BZOJ 3509 分块FFT

思路：

跟今年WC的题几乎一样 （但是这道题有重 不能用bitset水过去）

正解：分块FFT

http://blog.csdn.net/geotcbrl/article/details/50636401    from GEOTCBRL

可以看看hgr的题解..写得很详细

//By SiriusRen
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const double pi=acos(-);
const int N=;
int n,nn,num[N],R[N],L,Block,block[N],cnt[][N];
long long ans;
struct Complex{
    double x,y;Complex(){}
    Complex(double X,double Y){x=X,y=Y;}
}A[N],B[N],C[N];
Complex operator+(Complex a,Complex b){return Complex(a.x+b.x,a.y+b.y);}
Complex operator-(Complex a,Complex b){return Complex(a.x-b.x,a.y-b.y);}
Complex operator*(Complex a,Complex b){return Complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
Complex operator/(Complex a,int b){return Complex(a.x/b,a.y/b);}
void FFT(Complex *a,int f){
    for(int i=;i<n;i++)if(i<R[i])swap(a[i],a[R[i]]);
    for(int i=;i<n;i<<=){
        Complex wn=Complex(cos(pi/i),f*sin(pi/i));
        for(int j=;j<n;j+=(i<<)){
            Complex w=Complex(,);
            for(int k=;k<i;k++,w=w*wn){
                Complex 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<n;i++)a[i]=a[i]/n;
}
int main(){
    scanf("%d",&nn);
    for(int i=;i<=nn;i++)scanf("%d",&num[i]);
    for(n=;n<=;n<<=)L++;
    for(int i=;i<n;i++)R[i]=(R[i>>]>>)|((i&)<<(L-));
    Block=min(int(sqrt(nn)*),nn);
    for(int i=;i<=nn;i++)block[i]=(i-)/Block+;
    for(int i=;i<=nn;i++)cnt[block[i]][num[i]]++;
    for(int I=;I<=block[nn];I++){
        int L=lower_bound(block+,block++nn,I)-block,R=upper_bound(block+,block++nn,I)-block-;
        for(int j=L;j<=R;j++){
            cnt[I][num[j]]--;
            for(int i=L;i<j;i++)
                if(num[j]*-num[i]>=)ans+=cnt[I][num[j]*-num[i]];
        }
    }
    for(int i=;i<=nn;i++)cnt[][num[i]]++;
    for(int I=;I<=block[nn];I++){
        int L=lower_bound(block+,block++nn,I)-block,R=upper_bound(block+,block++nn,I)-block-;
        for(int i=L;i<=R;i++)cnt[][num[i]]--;
        for(int j=L;j<=R;j++)
            for(int i=L;i<j;i++)
                if(num[j]*-num[i]>=)ans+=cnt[][num[j]*-num[i]];
    }
    for(int i=;i<=nn;i++)cnt[][num[i]]++;
    for(int I=block[nn];I;I--){
        int L=lower_bound(block+,block++nn,I)-block,R=upper_bound(block+,block++nn,I)-block-;
        for(int i=L;i<=R;i++)cnt[][num[i]]--;
        for(int k=L;k<=R;k++)
            for(int j=k-;j>=L;j--)
                if(num[j]*-num[k]>=)ans+=cnt[][num[j]*-num[k]];
    }
    for(int I=;I<=block[nn];I++){
        for(int i=;i<n;i++)A[i].x=A[i].y=B[i].x=B[i].y=;
        int L=lower_bound(block+,block++nn,I)-block,R=upper_bound(block+,block++nn,I)-block-;
        for(int i=;i<L;i++)A[num[i]].x++;
        for(int i=R+;i<=nn;i++)B[num[i]].x++;
        FFT(A,),FFT(B,);
        for(int i=;i<n;i++)C[i]=A[i]*B[i];
        FFT(C,-);
        for(int i=L;i<=R;i++)ans+=(long long)(C[num[i]*].x+0.2);
    }
    printf("%lld\n",ans);
}

分块FFT哦~