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=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=1;ch=getchar();}
    while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}

int n,num,L;
int a[N],rev[N],f[N];
struct comp
{
    double r,v;
    comp(double _r=0,double _v=0){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=1;i<num;i++)
        if (i<rev[i]) swap(a[i],a[rev[i]]);
    for (int i=1;i<num;i<<=1)
    {
        comp wn=comp(cos(pi/i),f*sin(pi/i));
        for (int j=0;j<num;j+=(i<<1))
        {
            comp w=comp(1,0);
            for (int k=0;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==-1) for (int i=0;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)>>1;
    CDQ(l,mid);
    L=0;for (num=1;num<=(r-l+1);num<<=1,L++);if (L) L--;
    for (int i=0;i<num;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<L);
    for (int i=1;i<num;i++) x[i]=y[i]=comp(0,0);
    for (int i=l;i<=mid;i++) x[i-l]=comp(f[i],0);
    for (int i=1;i<=r-l;i++) y[i-1]=comp(a[i],0);
    FFT(x,1);FFT(y,1);
    for (int i=0;i<num;i++) x[i]=x[i]*y[i];
    FFT(x,-1);
    for (int i=mid+1;i<=r;i++)
        (f[i]+=x[i-l-1].r+0.5)%=mod;
    CDQ(mid+1,r);
}
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        for (int i=1;i<=n;i++) a[i]=read()%mod,f[i]=0;
        CDQ(1,n);
        printf("%d
",f[n]);
    }
}