Starry

发表于 更新于 分类于 本文字数： 190 阅读时长 ≈ 1 分钟

下降幂每次都重新计算会TLE :)

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#include <cstdio>
#include <assert.h>

using namespace std;

using ll = long long;

const int mod = 998244353;
ll s[(int)2e3 + 9][(int)2e3 + 9] = {1};
ll n, m, k, tt;

ll cal(ll x, ll t)
{
    ll res = 1;
    for (ll i = x; i >= x - t + 1; --i)
        res = i * res % mod;
    return res;
}

ll powmod(ll a, ll b)
{
    ll res = 1;
    for (; b > 0; b >>= 1, a = a * a % mod)
        if (b & 1)
            res = res * a % mod;
    return res;
}

void solve()
{
    scanf("%lld%lld%lld", &n, &m, &k);
    ll xjm = 1;
    ll ans = 0, x = (m + 1) / 2, y = m / 2;
    for (int i = 0; i <= k && n >= i; ++i)
        ans = (ans + s[k][i] * xjm % mod * powmod(x, i) % mod * powmod(m, n - i)) % mod, xjm = xjm * (n - i) % mod;
    printf("%lld\n", ans);
}

int main()
{
    for (int i = 1; i <= 2e3; ++i)
        for (ll j = 1; j <= 2e3; ++j)
            s[i][j] = (s[i - 1][j - 1] + s[i - 1][j] * j) % mod;
    scanf("%lld", &tt);
    for (int i = 1; i <= tt; ++i)
        solve();
}

发表于 更新于 分类于 本文字数： 366 阅读时长 ≈ 1 分钟 
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#include <cstdio>
#include<assert.h>

using namespace std;

using ll = long long;

const int mod = 998244353;

template <int MOD, int RT>
struct mint
{
    static const int mod = MOD;
    static constexpr mint rt() { return RT; } // primitive root for FFT
    int v;
    explicit operator int() const { return v; } // explicit -> don't silently convert to int
    mint() : v(0) {}
    mint(ll _v)
    {
        v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
        if (v < 0)
            v += MOD;
    }
    bool operator==(const mint &o) const
    {
        return v == o.v;
    }
    friend bool operator!=(const mint &a, const mint &b)
    {
        return !(a == b);
    }
    friend bool operator<(const mint &a, const mint &b)
    {
        return a.v < b.v;
    }

    mint &operator+=(const mint &o)
    {
        if ((v += o.v) >= MOD)
            v -= MOD;
        return *this;
    }
    mint &operator-=(const mint &o)
    {
        if ((v -= o.v) < 0)
            v += MOD;
        return *this;
    }
    mint &operator*=(const mint &o)
    {
        v = int((ll)v * o.v % MOD);
        return *this;
    }
    mint &operator/=(const mint &o) { return (*this) *= inv(o); }
    friend mint pow(mint a, ll p)
    {
        mint ans = 1;
        assert(p >= 0);
        for (; p; p /= 2, a *= a)
            if (p & 1)
                ans *= a;
        return ans;
    }
    friend mint inv(const mint &a)
    {
        assert(a.v != 0);
        return pow(a, MOD - 2);
    }

    mint operator-() const { return mint(-v); }
    mint &operator++() { return *this += 1; }
    mint &operator--() { return *this -= 1; }
    friend mint operator+(mint a, const mint &b) { return a += b; }
    friend mint operator-(mint a, const mint &b) { return a -= b; }
    friend mint operator*(mint a, const mint &b) { return a *= b; }
    friend mint operator/(mint a, const mint &b) { return a /= b; }
};

using mi = mint<mod, 3>; // 5 is primitive root for both common mods

mi s[(int)5e3 + 9][(int)5e3 + 9] = {1};
ll n, m, k;

ll cal(ll x, ll t)
{
    ll res = 1;
    for (ll i = x; i>=x-t+1; --i)
        res = i * res % mod;
    return res;
}

int main()
{
    scanf("%lld%lld%lld", &n, &m, &k);
    for(int i=1;i<=k;++i)for(int j=1;j<=k;++j)s[i][j]=s[i-1][j-1]+s[i-1][j]*j;
    mi ans=0,invm=inv(mi(m));
    for(int i=0;i<=k;++i)ans+=s[k][i]*cal(n,i)*pow(invm,i);
    printf("%d",ans.v);
}

发表于 更新于 分类于 本文字数： 589 阅读时长 ≈ 2 分钟

分治NTT，套了个dls自动取模的整数板子，但常数过大。

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#include <assert.h>
#include <cstdio>
#include <iostream>
using namespace std;
using ll = long long;

const int NR = 1 << 21;
const int G = 3, Gi = 332748118;
const int mod = 998244353;

template <int MOD, int RT>
struct mint
{
    static const int mod = MOD;
    static constexpr mint rt() { return RT; } // primitive root for FFT
    int v;
    explicit operator int() const { return v; } // explicit -> don't silently convert to int
    mint() : v(0) {}
    mint(ll _v)
    {
        v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
        if (v < 0)
            v += MOD;
    }
    bool operator==(const mint &o) const
    {
        return v == o.v;
    }
    friend bool operator!=(const mint &a, const mint &b)
    {
        return !(a == b);
    }
    friend bool operator<(const mint &a, const mint &b)
    {
        return a.v < b.v;
    }

    mint &operator+=(const mint &o)
    {
        if ((v += o.v) >= MOD)
            v -= MOD;
        return *this;
    }
    mint &operator-=(const mint &o)
    {
        if ((v -= o.v) < 0)
            v += MOD;
        return *this;
    }
    mint &operator*=(const mint &o)
    {
        v = int((ll)v * o.v % MOD);
        return *this;
    }
    mint &operator/=(const mint &o) { return (*this) *= inv(o); }
    friend mint pow(mint a, ll p)
    {
        mint ans = 1;
        assert(p >= 0);
        for (; p; p /= 2, a *= a)
            if (p & 1)
                ans *= a;
        return ans;
    }
    friend mint inv(const mint &a)
    {
        assert(a.v != 0);
        return pow(a, MOD - 2);
    }

    mint operator-() const { return mint(-v); }
    mint &operator++() { return *this += 1; }
    mint &operator--() { return *this -= 1; }
    friend mint operator+(mint a, const mint &b) { return a += b; }
    friend mint operator-(mint a, const mint &b) { return a -= b; }
    friend mint operator*(mint a, const mint &b) { return a *= b; }
    friend mint operator/(mint a, const mint &b) { return a /= b; }
};

using mi = mint<mod, 3>;

int rev[NR];

void NTT(mi *a, int n, int type)
{
    for (int i = 0; i < n; ++i)
        if (i < rev[i])
            swap(a[i], a[rev[i]]);
    for (int k = 1; k < n; k <<= 1)
    {
        mi wn = pow((type == 1 ? mi(G) : mi(Gi)), (mod - 1) / (k << 1));
        for (int i = 0; i < n; i += (k << 1))
        {
            mi w = mi(1);
            for (int j = 0; j < k; ++j, w = w * wn)
            {
                mi x = a[i + j], y = w * a[i + j + k];
                a[i + j] = x + y;
                a[i + j + k] = x - y;
            }
        }
    }
    if (type == 1)
        return;
    mi invn = inv(mi(n));
    for (int i = 0; i < n; ++i)
        a[i] = a[i] * invn;
}

int n;
mi A[NR], B[NR], C[NR],D[NR];

void cdq(int l, int r)
{
    if(l==r)return ;
    int mid = (l + r) >> 1;
    cdq(l, mid);
    int bit = 0, num = 1;
    while(num<((r-l+1)<<1))
        num <<= 1, bit += 1;
    for (int i = 0; i < num; ++i)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    for (int i = 0; i < num;++i)
        C[i] = D[i] = 0;
    for (int i = l; i <= r;++i)
        C[i - l] = A[i - l];
    for (int i = l; i <= mid;++i)
        D[i - l] = B[i];
    NTT(C, num, 1), NTT(D, num, 1);
    for (int i = 0; i < num;++i)
        C[i] = C[i] * D[i];
    NTT(C, num, 0);
    for (int i = mid + 1; i <= r;++i)
        B[i] += C[i - l];
    cdq(mid + 1, r);
}


int main()
{
    scanf("%d", &n);
    for (int i = 1,x; i <= n-1;++i){
        scanf("%d", &x);
        A[i] = mi(x);
    }
    B[0] = mi(1);
    cdq(0, n-1);
    for (int i = 0; i <= n - 1;++i)
        printf("%d ", B[i].v);
}

发表于 更新于 分类于 本文字数： 400 阅读时长 ≈ 1 分钟

只需把上下界最小流跑的残量网络最大流起始点换成$s->t$，和可行流相加即可。

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#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

const int N = 125100;
int n, m, s, t, tot = 1, hd[(int)6e4], cur[(int)6e4], ss, tt, dep[(int)6e4];
long long flow[(int)6e4], l[N];
struct Edge
{
    int to, nxt;
    long long val;
} e[N * 3 + 99];
inline void add(int u, int v, long long w)
{
    e[++tot] = {v, hd[u], w}, hd[u] = tot;
    e[++tot] = {u, hd[v], 0ll}, hd[v] = tot;
}

bool bfs(int st, int en)
{
    for (int i = 0; i <= n + 2; ++i)
        dep[i] = 0;
    queue<int> q;
    memcpy(cur, hd, sizeof(hd));
    q.push(st);
    dep[st] = 1;

    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        for (int eg = hd[u]; eg; eg = e[eg].nxt)
        {
            if (!dep[e[eg].to] && e[eg].val > 0)
            {
                dep[e[eg].to] = dep[u] + 1;
                q.push(e[eg].to);
            }
        }
    }
    return !!dep[en];
}

long long dfs(int u, int en, long long flow)
{
    if (u == en)
        return flow;
    long long r = flow;
    for (int eg = cur[u]; eg && r; eg = e[eg].nxt)
    {
        cur[u] = eg;
        if (e[eg].val > 0 && dep[e[eg].to] == dep[u] + 1)
        {
            long long c = dfs(e[eg].to, en, min(r, e[eg].val));
            r -= c;
            e[eg].val -= c;
            e[eg ^ 1].val += c;
        }
    }
    return flow - r;
}

long long Dinic(int st, int en)
{
    long long ret = 0;
    while (bfs(st, en))
        ret += dfs(st, en, 1ll << 53);
    return ret;
}

int main()
{
    long long tmp, sum = 0;
    scanf("%d%d%d%d", &n, &m, &s, &t);
    tt = n + 1;
    for (int i = 1, w, x; i <= m; ++i)
    {
        long long y, z;
        scanf("%d%d%lld%lld", &w, &x, &y, &z);
        l[i] = y;
        add(w, x, z - y);
        flow[w] -= l[i], flow[x] += l[i];
    }
    for (int i = 1; i <= n; ++i)
    {
        if (flow[i] > 0)
        {
            sum += flow[i];
            add(ss, i, flow[i]);
        }
        else if (flow[i] < 0)
            add(i, tt, -flow[i]);
    }
    add(t, s, 1ll << 53);
    if (sum == Dinic(ss, tt))
    {
        tmp = e[tot].val;
        e[tot].val = e[tot ^ 1].val = 0;
        printf("%lld\n", tmp + Dinic(s, t));
    }
    else
        puts("please go home to sleep");
    return 0;
}

发表于 更新于 分类于 本文字数： 631 阅读时长 ≈ 2 分钟

头晕。。。

所以即求

其中

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#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
using ll = long long;

const int N = 1 << 21;
const int P = 998244353;
const int G = 3, Gi = 332748118;
const int mod = 998244353;

int qmi(int a, int b)
{
    int v = 1;
    while (b)
    {
        if (b & 1)
            v = 1ll * v * a % P;
        a = 1ll * a * a % P;
        b >>= 1;
    }
    return v;
}
int rev[N];
void NTT(int *a, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(a[i], a[rev[i]]);

    for (int mid = 1; mid < n; mid <<= 1)
    {
        int Wn = qmi((inv == 1) ? G : Gi, (P - 1) / (mid << 1));
        for (int i = 0; i < n; i += (mid << 1))
        {
            int w = 1;
            for (int j = 0; j < mid; j++, w = 1ll * w * Wn % P)
            {
                int x = a[i + j], y = 1ll * w * a[i + j + mid] % P;
                a[i + j] = (x + y) % P;
                a[i + j + mid] = (x - y + P) % P;
            }
        }
    }
    if (inv == -1)
    {
        int invn = qmi(n, P - 2);
        for (int i = 0; i < n; i++)
            a[i] = 1ll * a[i] * invn % P;
    }
}
//=======================================================
int fact[N], infact[N];
void init()
{
    fact[0] = 1;
    for (int i = 1; i <= 100000; i++)
        fact[i] = 1ll * i * fact[i - 1] % mod;
    infact[100000] = qmi(fact[100000], mod - 2);

    for (int i = 99999; i; i--)
        infact[i] = 1ll * infact[i + 1] * (i + 1) % mod;
}
int n, m, A[N], B[N], C[N];
int Mul(int *a, int *b, int n, int m, int *ans)
{
    int bit = 0, num = 1;
    while (num < n + m + 1)
        num <<= 1, bit++;

    for (int i = 0; i < num; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    for (int i = 0; i <= n; i++)
        A[i] = a[i];
    for (int i = 0; i <= m; i++)
        B[i] = b[i];
    for (int i = n + 1; i < num; i++)
        A[i] = 0;
    for (int i = m + 1; i < num; i++)
        B[i] = 0;
    NTT(A, num, 1);
    NTT(B, num, 1);
    for (int i = 0; i < num; i++)
        C[i] = 1ll * A[i] * B[i] % mod;
    NTT(C, num, -1);
    for (int i = 0; i <= n + m; i++)
        ans[i] = C[i];
    return n + m;
}

int pool[N << 1], tot;
struct Node
{
    int *p, n;
    void init(int x)
    {
        this->n = 1;
        p = pool + tot;
        for (int i = 0; i <= n; i++)
            p[i] = 0;
        p[0] = 1;
        p[1] = x;
        tot += n + 1;
    }
    void mul(const Node &o)
    {
        n = Mul(p, o.p, n, o.n, p);
    }
};
Node solve(int l, int r)
{
    Node ans;
    if (l == r)
    {
        int x;
        cin >> x;
        ans.init(x);
        return ans;
    }
    int mid = (l + r) >> 1;
    ans = solve(l, mid);
    ans.mul(solve(mid + 1, r));
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    init();
    cin >> n >> m;
    ll invk = qmi(m, mod - 2);
    invk = qmi(invk, m / 2);
    Node res = solve(1, n);
    printf("%lld", 1ll * res.p[m] * fact[m] % mod * infact[n] % mod * fact[n - m] % mod * invk % mod);
    return 0;
}
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