HDU 5730-shell Necklace

Test instructions

Give a continuous coloring method of 1-n beads A[i] (1<=i<=n), ask how many coloring schemes are in the bead chain of length n

Analysis:

The DP equation can be obtained: Dp[n] =∑ (i=1,n) (Dp[n-i]*a[i]).

The equation is convolution form, so CDQ + FFT

　　

CDQ: Will [L,r] two points, first get [L,mid] answer, and then update [L,MID] contribution to [mid+1,r].

For any dp[j] (mid+1 <= J <= R), [L,mid] contributes to ∑ (I=l,mid) (Dp[i]*a[j-i]), that is, the polynomial DP and a phase after the number of J.

FFT: Optimizes the multiplication of polynomial.

(1 and L can not see the same broken blog park, the code is still sticky good, = =)

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5 using namespacestd;6 Const DoublePI =4* ATAN (1.0);7 Const intMAXN =200005;8 Const intMOD =313;9 structComplexTen { One     Doublex, y; AComplex (Doublexx =0.0,Doubleyy =0.0): X (xx), Y (yy) {} -Complexoperator- (ConstComplex &b)Const  -     { the         returnComplex (x-b.x, Y-b.y); -     } -Complexoperator+ (ConstComplex &b)Const -     { +         returnComplex (x + b.x, y +b.y); -     } +Complexoperator* (ConstComplex &b)Const A     { at         returnComplex (x*b.x-y*b.y, X*b.y + y*b.x); -     } - }; - voidChange (Complex y[],intlen) - { -     intI, J, K; in      for(i =1, j = len/2; I < len-1; i++) -     { to         if(I <j) Swap (Y[i], y[j]); +K = len/2; -          while(J >=k) the         { *J-=K; $K/=2;Panax Notoginseng         } -         if(J < k) J + =K; the     } + } A voidFFT (Complex y[],intLenintOn ) the { + Change (y, len); -      for(inth =2; H <= Len; H <<=1) $     { $Complex wn (cos (-on*2*pi/h), sin (-on*2*pi/h)); -          for(intj =0; J < Len; J + =h) -         { theComplex W (1,0); -              for(intK = J; K < J + h/2; k++)Wuyi             { theComplex U =Y[k]; -Complex t = w * y[k + h/2]; WuY[k] = U +T; -Y[k + h/2] = U-T; Aboutw = w *WN;  $             } -         } -     }  -     if(On =-1) A          for(inti =0; i < Len; i++) +y[i].x/=Len; the } - intt, N; $ Complex X[MAXN], Y[MAXN]; the inta[maxn/2], dp[maxn/2]; the voidCDQ (intLintR) the { the     if(L = = r) {Dp[l] = (Dp[l] + a[l])% MOD;return; }  -     intMid = (L + r) >>1;  inCDQ (L, mid);//processing the first half of the paragraph the     intLen =1, len1 = Mid-l +1, Len2 = R-l +1; the      while(Len < len2) Len <<=1; About      for(inti =0; i < len1; i++) X[i] = Complex (Dp[i + l],0); the      for(inti = len1; i < Len; i++) X[i] = Complex (0,0); the      for(inti =0; i < len2; i++) Y[i] = Complex (A[i],0); the      for(inti = len2; i < Len; i++) Y[i] = Complex (0,0);  +FFT (x, Len,1); -FFT (y, Len,1); the      for(inti =0; i < Len; i++) X[i] = x[i] *Y[i];BayiFFT (x, Len,-1); the      for(inti = mid+1; I <= R; i++)//Update Contributions the     { -Dp[i] = (int) (Dp[i] + x[i-l].x +0.5) %MOD; -     } theCDQ (Mid +1, R);//processing the latter half of the paragraph the } the intMain () the { -      while(~SCANF ("%d", &n) &&N) the     { the          for(inti =1; I <= N; i++) the         {94scanf"%d",&a[i]); theA[i]%=MOD; theDp[i] =0; the         }98CDQ (1, n); Aboutprintf"%d\n", Dp[n]); -     }101}

HDU 5730-shell Necklace