hdu Shell Necklace 5730 分治FFT

Description

Perhaps 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 is 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.

Input

There are multiple test cases(no more than 20 cases and no more than 1 in extreme case), ended by 0.

For each test cases, the first line contains an integer n, meaning the number of shells in this shell necklace, where 1≤n≤10⁵. Following line is a sequence with n non-negative integer a1,a2,…,an, and ai≤10⁷ meaning the number of schemes to decorate i continuous shells together with a declaration of love.

Output

For 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 Input

3
1 3 7
4
2 2 2 2 
0

Sample Output

14
54

题意：就是给定长度为n，然后长度为1-n的珠子有a[i]种，问拼成长度n的方案数，每种珠子有无限个。

题解：f[i]设为拼成i的方案数，发现转移时f[i]=∑f[j]*a[i-j]+a[i] (1<=j<i)发现是个卷积的形式。

　　　如何去优化。

　　　去分治解决，设 l,mid的已经处理好了，那么，考虑l,mid对mid+1,r的贡献，

　　　对于l,mid的贡献部分做卷积运算，得出其贡献，加到mid,r中。

　　　这个过程就是CDQ的过程。

 #include<cstring>
 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 
 #define N 200007
 #define pi acos(-1)
 #define mod 313
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(!isdigit(ch)){if(ch=='-')f=;ch=getchar();}
     while(isdigit(ch)){x=(x<<)+(x<<)+ch-'';ch=getchar();}
     return x*f;
 }
 
 int n,num,L;
 int a[N],rev[N],f[N];
 struct comp
 {
     double r,v;
     comp(double _r=,double _v=){r=_r;v=_v;}
     friend inline comp operator+(comp x,comp y){return comp(x.r+y.r,x.v+y.v);}
     friend inline comp operator-(comp x,comp y){return comp(x.r-y.r,x.v-y.v);}
     friend inline comp operator*(comp x,comp y){return comp(x.r*y.r-x.v*y.v,x.r*y.v+x.v*y.r);}
 }x[N],y[N];
 
 void FFT(comp *a,int f)
 {
     for (int i=;i<num;i++)
         if (i<rev[i]) swap(a[i],a[rev[i]]);
     for (int i=;i<num;i<<=)
     {
         comp wn=comp(cos(pi/i),f*sin(pi/i));
         for (int j=;j<num;j+=(i<<))
         {
             comp w=comp(,);
             for (int k=;k<i;k++,w=w*wn)
             {
                 comp x=a[j+k],y=w*a[j+k+i];
                 a[j+k]=x+y,a[j+k+i]=x-y;
             }
         }
     }
     if (f==-) for (int i=;i<num;i++) a[i].r/=num;
 }
 void CDQ(int l,int r)
 {
     if (l==r) {f[l]=(f[l]+a[l])%mod;return;}
     int mid=(l+r)>>;
     CDQ(l,mid);
     L=;for (num=;num<=(r-l+);num<<=,L++);if (L) L--;
     for (int i=;i<num;i++) rev[i]=(rev[i>>]>>)|((i&)<<L);
     for (int i=;i<num;i++) x[i]=y[i]=comp(,);
     for (int i=l;i<=mid;i++) x[i-l]=comp(f[i],);
     for (int i=;i<=r-l;i++) y[i-]=comp(a[i],);
     FFT(x,);FFT(y,);
     for (int i=;i<num;i++) x[i]=x[i]*y[i];
     FFT(x,-);
     for (int i=mid+;i<=r;i++)
         (f[i]+=x[i-l-].r+0.5)%=mod;
     CDQ(mid+,r);
 }
 int main()
 {
     while(~scanf("%d",&n)&&n)
     {
         for (int i=;i<=n;i++) a[i]=read()%mod,f[i]=;
         CDQ(,n);
         printf("%d\n",f[n]);
     }
 }