hdu5730 Shell Necklace

列出dp方程$dp_n=sum_{i=0}^{n-1}dp_ia_{n-i}$，发现可以用分治FFT优化。

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double pi=acos(-1);
const int p=313,maxn=800010;
struct Complex
{
    double a,b;
    Complex operator + (const Complex &c) const
    {
        return (Complex){a+c.a,b+c.b};
    }
    Complex operator - (const Complex &c) const
    {
        return (Complex){a-c.a,b-c.b};
    }
    Complex operator * (const Complex &c) const
    {
        return (Complex){a*c.a-b*c.b,a*c.b+b*c.a};
    }
}f[maxn],g[maxn],w[maxn];
int a[maxn],dp[maxn],rev[maxn],n;
void fft(Complex *a,int m,int t,int flag)
{
    int x;
    Complex t1,t2;
    for (int i=0;i<m;i++)
        if (rev[i]>i) swap(a[i],a[rev[i]]);
    for (int i=0;i<t;i++)
        for (int j=0;j<m;j+=1<<(i+1))
        {
            x=0;
            for (int k=j;k<j+(1<<i);k++)
            {
                t1=a[k];
                t2=a[k+(1<<i)];
                a[k]=t1+t2*w[x];
                a[k+(1<<i)]=t1-t2*w[x];
                x+=flag*(1<<(t-i-1));
                if (x<0) x+=m;
            }
        }
}
void divcon(int l,int r)
{
    if (l==r) return;
    int mid=(l+r)/2,m=1,t=0;
    divcon(l,mid);
    for (int i=0;i<=mid-l;i++) f[i]=(Complex){(double)dp[l+i],0.0};
    for (int i=0;i<=r-l-1;i++) g[i]=(Complex){(double)a[i+1],0.0};
    while (m<=2*(r-l-1)) m<<=1,t++;
    for (int i=mid-l+1;i<m;i++) f[i]=(Complex){0.0,0.0};
    for (int i=r-l;i<m;i++) g[i]=(Complex){0.0,0.0};
    for (int i=0;i<m;i++)
    {
        rev[i]=0;
        for (int j=0;j<t;j++)
            rev[i]|=((i>>j)&1)<<(t-j-1);
    }
    for (int i=0;i<m;i++) w[i]=(Complex){cos(2*pi*i/m),sin(2*pi*i/m)};
    fft(f,m,t,1);
    fft(g,1);
    for (int i=0;i<m;i++) f[i]=f[i]*g[i];
    fft(f,-1);
    for (int i=mid+1;i<=r;i++)
        dp[i]=(dp[i]+int(f[i-l-1].a/m+0.5))%p;
    divcon(mid+1,r);
}
void solve()
{
    for (int i=1;i<=n;i++) scanf("%d",&a[i]),a[i]%=p,dp[i]=0;
    dp[0]=1;
    divcon(0,n);
    printf("%dn",dp[n]);
}
int main()
{
    //freopen("e.in","r",stdin);
    while (scanf("%d",&n)&&n) solve();
}