传送门

题目大意

　　有\(n\)类物品，每种物品体积为\(V_i\)，且都有无数多件。

　　问你塞满容量为\(s\)的背包方案数，对于每个\(s\in [1,m]\)，\(m\)给定且\(\leq 10^5\)，都求出方案数。答案对\(998244353\)取模。

　　\(30\%\)的数据，\(n,m\leq 3000\)；

　　\(60\%\)的数据，纯随机生成；

　　\(100\%\)的数据，\(n,m\leq 10^5\)，且一定满足\(V_i \leq m\)

题解

30 %

　　普通背包随便搞搞，复杂度\(\text{O}(nm)\)。

100 %

　　显然对于这类问题我们有一个生成函数的解法：定义每种体积为\(k\)的物品的生成函数：

\[G_k(x)=1+x^k+x^{2k}+x^{3k}+\dotsb=\frac{1}{1-x^k}\]

　　那么答案的生成函数就是：

\[F(x)=\prod_i G_{V_i}(x)\]

　　容量为\(s\)的答案即为\([x^s]F(x)\)。

　　关键一次多项式乘法的复杂度很高。数据随机还可以乱搞，但不随机会被卡。

　　降低复杂度，除了换成点值表示，还可以乘法变加法：我们有\(e^a\cdot e^b=e^{a+b}\)。

　　所以将所有\(G(x)\)求\(ln\)得到指数，然后相加可行呢？可是求指数也是个复杂度高的东西，但发现我们实际上\(G(x)\)求的\(ln\)只与\(k\)有关，而且是有规律的！

　　推导一下：

\[ \begin{aligned} \ln\frac{1}{1-x^k}&=\int \frac{(\dfrac{1}{1-x^k})'}{\dfrac{1}{1-x^k}}\mathrm dx\\ &=\int (1+x^k+x^{2k}+\dotsb)'(1-x^k)\mathrm dx\\ &=\int (kx^{k-1}+2kx^{2k-1}+\dotsb)(1-x^k)\mathrm dx\\ &=k\int (x^{k-1}+x^{2k-1}+\dotsb)\mathrm dx\\ &=k(\frac{1}{k}x^k+\frac{1}{2k}x^{2k}+\dotsb)\\ &=x^k+\frac{1}{2}x^{2k}+\frac{1}{3}x^{3k}\dotsb \end{aligned} \]

　　我们可以在调和级数复杂度求出来所有\(\ln G_k(x)\)！

　　边求边加，最后再多项式\(exp\)一下就完事了。总复杂度：\(\text{O}(n\log n)\)

代码

#include <bits/stdc++.h>

using namespace std;

const int maxn = 130000 + 5;
const int P = 1004535809, g = 3;

int inc(int a, int b) { return (a += b) >= P ? a-P : a; }
int qpow(int a, int b) {
    int res = 1;
    for (int i = a; b; i = 1ll*i*i%P, b >>= 1)
        if (b & 1) res = 1ll*res*i%P;
    return res;
}

int W[maxn << 3], inv[maxn << 2], fac[maxn << 2], ifac[maxn << 2];
void prework(int n) {
    for (int i = 1; i < n; i <<= 1) {
        W[i] = 1;
        for (int j = 1, Wn = qpow(g, (P-1)/i>>1); j < i; j++) W[i+j] = 1ll*W[i+j-1]*Wn%P;
    }
    inv[1] = fac[0] = ifac[0] = 1;
    for (int i = 2; i < n; i++) inv[i] = 1ll*(P-P/i)*inv[P%i]%P;
    for (int i = 1; i < n; i++) fac[i] = 1ll*fac[i-1]*i%P, ifac[i] = 1ll*ifac[i-1]*inv[i]%P;
}

void ntt(int *a, int n, int opt) {
    static int rev[maxn << 2];
    for (int i = 1; i < n; i++)
        if ((rev[i] = rev[i>>1]>>1|(i&1?n>>1:0)) > i) swap(a[i], a[rev[i]]);
    for (int q = 1; q < n; q <<= 1)
        for (int p = 0; p < n; p += q << 1)
            for (int i = 0, t; i < q; i++)
                t = 1ll*W[q+i]*a[p+q+i]%P, a[p+q+i] = inc(a[p+i], P-t), a[p+i] = inc(a[p+i], t);
    if (~opt) return;
    for (int i = 0; i < n; i++) a[i] = 1ll*a[i]*inv[n]%P;
    reverse(a+1, a+n);
}

struct poly {
    vector<int> A;
    int len;
    poly(int a0 = 0) : len(1) { A.push_back(a0); }
    int &operator [] (int i) { return A[i]; }
    void write() {
        for (int i = 0; i < len; i++) printf("%d ", A[i]);
        putchar('\n');
    }
    void load(int *from, int n) {
        A.resize(len = n);
        memcpy(&A[0], from, sizeof(int) * len);
    }
    void cpyto(int *to, int n) {
        memcpy(to, &A[0], sizeof(int) * min(len, n));
    }
    void resize(int n = 0) {
        if (!n) { while (len > 1 && !A[len - 1]) len--; A.resize(len); } else A.resize(len = n, 0);
    }
} F, G;

poly poly_inv(poly A) {
    poly B = poly(qpow(A[0], P-2));
    for (int len = 1; len < A.len; len <<= 1) {
        static int x[maxn << 2], y[maxn << 2];
        for (int i = 0; i < len << 2; i++) x[i] = y[i] = 0;
        A.cpyto(x, len << 1), B.cpyto(y, len);
        ntt(x, len << 2, 1), ntt(y, len << 2, 1);
        for (int i = 0; i < len << 2; i++) x[i] = inc(y[i], inc(y[i], P-1ll*x[i]*y[i]%P*y[i]%P));
        ntt(x, len << 2, -1);
        B.load(x, len << 1);
    }
    return B.resize(A.len), B;
}

poly operator + (poly A, poly B) {
    if (A.len < B.len) A.resize(B.len);
    for (int i = 0; i < B.len; i++) A[i] = inc(A[i], B[i]);
    return A.resize(), A;
}

poly operator - (poly A, poly B) {
    if (A.len < B.len) A.resize(B.len);
    for (int i = 0; i < B.len; i++) A[i] = inc(A[i], P-B[i]);
    return A.resize(), A;
}

int getsize(int n) { int N = 1; while (N < n) N <<= 1; return N; }

poly operator * (poly A, poly B) {
    static int x[maxn << 2], y[maxn << 2];
    int len = getsize(A.len + B.len - 1);
    for (int i = 0; i < len; i++) x[i] = y[i] = 0;
    A.cpyto(x, A.len), B.cpyto(y, B.len);
    ntt(x, len, 1), ntt(y, len, 1);
    for (int i = 0; i < len; i++) x[i] = 1ll*x[i]*y[i]%P;
    ntt(x, len, -1);
    return A.load(x, A.len + B.len - 1), A.resize(), A;
}

poly poly_deri(poly A) {
    for (int i = 0; i < A.len - 1; i++) A[i] = 1ll*A[i+1]*(i+1)%P;
    return A[A.len - 1] = 0, A.resize(), A;
}

poly poly_int(poly A) {
    for (int i = A.len - 1; i; i--) A[i] = 1ll*A[i-1]*inv[i]%P;
    return A[0] = 0, A;
}

poly poly_ln(poly A) {
    poly B = poly_deri(A) * poly_inv(A);
    return B.resize(A.len), poly_int(B);
}

int n;

int main() {
    scanf("%d", &n); prework((n+1)<<2);
    G.resize(n+1);
    for (int i = 1; i <= n; i++) G[i] = 1ll*qpow(2, 1ll*i*(i-1)/2%(P-1))*ifac[i]%P;
    G[0] = 1;
    F = poly_ln(G);
    printf("%d", 1ll*F[n]*fac[n]%P);
    return 0;
}

标签：背包,洛谷,int,复杂度,dotsb,len,P4389,frac,2k
来源： https://www.cnblogs.com/ac-evil/p/12089805.html