HDU - 5730 Shell Necklace CDQ分治+fft

題目鏈接點這裏

可以看這位大爺的題解：http://blog.csdn.net/snowy_smile/article/details/52020971

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f3f
#define FIN freopen("input.txt","r",stdin)
#define mem(x,y) memset(x,y,sizeof(x))
typedef unsigned long long ULL;
typedef long long LL;
#define fuck(x) cout<<"q"<<endl;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef pair<pair<int,int>,int> PIII;
typedef pair<int,int> PII;
const double eps=1e-6;
const int MX=4e5+5;
const int P=313;
int n;
int a[MX];

const double pi = acos(-1.0);
int len,mx;//開大4倍
int res[MX];
struct Complex
{
    double r,i;
    Complex(double r=0,double i=0):r(r),i(i) {};
    Complex operator+(const Complex &rhs)
    {
        return Complex(r + rhs.r,i + rhs.i);
    }
    Complex operator-(const Complex &rhs)
    {
        return Complex(r - rhs.r,i - rhs.i);
    }
    Complex operator*(const Complex &rhs)
    {
        return Complex(r*rhs.r - i*rhs.i,i*rhs.r + r*rhs.i);
    }
} va[MX],vb[MX];
void rader(Complex F[],int len)   //len = 2^M,reverse F[i] with  F[j] j爲i二進制反轉
{
    int j = len >> 1;
    for(int i = 1; i < len - 1; ++i)
    {
        if(i < j) swap(F[i],F[j]);  // reverse
        int k = len>>1;
        while(j>=k)
        {
            j -= k;
            k >>= 1;
        }
        if(j < k) j += k;
    }
}
void FFT(Complex F[],int len,int t)
{
    rader(F,len);
    for(int h=2; h<=len; h<<=1)
    {
        Complex wn(cos(-t*2*pi/h),sin(-t*2*pi/h));
        for(int j=0; j<len; j+=h)
        {
            Complex E(1,0); //旋轉因子
            for(int k=j; k<j+h/2; ++k)
            {
                Complex u = F[k];
                Complex v = E*F[k+h/2];
                F[k] = u+v;
                F[k+h/2] = u-v;
                E=E*wn;
            }
        }
    }
    if(t==-1)   //IDFT
        for(int i=0; i<len; ++i)
            F[i].r/=len;
}
void Conv(Complex a[],Complex b[],int len)   //求卷積
{
    FFT(a,len,1);
    FFT(b,len,1);
    for(int i=0; i<len; ++i) a[i] = a[i]*b[i];
    FFT(a,len,-1);
}
void cdq(int l,int r)
{
    if(l==r)
    {
        res[l]=(res[l]+a[l])%P;
        return ;
    }
    int mid=(l+r)>>1;
    cdq(l,mid);
    //fft
    int len=1;
    while(len<mid-2*l+r-1)len<<=1;
    for(int i=0; i<len; i++)va[i].i=va[i].r=vb[i].i=vb[i].r=0;
    for(int i=0; i+l<=mid; i++)
    {
        va[i].r=res[i+l];
        va[i].i=0;
    }
    for(int i=0; i+1<=r-l; i++)
    {
        vb[i].r=a[i+1];
        vb[i].i=0;
    }
    Conv(va,vb,len);
    for(int i=mid+1; i<=r; i++)
        res[i]=(res[i]+(int)(va[i-1-l].r+0.5))%P;
    cdq(mid+1,r);
}
int main()
{
    FIN;
    while(cin>>n&&n)
    {
        for(int i=1; i<=n; i++)res[i]=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            a[i]%=P;
        }
        cdq(1,n);
        cout<<res[n]<<endl;
    }
    return 0;
}