hdu5730 Shell Necklace

列出dp方程dpn=∑n−1i=0dpian−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,m,t,1);    for (int i=0;i<m;i++) f[i]=f[i]*g[i];    fft(f,m,t,-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("%d\n",dp[n]);}int main(){    //freopen("e.in","r",stdin);    while (scanf("%d",&n)&&n) solve();}