HDU5730 Shell Necklace

Time limit:16000/8000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 999 Accepted Submission (s): 434

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.

Authorhit

Source2016 multi-university Training Contest 1

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A long x necklace as a whole, with a[x] kind of decoration scheme. It is possible to connect the different whole together. Ask how many options there are for a necklace of length n.

Dynamic programming for the division and treatment of FFT

Set F[i] As the number of schemes long as I, obviously $ f[i]=\sum_{j=1}^{i} A[j]*f[i-j] $

Modulus is 313, what is the original root of this number? I don't know. The modulus is so small that it can be used with an FFT.

PS1 attention to read into the a[] when it is easy to take the mold, or it will easily explode

PS2 I don't know why I want to output a newline character, White WA took three times to see

1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <cstring>5#include <cmath>6 #defineLL Long Long7 using namespacestd;8 Const DoublePi=acos (-1.0);9 Const intMod=313;Ten Const intmxn=200010; One intRead () { A     intx=0, f=1;CharCh=GetChar (); -      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} -      while(ch>='0'&& ch<='9') {x=x*Ten-'0'+ch;ch=GetChar ();} the     returnx*F; - } - structcom{ -     Doublex, y; +Comoperator+ (Constcom &b) {return(COM) {x+b.x,y+b.y};} -Comoperator- (Constcom &b) {return(COM) {x-b.x,y-b.y};} +Comoperator* (Constcom &b) {return(COM) {x*b.x-y*b.y,x*b.y+y*b.x};} AComoperator/ (Const DoubleV) {return(COM) {x/v,y/v};} at}a[mxn<<2],b[mxn<<2]; - intn,len,rev[mxn<<2]; - voidFFT (COM *a,intflag) { -      for(intI=0; i<n;i++)if(i<Rev[i]) swap (a[i],a[rev[i]); -      for(intI=1; i<n;i<<=1){ -COM wn= (COM) {cos (pi/i), Flag*sin (pi/i)}; in         intp=i<<1; -          for(intj=0; j<n;j+=p) { toCOM w= (COM) {1,0}; +              for(intk=0; k<i;k++,w=w*WN) { -COM x=a[j+k],y=w*a[j+k+i]; thea[j+k]=x+y; *a[j+k+i]=x-y; $             }Panax Notoginseng         } -     } the     if(flag==-1) +          for(intI=0; i<n;i++) a[i].x/=N; A     return; the } + intN,W[MXN]; - LL F[MXN]; $ voidSolveintLintR) { $     if(L==R) {(f[l]+=w[l])%=mod;return;} -     intMid= (l+r) >>1; - solve (l,mid); the     intI,j,m= (r-l+1); -      for(n=1, len=0; n<=m; n<<=1)++Len;Wuyi      for(i=0; i<n;i++) rev[i]= (rev[i>>1]>>1)| ((i&1) << (len-1)); the     // -      for(i=l;i<=mid;i++) {a[i-l]= (COM) {f[i],0};} Wu      for(i=mid-l+1; i<n;i++) a[i]= (COM) {0,0}; -      for(i=0; i<n;i++) b[i]= (COM) {w[i+1],0}; About     // $FFT (A,1); FFT (b,1); -      for(i=0; i<n;i++) a[i]=a[i]*B[i]; -FFT (a,-1); -      for(i=mid+1; i<=r;i++){ A(f[i]+= (LL) (a[i-l-1].x+0.5))%mod)%=MoD; +     } theSolve (mid+1, R); -     return; $ } the intMain () { the     inti,j; the      while(SCANF ("%d", &n)!=eof &&N) { theMemset (F,0,sizeoff); -          for(i=1; i<=n;i++) W[i]=read ()%mod;// inSolve1, n); theprintf"%lld\n", f[n]%MoD); the     } About     return 0; the}

HDU5730 Shell Necklace