hdu 4656 Evaluation [任意模数fft trick]
2017-04-23 22:17
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hdu 4656 Evaluation
题意:给出\(n,b,c,d,f(x) = \sum_{i=1}^{n-1} a_ix^i\),求\(f(b\cdot c^{2k}+d):0\le k < n\)取模\(10^6+3\)
昨天刚看过《具体数学》上求和一章
代入\(b\cdot c^{2k}+d\)然后展开,交换求和顺序,得到
\[
f(k) = \sum_{j=0}^n \frac{b^j c^{2kj}}{j!} \sum_{i=j}^n a_i i! \frac{d^{i-j}}{(i-j)!}
\]
后面的式子反转后变成卷积,fft预处理\(p_j = \sum_{i=j}^n a_i i! \frac{d^{i-j}}{(i-j)!}\)
那么我们要求
\[
f(k) = \sum_{j=0}^n \frac{b^j c^{2kj} p_j}{j!}
\]
到这里就不会了,看了叉姐的课件,很神的一步来转化成卷积
\[
(k-j)^2 = k^2 + j^2 - 2kj
\]
代入之后得到
\[
f_k = c^{k^2}\sum_{j=0}^n \frac{b^j c^{j^2}p_j}{j!c^{(k-j)^2}}
\]
再次fft就行了
注意,指数带着平方可以为负,所以我们先乘上\(x^n\)
然后还要用拆系数fft.... 完虐ntt怒拿rank1
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <ctime> using namespace std; typedef long long ll; const int N = (1<<18) + 5, mo = 1e6+3, P = 1e6+3; const double PI = acos(-1.0); inline int read() { char c=getchar(); int x=0,f=1; while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();} return x*f; } struct meow{ double x, y; meow(double a=0, double b=0):x(a), y(b){} }; meow operator +(meow a, meow b) {return meow(a.x+b.x, a.y+b.y);} meow operator -(meow a, meow b) {return meow(a.x-b.x, a.y-b.y);} meow operator *(meow a, meow b) {return meow(a.x*b.x-a.y*b.y, a.x*b.y+a.y*b.x);} meow conj(meow a) {return meow(a.x, -a.y);} typedef meow cd; namespace fft { int maxlen, rev ; cd omega , omegaInv ; void init(int lim) { maxlen = 1; while(maxlen < lim) maxlen<<=1; for(int i=0; i<maxlen; i++) { omega[i] = cd(cos(2*PI/maxlen*i), sin(2*PI/maxlen*i)); omegaInv[i] = conj(omega[i]); } } void dft(cd *a, int n, int flag) { cd *w = flag == 1 ? omega : omegaInv; int k = 0; while((1<<k) < n) k++; for(int i=0; i<n; i++) { rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1)); if(i < rev[i]) swap(a[i], a[rev[i]]); } for(int l=2; l<=n; l<<=1) { int m = l>>1; for(cd *p = a; p != a+n; p += l) for(int k=0; k<m; k++) { cd t = w[maxlen/l*k] * p[k+m]; p[k+m] = p[k] - t; p[k] = p[k] + t; } } if(flag == -1) for(int i=0; i<n; i++) a[i].x /= n; } cd a , b , c , d ; void mul_any(int *x, int *y, int lim) { int n = maxlen; for(int i=0; i<lim; i++) { a[i] = cd(x[i]>>15), b[i] = cd(x[i]&32767); c[i] = cd(y[i]>>15), d[i] = cd(y[i]&32767); } for(int i=lim; i<n; i++) a[i] = b[i] = c[i] = d[i] = cd(); dft(a, n, 1); dft(b, n, 1); dft(c, n, 1); dft(d, n, 1); for(int i=0; i<n; i++) { cd _a = a[i], _b = b[i], _c = c[i], _d = d[i]; a[i] = _a * _c; b[i] = _a * _d + _b * _c; c[i] = _b * _d; } dft(a, n, -1); dft(b, n, -1); dft(c, n, -1); for(int i=0; i<lim; i++) x[i] = ( (ll(a[i].x + 0.5) %mo <<30) + (ll(b[i].x + 0.5) %mo <<15) + ll(c[i].x + 0.5)%mo) %mo; } } inline ll Pow(ll a, int b) { ll ans = 1; for(; b; b>>=1, a=a*a%P) if(b&1) ans=ans*a%P; return ans; } int n, a , b, c, d, p , g , f , c2 ; ll inv , fac , facInv ; int main() { freopen("in", "r", stdin); n=read(); b=read(); c=read(); d=read(); for(int i=0; i<n; i++) a[i] = read(); fft::init(n+n+1); inv[1] = fac[0] = facInv[0] = 1; for(int i=1; i<=n; i++) { if(i != 1) inv[i] = (P-P/i) * inv[P%i] %P; fac[i] = fac[i-1] * i %P; facInv[i] = facInv[i-1] * inv[i] %P; } ll mi = 1; for(int i=0; i<=n; i++, mi = mi * d %P) p[i] = a[n-i] * fac[n-i] %P, g[i] = mi * facInv[i] %P; fft::mul_any(p, g, n+1); for(int i=0; i<=n>>1; i++) swap(p[i], p[n-i]); mi = 1; memset(g, 0, sizeof(g)); for(int i=0; i<=n; i++, mi = mi * b %P) { int t = Pow(c, (ll) i * i %(P-1)); f[i] = mi * t %P * p[i] %P * facInv[i] %P; if(i != n) g[i + n] = g[n - i] = Pow(t, P-2); //if(i != n) g[i + n-1] = g[n-1 - i] = Pow(t, P-2); c2[i] = t; } fft::mul_any(f, g, n+n+1); for(int i=0; i<n; i++) f[i+n] = (ll) f[i+n] * c2[i] %P, printf("%d\n", f[i+n]); //for(int i=0; i<n; i++) f[i+n-1] = (ll) f[i+n-1] * c2[i] %P, printf("%d\n", f[i+n-1]); }
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