RiverSheepSHEEP

付公主的背包

奔跑的心634·2023-03-15 21:53·38 次阅读

付公主的背包

P4389 付公主的背包#

对于一种商品,设其OGFfi(x)=xvij=11xvi
那么答案就为fi(x),套路的,求 ln,则原式为

lnfi(x)=ln(1xvi)

其中

ln(1xvi)=j1xvijj

所以答案就为

exp(ij1xvijj)

发现vijm,直接枚举,时间复杂度o(mlnm)
总时间为O(nlogn).

Code
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#include<cstdio> #include<iostream> #include<cstring> #define IN inline #define LL long long using namespace std; const int N = 2e5 + 5, P = 998244353, G = 3; int n, m, a[N << 2], b[N << 2]; IN int read() { int t = 0,res = 0; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) t |= (ch == '-'); for (; isdigit(ch); ch = getchar()) res = (res << 3) + (res << 1) + (ch ^ 48); return t ? -res : res; } LL fpow(LL x, LL y) { LL res = 1; for (; x; x >>= 1, y = y * y % P) if (x & 1) res = res * y % P; return res; } const int invG = fpow(P - 2, G); namespace Poly{ int rev[N << 2], insF[N << 2], sginv[N << 2]; void NTT(int *f, int len, int fl) { static int W[N << 2] = {1}; if (len == 1) return; for (int i = 0; i < len; i++) if (i < rev[i]) swap(f[i], f[rev[i]]); for (int l = 1; l < len; l <<= 1) { int I = fpow((P - 1) / (l << 1), fl == 1 ? G : invG); for (int i = 1; i < l; i++) W[i] = (LL)W[i - 1] * I % P; for (int i = 0; i < len; i += (l << 1)) for (int j = 0; j < l; j++) { int x = f[i | j], y = (LL)f[i | j | l] * W[j] % P; f[i | j] = (x + y >= P ? x + y - P : x + y); f[i | j | l] = (x - y < 0 ? x - y + P : x - y); } } if (fl == -1) { int IV = fpow(P - 2, len); for (int i = 0; i < len; i++) f[i] = (LL)f[i] * IV % P; } } void Mulpoly(int *f, int *g, int lenF, int lenG) { int len = 1, bit = 0; while (len <= lenF + lenG) len <<= 1, bit++; for (int i = 1; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1); NTT(f, len, 1), NTT(g, len, 1); for (int i = 0; i < len; i++) f[i] = (LL)f[i] * g[i] % P; NTT(f, len, -1); for (int i = lenF + lenG + 1; i < len; i++) f[i] = 0; } void Invpoly(int *f, int lenF, int *g) { // [0, lenF); static int s2[N << 2], s1[N << 2]; int limit = 1; while (limit < lenF) limit <<= 1; g[0] = fpow(P - 2, f[0]); for (int len = 2, bit = 1; len <= limit; len <<= 1, bit++) { for (int i = 1; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1); for (int i = 0; i < (len >> 1); i++) s2[i] = g[i]; for (int i = (len >> 1); i < len; i++) s2[i] = 0; for (int i = 0; i < len; i++) s1[i] = f[i]; NTT(s2, len, 1), NTT(s1, len, 1); for (int i = 0; i < len; i++) s1[i] = (LL)s2[i] * s1[i] % P; NTT(s1, len, -1); for (int i = 0; i < (len >> 1); i++) s1[i] = 0; NTT(s1, len, 1); for (int i = 0; i < len; i++) s1[i] = (LL)s1[i] * s2[i] % P; NTT(s1, len, -1); for (int i = (len >> 1); i < len; i++) g[i] = (s1[i] == 0 ? 0 : P - s1[i]); } for (int i = lenF; i < limit; i++) g[i] = 0; for (int i = 0; i < limit; i++) s2[i] = s1[i] = 0; } void Getsginv(int len) { sginv[0] = sginv[1] = 1; for (int i = 2; i <= len; i++) sginv[i] = (LL)sginv[P % i] * (P - P / i) % P; } void Dpoly(int *f, int lenF) { for (int i = 0; i < lenF; i++) f[i] = (LL)f[i + 1] * (i + 1) % P; f[lenF] = 0; } void Jfpoly(int *f, int lenF) { for (int i = lenF; i >= 0; i--) f[i + 1] = (LL)f[i] * sginv[i + 1] % P; f[0] = 0; } void Lnpoly(int *f, int lenF) { // [0, lenF) static int Inv_f[N << 2]; Invpoly(f, lenF, Inv_f), Dpoly(f, lenF - 1), Mulpoly(f, Inv_f, lenF - 1, lenF - 1), Jfpoly(f, lenF - 1); for (int i = lenF; i <= (lenF << 1); i++) f[i] = Inv_f[i] = 0; for (int i = 0; i < lenF; i++) Inv_f[i] = 0; } void Exppoly(int *f, int lenF, int *g) { static int s[N << 2]; int limit = 1; while (limit < lenF) limit <<= 1; g[0] = 1; for (int len = 2; len <= limit; len <<= 1) { for (int i = (len >> 1); i < len; i++) g[i] = 0; for (int i = 0; i < (len >> 1); i++) s[i] = g[i]; for (int i = (len >> 1); i < len; i++) s[i] = 0; Lnpoly(s, len); for (int i = 0; i < len; i++) s[i] = (f[i] - s[i] + P) % P; s[0] = (s[0] + 1) % P, Mulpoly(g, s, len - 1, len - 1); } for (int i = lenF; i < limit; i++) g[i] = 0; for (int i = 0; i < limit; i++) s[i] = 0; } } void write(int x) { if (!x) return; write(x / 10), putchar((x % 10) + '0'); } int main() { n = read(), m = read(); for (int i = 1; i <= n; i++) a[read()]++; Poly::Getsginv(m << 2); for (int i = 1; i <= m; i++) for (int j = i, k = 1; j <= m; j += i, k++) (b[j] += (LL)a[i] * Poly::sginv[k] % P) %= P; memset(a, 0, sizeof a); Poly::Exppoly(b, m + 1, a); for (int i = 1; i <= m; i++) if (!a[i]) puts("0"); else write(a[i]), puts(""); }
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