HDU Shell Necklace CDQ Division +fft

Shell Necklace

problem DescriptionPerhaps the sea ' s definition of a shell is the pearl. However, in my view, a shell necklace with n beautiful shells contains the most sincere feeling for my best lover Arrietty , but even that's not enough.

Suppose the shell necklace is a sequence of shells (not a chain end to end). Considering I continuous shells in the shell necklace, I know that there exist different schemes to decorate the I shells Together with one declaration of love.

I want to decorate all the shells with some declarations of love and decorate each shell just one time. As a problem, I want to know the total number of schemes. InputThere is multiple test cases (no more than -Cases and no more than 1 in extreme case), ended by 0.

For each test cases, the first line contains an integerN, meaning the number of shells in this shell necklace, where1≤n≤5 . Following line was a sequence withNNon-negative integera1,< Span id= "mathjax-span-29" class= "Msubsup" >a2, ... , An , andai≤7 Meaning the number of schemes to decoratei continuous shells together with a declaration of love.
OutputFor each test case, print one line containing the total number of schemes module 313(three hundred and thirteen implies the March 13th, a special and purposeful day). Sample Input31 3 742 2 2 2 0 Sample Output1454 HintFor the first test case in Sample Input, the Figure 1 provides all schemes about it. The total number of schemes is 1 + 3 + 3 + 7 = 14.

Exercises

F[i] =∑f[n-i] * A[i]

Divide and conquer the FFT classics without repeating them

#include <bits/stdc++.h>using namespacestd;#pragmaComment (linker, "/stack:102400000,102400000")#defineLS i<<1#defineRS ls | 1#defineMid ((LL+RR) >>1)#definePII pair<int,int>#defineMP Make_pairtypedefLong LongLL;Const Long LongINF = 1e18+1LL;Const DoublePI = ACOs (-1.0);Const intN = 3e5+Ten, M = 1e3+ -, INF =2e9;ConstLL mod =313LL;structComplex {DoubleR, I; Complex () {} Complex (DoubleRDoublei): R (R), I (i) {} Complexoperator+ (Constcomplex& t)Const {        returnComplex (R + T.R, i +t.i); } Complexoperator- (Constcomplex& t)Const {        returnComplex (R-T.R, I-t.i); } Complexoperator* (Constcomplex& t)Const {        returnComplex (R * t.r-i * t.i, R * t.i + i *T.R); }} ;voidFFT (Complex y[],intNintrev) {     for(inti =1, J, T, K; I < n; ++i) { for(j =0, t = i, k = n >>1; K K >>=1, T >>=1) J = J <<1| T &1 ; if(I <j) Swap (Y[i], y[j]); }     for(ints =2, ds =1; s <= N; DS = s, S <<=1) {Complex WN= Complex (cos (rev *2* pi/s), sin (rev *2* pi/s)), W (1,0), t;  for(intK =0; K < DS; + + K, w = w *WN) {             for(inti = k; I < n; i + =s) {y[i+ ds] = Y[i]-(t = w * y[i +DS])                ; Y[i]= Y[i] +T; }        }    }    if(Rev = =-1) for(inti =0; I < n; + + i) Y[I].R/=N;} Complex Y[n],s[n]; LL Dp[n],a[n];intN;voidCdqintllintRR) {     if(LL = =RR) {Dp[ll]+=A[ll]; DP[LL]%=MoD; return ;    } CDQ (Ll,mid); intLen =1;  while(Len <= (rr-ll+2)) len<<=1;  for(inti =0; i < Len; ++i) Y[i] = Complex (0,0), s[i] =Y[i];  for(inti =0; i < Len; ++i) Y[i] = Complex (a[i+1],0);  for(inti = ll; I <= mid; ++i) S[i-ll] = Complex (Dp[i],0); FFT (S,len,1); FFT (Y,len,1);  for(inti =0; i < Len; ++i) S[i] = y[i] *S[i]; FFT (S,len,-1);  for(inti = mid; I < RR; ++i) dp[i+1] + = LL (s[i-ll].r+0.5), Dp[i]%=MoD; CDQ (Mid+1, RR);}intMain () { while(SCANF ("%d", &n)! =EOF) {        if(!n)return 0; Memset (A,0,sizeof(a));  for(inti =1; I <= N; ++i) scanf ("%lld", &a[i]), dp[i] =0, a[i] = a[i]%MoD; CDQ (1, N); printf ("%lld\n", Dp[n]); }    return 0;}

HDU Shell Necklace CDQ Division +fft