BZOJ4555 求和

题面

解：把第二类斯特林数和组合数拆开。

惊讶的发现推不下去了......后面那个i的∑跟j有关，不能卷积。

这时发现一开始的式子，j的上界可以改为n。

于是这里i的下界改为1。就是一个常数，可以卷积。

卷积出来之后枚举j即可。

#include <bits/stdc++.h>

typedef long long LL;

const int N = 100010, MO = 998244353;

int A[N << 2], B[N << 2], r[N << 2];
int fac[N], inv[N], invn[N];

inline int qpow(int a, int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1ll * ans * a % MO;
        a = 1ll * a * a % MO;
        b = b >> 1;
    }
    return ans;
}

inline void prework(int n) {
    static int R = 0;
    if(R == n) return;
    R = n;
    int lm = 1;
    while((1 << lm) < n) lm++;
    for(int i = 1; i < n; i++) {
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (lm - 1));
    }
    return;
}

inline void NTT(int *a, int n, int f) {
    prework(n);
    for(int i = 0; i < n; i++) {
        if(i < r[i]) std::swap(a[i], a[r[i]]);
    }
    for(int len = 1; len < n; len <<= 1) {
        int Wn = qpow(3, (MO - 1) / (len << 1));
        if(f == -1) Wn = qpow(Wn, MO - 2);
        for(int i = 0; i < n; i += (len << 1)) {
            int w = 1;
            for(int j = 0; j < len; j++) {
                int t = 1ll * a[i + len + j] * w % MO;
                a[i + len + j] = (a[i + j] - t) % MO;
                a[i + j] = (a[i + j] + t) % MO;
                w = 1ll * w * Wn % MO;
            }
        }
    }
    if(f == -1) {
        int inv = qpow(n, MO - 2);
        for(int i = 0; i < n; i++) {
            a[i] = 1ll * a[i] * inv % MO;
        }
    }
    return;
}

inline int C(int n, int m) {
    if(n < 0 || m < 0 || n < m) return 0;
    return 1ll * fac[n] * invn[m] % MO * invn[n - m] % MO;
}

inline int getQ(int a, int q, int n) {
    if(q == 1) {
        return 1ll * a * n % MO;
    }
    else {
        return 1ll * a * (qpow(q, n) - 1) % MO * qpow(q - 1, MO - 2) % MO;
    }
}

int main() {
    int n;
    scanf("%d", &n);
    fac[0] = inv[0] = invn[0] = 1;
    fac[1] = inv[1] = invn[1] = 1;
    for(int i = 2; i <= n; i++) {
        fac[i] = 1ll * fac[i - 1] * i % MO;
        inv[i] = 1ll * inv[MO % i] * (MO - MO / i) % MO;
        invn[i] = 1ll * invn[i - 1] * inv[i] % MO;
    }

    /// solve
    for(int i = 0; i <= n; i++) {
        A[i] = invn[i] * ((i & 1) ? -1 : 1);
        B[i] = 1ll * invn[i] * getQ(1, i, n + 1) % MO;
    }
    int len = 1;
    while(len <= n * 2) len <<= 1;
    NTT(A, len, 1); NTT(B, len, 1);
    for(int i = 0; i < len; i++) {
        A[i] = 1ll * A[i] * B[i] % MO;
    }
    NTT(A, len, -1);

    int ans = 0;
    for(int i = 0; i <= n; i++) {
        ans = (ans + 1ll * fac[i] * qpow(2, i) % MO * A[i] % MO) % MO;
    }
    printf("%d
", (ans + MO) % MO);
    return 0;
}

 解法二：资料1，资料2。