做网站很烧钱,爱南宁app下载官网最新,多平台网站建设,什么可以用手机做网站P4389 付公主的背包
考虑生成函数有#xff1a; ∏i1n11−xvi对其取对数得#xff0c;∑i1nln11−xviF(x)11−xv,G(x)lnF(x)G(x)∫F′(x)F(x)dxG(x)∫vxv−11−xvdxG(x)∫∑n≥0vxvnv−1dxG(x)∑n≥0vxvnvvnvG(x)∑n≥0xv(n1)n1G(x)∑n≥1xvnn对于原式:∑i1n∑j1∞xvijj…P4389 付公主的背包
考虑生成函数有 ∏i1n11−xvi对其取对数得∑i1nln11−xviF(x)11−xv,G(x)lnF(x)G(x)∫F′(x)F(x)dxG(x)∫vxv−11−xvdxG(x)∫∑n≥0vxvnv−1dxG(x)∑n≥0vxvnvvnvG(x)∑n≥0xv(n1)n1G(x)∑n≥1xvnn对于原式:∑i1n∑j1∞xvijj\prod_{i 1} ^{n} \frac{1}{1 - x ^{v_i}}\\ 对其取对数得\sum_{i 1} ^{n} \ln \frac{1}{1 - x ^{v_i}}\\ F(x) \frac{1}{1 - x ^ v}, G(x) \ln F(x)\\ G(x) \int \frac{F(x)}{F(x)} dx\\ G(x) \int \frac{v x ^{v - 1}}{1 - x ^ v} dx\\ G(x) \int \sum_{n \geq 0} vx ^{vn v - 1} dx\\ G(x) \sum_{n \geq 0} \frac{v x ^{vn v}}{vn v}\\ G(x) \sum_{n \geq 0} \frac{x ^{v(n 1)}}{n 1}\\ G(x) \sum_{n \geq 1} \frac{x ^{vn}}{n}\\ 对于原式:\sum_{i 1} ^{n} \sum_{j 1} ^{\infty} \frac{x ^{v_i j}}{j}\\ i1∏n1−xvi1对其取对数得i1∑nln1−xvi1F(x)1−xv1,G(x)lnF(x)G(x)∫F(x)F′(x)dxG(x)∫1−xvvxv−1dxG(x)∫n≥0∑vxvnv−1dxG(x)n≥0∑vnvvxvnvG(x)n≥0∑n1xv(n1)G(x)n≥1∑nxvn对于原式:i1∑nj1∑∞jxvij 提前预处理出invinvinv然后O(nlogn)O(n \log n)O(nlogn)对vijv_i jvij项加上inv[j]inv[j]inv[j]再做一次(modxm1)\pmod{x ^{m 1}}(modxm1)的多项式expexpexp即可。
#include bits/stdc.husing namespace std;const int mod 998244353, inv2 mod 1 1;namespace Quadratic_residue {struct Complex {int r, i;Complex(int _r 0, int _i 0) : r(_r), i(_i) {}};int I2;Complex operator * (const Complex a, Complex b) {return Complex((1ll * a.r * b.r % mod 1ll * a.i * b.i % mod * I2 % mod) % mod, (1ll * a.r * b.i % mod 1ll * a.i * b.r % mod) % mod);}Complex quick_pow(Complex a, int n) {Complex ans Complex(1, 0);while (n) {if (n 1) {ans ans * a;}a a * a;n 1;}return ans;}int get_residue(int n) {mt19937 e(233);if (n 0) {return 0;}if(quick_pow(n, (mod - 1) 1).r mod - 1) {return -1;}uniform_int_distributionint r(0, mod - 1);int a r(e);while(quick_pow((1ll * a * a % mod - n mod) % mod, (mod - 1) 1).r 1) {a r(e);}I2 (1ll * a * a % mod - n mod) % mod;int x quick_pow(Complex(a, 1), (mod 1) 1).r, y mod - x;if(x y) swap(x, y);return x;}
}const int N 1e6 10;int r[N], inv[N], a[N], b[N], c[N], d[N], e[N], t[N], n, m;//a是输入数组b存放多项式逆c存放多项式开根d存放多项式对数lne存放多项式指数expt作为中间转移数组int quick_pow(int a, int n) {int ans 1;while (n) {if (n 1) {ans 1ll * a * ans % mod;}a 1ll * a * a % mod;n 1;}return ans;
}void get_r(int lim) {for (int i 0; i lim; i) {r[i] (i 1) * (lim 1) (r[i 1] 1);}
}void get_inv(int n) {inv[1] 1;for (int i 2; i n; i) {inv[i] 1ll * (mod - mod / i) * inv[mod % i] % mod;}
}void NTT(int *f, int lim, int rev) {for (int i 0; i lim; i) {if (i r[i]) {swap(f[i], f[r[i]]);}}for (int mid 1; mid lim; mid 1) {int wn quick_pow(3, (mod - 1) / (mid 1));for (int len mid 1, cur 0; cur lim; cur len) {int w 1;for (int k 0; k mid; k, w 1ll * w * wn % mod) {int x f[cur k], y 1ll * w * f[cur mid k] % mod;f[cur k] (x y) % mod, f[cur mid k] (x - y mod) % mod;}}}if (rev -1) {int inv quick_pow(lim, mod - 2);reverse(f 1, f lim);for (int i 0; i lim; i) {f[i] 1ll * f[i] * inv % mod;}}
}void polyinv(int *f, int *g, int n) {/* 保证数组清零: 用了数组(a, b, t), a数组不变, b数组只有前n个不为零后面全为零, t数组用完后清零了。*/if (n 1) {g[0] quick_pow(f[0], mod - 2);return ;}polyinv(f, g, n 1 1);for (int i 0; i n; i) {t[i] f[i];}int lim 1;while (lim 2 * n) {lim 1;}get_r(lim);NTT(t, lim, 1);NTT(g, lim, 1);for (int i 0; i lim; i) {int cur (2 - 1ll * g[i] * t[i] % mod mod) % mod;g[i] 1ll * g[i] * cur % mod;t[i] 0;}NTT(g, lim, -1);for (int i n; i lim; i) {g[i] 0;}
}void polysqrt(int *f, int *g, int n) {if (n 1) {g[0] Quadratic_residue::get_residue(f[0]);return ;}polysqrt(f, g, n 1 1);polyinv(g, b, n);int lim 1;while (lim 2 * n) {lim 1;}get_r(lim);for (int i 0; i n; i) {t[i] f[i];}NTT(g, lim, 1);NTT(b, lim, 1);NTT(t, lim, 1);for (int i 0; i lim; i) {g[i] (1ll * inv2 * g[i] % mod 1ll * inv2 * b[i] % mod * t[i] % mod) % mod;b[i] t[i] 0;}NTT(g, lim, -1);for (int i n; i lim; i) {g[i] 0;}
}void derivative(int *a, int *b, int n) {for (int i 0; i n; i) {b[i] 1ll * a[i 1] * (i 1) % mod;}
}void integrate(int *a, int n) {for (int i n - 1; i 1; i--) {a[i] 1ll * a[i - 1] * inv[i] % mod;}a[0] 0;
}void polyln(int *f, int *g, int n) {polyinv(f, b, n);derivative(f, g, n);int lim 1;while (lim 2 * n) {lim 1;}get_r(lim);NTT(g, lim, 1);NTT(b, lim, 1);for (int i 0; i lim; i) {g[i] 1ll * g[i] * b[i] % mod;b[i] 0;}NTT(g, lim, -1);for (int i n; i lim; i) {g[i] 0;}integrate(g, n);
}void polyexp(int *f, int *g, int n) {if (n 1) {g[0] 1;return ;}polyexp(f, g, n 1 1);int lim 1;while (lim 2 * n) {lim 1;}polyln(g, d, n);for (int i 0; i n; i) {t[i] (f[i] - d[i] mod) % mod;}t[0] (t[0] 1) % mod;get_r(lim);NTT(g, lim, 1);NTT(t, lim, 1);for (int i 0; i lim; i) {g[i] 1ll * g[i] * t[i] % mod;t[i] d[i] 0;}NTT(g, lim, -1);for (int i n; i lim; i) {g[i] 0;}
}int vis[N];int main() {// freopen(in.txt, r, stdin);// freopen(out.txt, w, stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf(%d %d, n, m);get_inv(4 * m);for (int i 1, v; i n; i) {scanf(%d, v);vis[v];}for (int i 1; i m; i) {for (int j i; j m; j i) {a[j] (a[j] 1ll * inv[i] * vis[j / i] % mod) % mod;}}polyexp(a, e, m 1);for (int i 1; i m; i) {printf(%d\n, e[i]);}return 0;
}
