我是靠谱客的博主 勤劳胡萝卜,这篇文章主要介绍【学习总结】生成函数题目,多项式模板,现在分享给大家,希望可以做个参考。

学习资料:
王乐平、策爷冬令营讲义。

多项式技巧

牛顿迭代(泰勒展开)

很多时候推式子就是取对数、积分,exp和泰勒展开的结合
泰勒展开在mod xn意义下只需要保留前n项,性质非常优美。
而插值,对于一个k次的多项式,必须要k + 1个值的代入,如果只求该多项式的前n项,也不能只用
n + 1个点代入

这是一篇非常好的博客。特别是把所有多项式操作都用泰勒展开推导,以后就不用再死记,也不用担心推错了!from yyb

复合逆

拉格朗日反演
在这里插入图片描述

注意当f(x)本身不存在逆元的时候,可以求f(x) / x的逆元
否则(x / f(x))n 的第n - 1
项为0

例题

直接推生成函数

【BZOJ3625】小朋友和二叉树

from cz_xuyixuan
注意这道题是普通生成函数,不是指数型生成函数

代码只有main函数,多项式模板在下面

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int n,m; vector <int> fact,inv_fact; vector <int> f,g,h,tmp; void pre_calc(){ // g[0] = 1; rep(i,0,m){ // g[i] = mul(g[i],inv_fact[i]); sub(f[i],mul(g[i],4)); // cout<<f[i]<<" "; } f[0] = 1; // cout<<endl; f = sqrt(f); add(f[0],1); f = inverse(f); rep(i,0,m) f[i] = mul(f[i],2); } int main(){ scanf("%d %d",&n,&m); f.resize(m + 1) , g.resize(m + 1); rep(i,1,n){ int x; scanf("%d",&x); if ( x <= m ) g[x]++; } pre_calc(); rep(i,1,m){ // f[i] = mul(f[i],fact[i]); printf("%dn",f[i]); } }

bzoj 3684 大朋友和多叉树

复合逆裸题。
注意F(x) / x才可以求逆

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int n,m; poly fact,inv_fact; poly f,g,h,tmp; void pre_calc(){ add(g[0],1); g = inverse(g); g = power(g,n); int ans = mul(power(n,mod - 2),g[n - 1]); cout<<ans<<endl; } int main(){ scanf("%d %d",&n,&m); f.resize(n + 1) , g.resize(n + 1); rep(i,1,m){ int x; scanf("%d",&x); g[x - 1] = mod - 1; } pre_calc(); }

多项式模板

注意事项:
求导和积分多项式的次数变化
开根号如果常数项不为完全平方数,需要用二次剩余开根号的模板。详见或这位大佬
求逆常数项不能为0 , 否则不存在逆元
special thanks to wxh010910

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#include<bits/stdc++.h> using namespace std; #define rep(i,l,r) for(register int i = l ; i <= r ; i++) #define repd(i,r,l) for(register int i = r ; i >= l ; i--) #define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++) #define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--) #define fore(i,x)for (register int i = head[x] ; i ; i = e[i].next) #define forup(i,l,r) for (register int i = l ; i <= r ; i += lowbit(i)) #define fordown(i,id) for (register int i = id ; i ; i -= lowbit(i)) #define pb push_back #define prev prev_ #define stack stack_ #define mp make_pair #define fi first #define se second #define lowbit(x) ((x)&(-(x))) typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair<int,int> pr; const int maxn = 200020; typedef vector <int> poly; const int mod = 950009857; //NOTES: 任意乘法需要用mul,或者强制用long long。 //注意取模 //====================================basic operation=============================== inline void add(int &x, int y) { x += y; if (x >= mod) { x -= mod; } } inline void sub(int &x, int y) { x -= y; if (x < 0) { x += mod; } } inline int mul(int x, int y) { return (int) ((long long) x * y % mod); } inline int power(int x, int y) { int res = 1; while (y) { if (y & 1) { res = mul(res, x); } x = mul(x, x); y >>= 1; } return res; } inline int inv(int a) { int b = mod, u = 0, v = 1; while (a) { int t = b / a; b -= t * a; swap(a, b); u -= t * v; swap(u, v); } if (u < 0) { u += mod; } return u; } //======================================================================================= namespace ntt { int base = 1, root = -1, max_base = -1; poly rev = {0, 1}, roots = {0, 1}; void init() { int temp = mod - 1; max_base = 0; while (temp % 2 == 0) { temp >>= 1; ++max_base; } root = 2; while (true) { if (power(root, 1 << max_base) == 1 && power(root, 1 << (max_base - 1)) != 1) { break; } ++root; } } void ensure_base(int nbase) { //所有dft需要的预处理 if (max_base == -1) { init(); } if (nbase <= base) { return; } assert(nbase <= max_base); rev.resize(1 << nbase); for (int i = 0; i < 1 << nbase; ++i) { //预处理翻转位 rev[i] = rev[i >> 1] >> 1 | (i & 1) << (nbase - 1); } roots.resize(1 << nbase); while (base < nbase) { //预处理单位根 int z = power(root, 1 << (max_base - 1 - base)); for (int i = 1 << (base - 1); i < 1 << base; ++i) { roots[i << 1] = roots[i]; roots[i << 1 | 1] = mul(roots[i], z); } ++base; } } void dft(poly &a) { int n = a.size(), zeros = __builtin_ctz(n); ensure_base(zeros); int shift = base - zeros; for (int i = 0; i < n; ++i) { if (i < rev[i] >> shift) { swap(a[i], a[rev[i] >> shift]); } } for (int i = 1; i < n; i <<= 1) { for (int j = 0; j < n; j += i << 1) { for (int k = 0; k < i; ++k) { int x = a[j + k], y = mul(a[j + k + i], roots[i + k]); a[j + k] = (x + y) % mod; a[j + k + i] = (x + mod - y) % mod; } } } } poly multiply(poly a, poly b) { int need = a.size() + b.size() - 1, nbase = 0; while (1 << nbase < need) { ++nbase; } ensure_base(nbase); int sz = 1 << nbase; a.resize(sz); b.resize(sz); bool equal = a == b; dft(a); if (equal) { b = a; } else { dft(b); } int inv_sz = inv(sz); for (int i = 0; i < sz; ++i) { a[i] = mul(mul(a[i], b[i]), inv_sz); } reverse(a.begin() + 1, a.end()); //相当于NTT(a,-1) dft(a); a.resize(need); return a; } poly inverse(poly a) { //常数项不能为0,否则不存在逆元! int n = a.size(), m = (n + 1) >> 1; if (n == 1) { return poly(1, inv(a[0])); } else { poly b = inverse(poly(a.begin(), a.begin() + m)); int need = n << 1, nbase = 0; while (1 << nbase < need) { ++nbase; } ensure_base(nbase); int sz = 1 << nbase; a.resize(sz); b.resize(sz); dft(a); dft(b); int inv_sz = inv(sz); for (int i = 0; i < sz; ++i) { a[i] = mul(mul(mod + 2 - mul(a[i], b[i]), b[i]), inv_sz); } reverse(a.begin() + 1, a.end()); dft(a); a.resize(n); return a; } } } using ntt::multiply; using ntt::inverse; poly& operator += (poly &a, const poly &b) { if (a.size() < b.size()) { a.resize(b.size()); } for (int i = 0; i < (int) b.size(); ++i) { add(a[i], b[i]); } return a; } poly operator + (const poly &a, const poly &b) { poly c = a; return c += b; } poly& operator -= (poly &a, const poly &b) { if (a.size() < b.size()) { a.resize(b.size()); } for (int i = 0; i < (int) b.size(); ++i) { sub(a[i], b[i]); } return a; } poly operator - (const poly &a, const poly &b) { poly c = a; return c -= b; } poly& operator *= (poly &a, const poly &b) { if ((int) min(a.size(), b.size()) < 128) { poly c = a; a.assign(a.size() + b.size() - 1, 0); for (int i = 0; i < (int) c.size(); ++i) { for (int j = 0; j < (int) b.size(); ++j) { add(a[i + j], mul(c[i], b[j])); } } } else { a = multiply(a, b); } return a; } poly operator * (const poly &a, const poly &b) { poly c = a; return c *= b; } poly& operator /= (poly &a, const poly &b) { int n = a.size(), m = b.size(); if (n < m) { a.clear(); } else { poly c = b; reverse(a.begin(), a.end()); reverse(c.begin(), c.end()); c.resize(n - m + 1); a *= inverse(c); a.erase(a.begin() + n - m + 1, a.end()); reverse(a.begin(), a.end()); } return a; } poly operator / (const poly &a, const poly &b) { poly c = a; return c /= b; } poly& operator %= (poly &a, const poly &b) { int n = a.size(), m = b.size(); if (n >= m) { poly c = (a / b) * b; a.resize(m - 1); for (int i = 0; i < m - 1; ++i) { sub(a[i], c[i]); } } return a; } poly operator % (const poly &a, const poly &b) { poly c = a; return c %= b; } poly derivative(const poly &a) { int n = a.size(); poly b(n - 1); for (int i = 1; i < n; ++i) { b[i - 1] = mul(a[i], i); } return b; } poly primitive(const poly &a) { int n = a.size(); poly b(n + 1), invs(n + 1); for (int i = 1; i <= n; ++i) { invs[i] = i == 1 ? 1 : mul(mod - mod / i, invs[mod % i]); b[i] = mul(a[i - 1], invs[i]); } return b; } poly logarithm(const poly &a) { poly b = primitive(derivative(a) * inverse(a)); b.resize(a.size()); return b; } poly exponent(const poly &a) { poly b(1, 1); while (b.size() < a.size()) { poly c(a.begin(), a.begin() + min(a.size(), b.size() << 1)); add(c[0], 1); poly old_b = b; b.resize(b.size() << 1); c -= logarithm(b); c *= old_b; for (int i = b.size() >> 1; i < (int) b.size(); ++i) { b[i] = c[i]; } } b.resize(a.size()); return b; } poly power(const poly &a, int m) { //高端的power写法 int n = a.size(), p = -1; poly b(n); for (int i = 0; i < n; ++i) { if (a[i]) { p = i; break; } } if (p == -1) { b[0] = !m; return b; } if ((long long) m * p >= n) { return b; } int mu = power(a[p], m), di = inv(a[p]); poly c(n - m * p); for (int i = 0; i < n - m * p; ++i) { c[i] = mul(a[i + p], di); } c = logarithm(c); for (int i = 0; i < n - m * p; ++i) { c[i] = mul(c[i], m); } c = exponent(c); for (int i = 0; i < n - m * p; ++i) { b[i + m * p] = mul(c[i], mu); } return b; } poly sqrt(const poly &a) { poly b(1,(int)sqrt(a[0])); //常数项是完全平方数,如果不是,则需要BSGS开根号 while (b.size() < a.size()) { poly c(a.begin(), a.begin() + min(a.size(), b.size() << 1)); poly old_b = b; b.resize(b.size() << 1); c *= inverse(b); for (int i = b.size() >> 1; i < (int) b.size(); ++i) { b[i] = mul(c[i], (mod + 1) >> 1); } } b.resize(a.size()); return b; } poly multiply_all(int l, int r, vector<poly > &all) { if (l > r) { return poly(); } else if (l == r) { return all[l]; } else { int y = (l + r) >> 1; return multiply_all(l, y, all) * multiply_all(y + 1, r, all); } } poly evaluate(const poly &f, const poly &x) { int n = x.size(); if (!n) { return poly(); } vector<poly> up(n * 2); for (int i = 0; i < n; ++i) { up[i + n] = poly{(mod - x[i]) % mod, 1}; } for (int i = n - 1; i; --i) { up[i] = up[i << 1] * up[i << 1 | 1]; } vector<poly> down(n * 2); down[1] = f % up[1]; for (int i = 2; i < n * 2; ++i) { down[i] = down[i >> 1] % up[i]; } poly y(n); for (int i = 0; i < n; ++i) { y[i] = down[i + n][0]; } return y; } poly interpolate(const poly &x, const poly &y) { int n = x.size(); vector<poly> up(n * 2); for (int i = 0; i < n; ++i) { up[i + n] = poly{(mod - x[i]) % mod, 1}; } for (int i = n - 1; i; --i) { up[i] = up[i << 1] * up[i << 1 | 1]; } poly a = evaluate(derivative(up[1]), x); for (int i = 0; i < n; ++i) { a[i] = mul(y[i], inv(a[i])); } vector<poly> down(n * 2); for (int i = 0; i < n; ++i) { down[i + n] = poly(1, a[i]); } for (int i = n - 1; i; --i) { down[i] = down[i << 1] * up[i << 1 | 1] + down[i << 1 | 1] * up[i << 1]; } return down[1]; }

cogs 2189

只有main函数的部分

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int n,k; poly f; int main(){ freopen("polynomial.in","r",stdin); freopen("polynomial.out","w",stdout); // freopen("input.txt","r",stdin); scanf("%d %d",&n,&k); f.resize(n); rep(i,0,n - 1) scanf("%d",&f[i]); f = sqrt(f); f = inverse(f); f = primitive(f); f = exponent(f); f = inverse(f); add(f[0],1); f = logarithm(f); add(f[0],1); f = power(f,k); f = derivative(f); rep(i,0,n - 2) printf("%d ",f[i]); puts("0"); }

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