize(deg + 1);  // 记得更改度
    }

    array<Poly, 2> DivMod(const Poly & b) {
        if (deg < b.deg) return {Poly(), *this};  // 小于直接返回
        Poly r(*this), q(b);    // 要先取反再 Resize
        reverse(r.begin(), r.begin() + r.deg + 1), reverse(q.begin(), q.begin() + q.deg + 1);
        q = (q.Resize(deg - b.deg + 1).inv() * r).Resize(deg - b.deg + 1);
        reverse(q.begin(), q.begin() + q.deg + 1);
        return {q, (*this - q * b).Resize(b.deg)};
    }

    int GetDelta() {   // at(0) == 0 的转化
        return find_if(begin(), end(), [](auto & x) { return x != 0; }) - begin();
    }

    Poly EasySqrt(int delta = 0, int root0 = 1) {   // 只用于 at(0) = 1,使用时用临时变量开至模 x^n 大小
        Poly f(1, root0);
        for (int bit = 2; (bit >> 1) + delta <= deg; bit <<= 1) {
            Poly h(begin() + delta, begin() + bit + delta, bit << 1), invf = f.Resize(bit).inv();
            f.Resize(h.size()), invf.Resize(h.size());
            f.NTT(1), h.NTT(1), invf.NTT(1);
            for (int i = 0; i < h.size(); i++) {
                (f[i] = Inv[2] * (f[i] + h[i] * invf[i] % mod) % mod) %= mod;
            }
            f.NTT(0), f.Resize(bit);
        }
        return f.Resize(deg + 1);  // 记得更改度
    } vector<Poly> sqrt(int delta = 0, vector<Poly> res = {}) { // 考虑所以情况和所有解,size为 0 表示无解
        if ((delta = GetDelta()) > deg) return {*this};   // 全 0
        if (delta & 1) return {};  // 无解
        auto root = Cipolla(at(delta), mod);    // 求字典序最小可以拿出来只跑一次
        for (auto & x : root) {
            res.push_back(EasySqrt(delta, x));
        }
        return res;
    }

    Poly dev() {    // 求导
        Poly f(deg);
        for (int i = 1; i <= deg; i++) {
            (f[i - 1] = i * at(i)) %= mod;
        }
        return f;
    } Poly indev() {    // 积分
        Poly f(deg + 2);
        for (int i = 0; i <= deg; i++) {
            (f[i + 1] = Inv[i + 1] * at(i)) %= mod;
        }
        return f;
    }

    Poly ln() { // 存在当且仅当 at(0) = 1
        return (dev() * inv()).Resize(deg).indev();
    } Poly exp() {    // 存在当且仅当 at(0) = 0
        Poly f(1, 1);
        for (int bit = 2; bit >> 1 <= deg; bit <<= 1) {
            Poly h(begin(), begin() + bit, bit << 1), lnf = f.Resize(bit).ln();
            f.Resize(h.size()), lnf.Resize(h.size());
            f.NTT(1), h.NTT(1), lnf.NTT(1);
            for (int i = 0; i < h.size(); i++) {
                (f[i] *= (1 - lnf[i] + mod + h[i])) %= mod;
            }
            f.NTT(0), f.Resize(bit);
        }
        return f.Resize(deg + 1);
    }

    Poly pow(string s, int delta = 0, int k0 = 0, int k1 = 0) { // 使用的时候 to_string
        if ((delta = GetDelta()) > deg) return *this;   // 全 0
        for (auto & x : s) {
            (k0 = k0 * 10 + (x - '0')) %= mod, (k1 = k1 * 10 + (x - '0')) %= mod - 1;
            if (delta * k1 > deg) return Poly(deg + 1); // mod x^sz 后为 0
        }
        auto f = ((*this) >> delta).Resize(deg - delta * k1 + 1) * binpow(at(delta), mod - 2, mod);
        return (f.ln() * k0).exp() * binpow(at(delta), k1, mod) << delta * k1;
    }
    // 使用的点都要互不相同
    static Poly & DivFFT(iterator l, iterator r, vector<Poly> & tr, int op = 1, int p = 1) {    // 求横坐标轴上插值, op = 1 正常,op = 0 求转置
        if (next(l) == r) return op ? tr[p] = {mod - *l, 1} : tr[p] = {1, mod - *l};    // 左闭右开
        auto mid = next(l, r - l + 1 >> 1);
        return tr[p] = DivFFT(l, mid, tr, op, p << 1) * DivFFT(mid, r, tr, op, p << 1 | 1);
    }
    Poly T_Mul(Poly & b, Poly c = {}) { // 辅助函数,转置乘法,会修改 b
        reverse(b.begin(), b.begin() + b.deg + 1), c = *this * b;
        return Poly(c.begin() + b.deg, c.begin() + c.deg + 1);
    }
    vector MultiPoint(iterator L, iterator R, vector<int> res = {}) {   // 多点求值,左闭右开
        vector<Poly> tr(max<int>(deg + 1, R - L) << 2);    // 传入 n 个点,返回 n 个对应点
        Poly x(L, R, max<int>(deg + 1, R - L));
        auto Find = [&](auto && self, auto l, auto r, int p = 1) {  // 求值过程
            if (next(l) == r) return res.push_back(tr[p][0]);
            auto mid = next(l, r - l + 1 >> 1);
            tr[p].Resize(r - l + 1);
            tie(tr[p << 1], tr[p << 1 | 1]) = pair{tr[p].T_Mul(tr[p << 1 | 1]), tr[p].T_Mul(tr[p << 1])};
            self(self, l, mid, p << 1), self(self, mid, r, p << 1 | 1);
        };
        tr[1] = T_Mul(tr[1] = DivFFT(x.begin(), x.begin() + x.deg + 1, tr, 0).inv());
        return Find(Find, x.begin(), x.begin() + x.deg + 1), res.resize(R - L), res;
    }
    static Poly QuickPolate(vector<array<int, 2>>::iterator L, vector<array<int, 2>>::iterator R) {
        vector<Poly> tr(R - L << 2);    // 快速插值,给二维点求函数,左闭右开
        Poly x(R - L);
        for (auto i = L; i != R; i++) {
            x[i - L] = i -> at(0);
        }
        auto ty = DivFFT(x.begin(), x.begin() + x.deg + 1, tr, 1).dev().MultiPoint(x.begin(), x.begin() + x.deg + 1);
        auto Find = [&](auto && self, auto l, auto r, int p = 1) {  // 求值过程
            if (next(l) == r) return Poly(1, l -> at(1) * binpow(ty[l - L], mod - 2, mod) % mod);
            auto mid = next(l, r - l + 1 >> 1);
            return self(self, l, mid, p << 1) * tr[p << 1 | 1] + self(self, mid, r, p << 1 | 1) * tr[p << 1];
        };
        return Find(Find, L, R);
    }
};
template <int mod, const int* G>
vector<int> Poly<mod, G>::Inv(2, 1);
template <int mod, const int* G>
array<vector<int>, 2> Poly<mod, G>::W;

vector<int> inv, fact, invfact;
struct Combine {
    Combine(int n) {
        inv = fact = invfact = vector<int>(n + 1, 1);
        for (int i = 2; i <= n; i++) {
            (inv[i] = (mod - mod / i) * inv[mod % i]) %= mod, (fact[i] = fact[i - 1] * i) %= mod, (invfact[i] = invfact[i - 1] * inv[i]) %= mod;
        }
    }
    int operator()(int n, int m) { return m > n || n < 0 || m < 0 ? 0 : fact[n] * invfact[m] % mod * invfact[n - m] % mod; }
    int P(int n, int m) { return m > n || n < 0 || m < 0 ? 0 : fact[n] * invfact[n - m] % mod; }
    int Lucas(int n, int m) { return m == 0 ? 1 : (*this)(n % mod, m % mod) * Lucas(n / mod, m / mod) % mod; }
} C(200);


void QAQ() {
    int n, k;
    cin >> n >> k;

    vector<int> a(n), b(n);

    for (int i = 0; i < n; cin >> a[i++]);
    for (int i = 0; i < n; cin >> b[i++]);

    auto F = [&](auto && self, vector<int> & a, vector<int> & b, int now) -> Poly<> {
        if (a.size() == 0 || b.size() == 0) return {1};
        auto get = [&](int a, int b) {
            Poly t(min(a, b) + 1);
            for (int i = 0; i <= t.deg; i++) {
                t[i] = C.P(a, i) * C(b, i) % mod;
            }
            return t;
        };
        if (now == -1) return get(a.size(), b.size());
        vector<int> a0, a1, b0, b1;
        for (auto & x : a) {
            if (x >> now & 1) a1.push_back(x ^ (1ll << now));
            else a0.push_back(x);
        }
        for (auto & x : b) {
            if (x >> now & 1) b1.push_back(x ^ (1ll << now));
            else b0.push_back(x);
        }

        if (k >> now & 1) {
            auto t1 = self(self, a0, b1, now - 1), t2 = self(self, a1, b0, now - 1);
            auto t = t1 * t2;
            t.Resize(min(a.size(), b.size()) + 1);
            return t;
        } else {
            auto t1 = self(self, a0, b0, now - 1), t2 = self(self, a1, b1, now - 1);
            Poly t(min(a.size(), b.size()) + 1);
            for (int i = 0; i <= t1.deg; i++) {
                for (int j = 0; j <= t2.deg; j++) {
                    auto tmp = get(a0.size() - i, b1.size() - j) * get(a1.size() - j, b0.size() - i);
                    for (int k = 0; k <= tmp.deg; k++) {
                        (t[i + j + k] += t1[i] * t2[j] % mod * tmp[k]) %= mod;
                    }
                }
            }
            return t;
        }
    };

    auto res = F(F, a, b, 61);

    int ans = 0;
    for (int i = 1; i <= res.deg; i++) {
        cout << res[i] << "\n";
    }
}

signed main() {
    cin.tie(0) -> sync_with_stdio(0);
    int t = 1;
   // cin >> t;

    while (t--) {
        QAQ();
    }
}

这程序好像有点Bug,我给组数据试试?

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 3720kb

input:

9 5
1 7 2 8 4 9 2 5 10
1 3 2 4 5 8 8 8 9

output:

51
1034
10768
62195
200965
348924
294444
97344
7200

result:

ok 9 lines

Test #2:

score: 0
Accepted
time: 67ms
memory: 3828kb

input:

200 569102225177443347
1103663383682482176 1103381908705771520 1099441259031822336 1098878309078401024 1089871109823660032 1080863910568919040 1074108511127863296 1071856711314178048 1069041961547071488 1068479011593650176 1067353111686807552 1062849512059437056 1049338713177325568 10448351135499550...

output:

20506
208107723
47878304
53020813
972282728
933586852
658157196
670189811
957980024
366179738
217980591
967482558
833450149
987731802
260904367
5263881
600332344
906061351
658256294
93700706
421323952
178075016
219871690
986880524
848776106
191185484
641917326
576497440
908609746
349728876
606714342...

result:

ok 200 lines

Test #3:

score: 0
Accepted
time: 11ms
memory: 3736kb

input:

200 1098723432450995412
972777519512027136 936748722493063168 864691128455135232 860187528827764736 857935729014079488 856246879153815552 855683929200394240 851180329573023744 848928529759338496 847802629852495872 847310048643252224 847239679899074560 846676729945653248 828662331436171264 8280993814...

output:

8320
34035512
428014253
916072411
696504298
748377440
32424677
188402417
233587075
130639672
759476380
546285625
187068736
159002787
131866334
381530906
133126344
373727612
923311054
725293681
718162548
39511135
322320638
44709653
156884882
285926751
787179409
282967016
153092344
615721347
950855850...

result:

ok 200 lines

Test #4:

score: 0
Accepted
time: 3ms
memory: 4200kb

input:

200 1
1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...

output:

3072
4412928
940025853
665370743
752672705
581549490
41990996
541401698
170451802
26584141
220983766
844620126
64506869
137621326
418866920
351049174
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
...

result:

ok 200 lines

Test #5:

score: 0
Accepted
time: 4ms
memory: 3992kb

input:

200 0
1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...

output:

40000
792020000
319908096
286630939
515085548
95626140
472749362
800008042
594973486
139901984
847967250
125664877
254232250
419486325
204481923
979660340
41873095
540955890
585129047
205968776
196086667
674391130
737319172
669580034
973843015
207125612
331414909
119844754
24879477
990932807
8476927...

result:

ok 200 lines

Extra Test:

score: 0
Extra Test Passed