又是一道FFT 好题。
首先来看一看求前缀和。
求一次前缀和就先当于卷上一个系数全为1的多项式,即\(\displaystyle \sum_{i=0}^{\infin}x^i\)(想一想,为什么),这个东西就等于 \(\displaystyle \frac{1}{1-x}\),简单证明一下。
\[
S=1+x+x^2+...\\
xS=x+x^2+x^3...\\
上边的式子减去下边的式子得到\\
S-xS=1\\
S=\frac{1}{1-x}
\]
是不是感觉天衣无缝但又十分扯淡?没错,这只有在\(-1<x<1\)时才成立。但生成函数是形式幂级数,我们不用关心x的具体取值。
回到我们刚才的问题,因为卷积具有结合律,所以我们要求的就是\(\displaystyle\frac{1}{(1-x)^k}\)
结论:这个式子的n次系数是 \(C_{n+k-1}^{k-1}\)(C是组合数)。
证明:回想一下\(\displaystyle \sum_{i=0}^{\infin}x^i\)的每一次相乘的含义,可知\(\displaystyle (\sum_{i=0}^{\infin}x^i)^k\)中n次系数的含义就是经过k次组成n的方案数,我们可以将n看成是n个小球,k看成是k个盒子,因为组成n的每个 “1”是一样的,每个多项式是不一样的,所以球相同,盒子不同,方案数就是\(C_{n+k-1}^{k-1}\)。
因为k很大,所以我们需要先让k对1004535809取模。至于为什么是可以的,可以回想一下P5245 多项式快速幂 这道题,考虑在模的意义下,对\(k \times lnF(x)\)做多项式指数函数,显然是可以的。
但是取模后的k依然很大,我们将 \(C_n^m\)拆开\(\displaystyle C_n^m=\frac{n!}{m!(n-m)!}=\frac{\prod_{i=n-m+1}^{n}i}{\prod_{i=1}^{m}i}\), 又因为\(C_n^0=1\)然后我们就能递推组合数啦。\(\displaystyle C_k^i=C_k^{i-1}\times \frac{k-i+1}{i}\)
接下来看一看求差分。
其实差分一次就相当于卷上\(1-x\)(想一想,为什么),所以我们要求的就是\((1-x)^k\).
求法1: 是不是感觉这个式子似曾相识?,没错,我们其实只用对刚刚求\(\displaystyle\frac{1}{(1-x)^k}\)求个逆就行了,
求法2:根据二项式定理\(\displaystyle (1-x)^k=\sum_{i=0}^kC_{k}^{i}(-x)^i=\sum_{i=0}^kC_{k}^{i}(-1)^ix^i\).然后就能求啦。
1004535809的原根是3.
#include<bits/stdc++.h> #define LL long long using namespace std; int n, opt, len; LL k; const int N = 400010, mod = 1004535809, G = 3, Ginv = (mod + 1) / 3; int r[N]; LL a[N], b[N], c[N], Y[N]; char ch[2501]; int read() { int x = 0; int f = 0; char c = getchar(); while (!isdigit(c)) f |= c == '-', c = getchar(); while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar(); return f ? -x : x; } LL ksm(LL a, LL b, LL mod) { LL res = 1; for (; b; b >>= 1, a = a * a % mod) if (b & 1)res = res * a % mod; return res; } void NTT(LL *A, int lim, int opt) { for (int i = 0; i < lim; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) ? (lim >> 1) : 0); for (int i = 0; i < lim; ++i) if (i < r[i])swap(A[i], A[r[i]]); int len; LL wn, w, x, y; for (int mid = 1; mid < lim; mid <<= 1) { len = mid << 1; wn = ksm(opt == 1 ? G : Ginv, (mod - 1) / len, mod); for (int j = 0; j < lim; j += len) { w = 1; for (int k = j; k < j + mid; ++k, w = w * wn % mod) { x = A[k]; y = A[k + mid] * w % mod; A[k] = (x + y) % mod; A[k + mid] = (x - y + mod) % mod; } } } if (opt == 1)return; int ni = ksm(lim, mod - 2, mod); for (int i = 0; i < lim; ++i)A[i] = A[i] * ni % mod; } void MUL(LL *A, int n, LL *B, int m) { int lim = 1; while (lim < (n + m))lim <<= 1; NTT(A, lim, 1); NTT(B, lim, 1); for (int i = 0; i < lim; ++i)A[i] = A[i] * B[i] % mod; NTT(A, lim, -1); } void INV(int siz, LL *A, LL *B) { if (siz == 1) { B[0] = ksm(A[0], mod - 2, mod); return; } INV((siz + 1) >> 1, A, B); int lim = 1; while (lim < (siz << 1))lim <<= 1; for (int i = 0; i < siz; ++i)c[i] = A[i]; for (int i = siz; i < lim; ++i)c[i] = 0; NTT(c, lim, 1); NTT(B, lim, 1); for (int i = 0; i < lim; ++i)B[i] = B[i] * (2 - c[i] * B[i] % mod + mod) % mod; NTT(B, lim, -1); for (int i = siz; i < lim; ++i)B[i] = 0; } int main() { cin >> n; scanf("%s", ch + 1); cin >> opt; len = strlen(ch + 1); for (int i = 1; i <= len; ++i)k = k * 10 + ch[i] - '0', k %= mod; for (int i = 1; i <= n; ++i)a[i] = read(); b[0] = 1; for (int i = 1; i <= n; ++i)b[i] = b[i - 1] * (i + k - 1) % mod * ksm(i, mod - 2, mod) % mod; if (opt == 0) { MUL(a, n, b, n); for (int i = 1; i <= n; ++i)printf("%lld ", a[i]); } else { INV(n + 1, b, Y); MUL(a, n, Y, n); for (int i = 1; i <= n; ++i)printf("%lld ", a[i]); } return 0; }
来源:https://www.cnblogs.com/wljss/p/12020771.html