ifrog 1044 - H. Quailty and Binary Operation cdq分治+fft

题目链接点这里

应该想一下就知道需要用fft，，但是怎么用那？

在cdq分治中我们发现，他在分治的将[l,mid对[mid+1，r]的贡献计算出来。。所以对于这题在分治的时候计算x[l,mid对y[mid+1，r]的贡献（x<y），x[mid+1,r对y[l，mid]的贡献(x>=y)就好了。。然而对于x==y的时候，无法在分治中计算我们可以在l==r的时候单独计算

#include<iostream>
#include<cstdio>
#include<math.h>
#include<algorithm>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<queue>
#include<string.h>
#include<cstring>
#include<vector>
#include<time.h>
#include<stdlib.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<<"x"<<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-100;
const int MX=222222;
int n,m,q;
int x[MX],y[MX];

const double pi = acos(-1.0);
//开大4倍
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为i二进制反转
{
    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); //旋转因子
            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)   //求卷积
{
    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);
}
LL ans[MX];
void cdq(int l,int r)
{
    if(l==r)
    {
        ans[0]+=(LL)x[l]*y[l];
        return ;
    }
    int mid=(l+r)>>1;
    cdq(l,mid);
    cdq(mid+1,r);

    int len=1;
    while(len<r-l)len<<=1;
    for(int i=l; i<=mid; i++)
    {
        va[i-l].r=x[i];
        va[i-l].i=0;
    }
    for(int i=mid-l+1; i<len; i++)va[i].i=va[i].r=0;
    for(int i=mid+1; i<=r; i++)
    {
        vb[i-mid-1].r=y[i];
        vb[i-mid-1].i=0;
    }
    for(int i=r-mid; i<len; i++)vb[i].i=vb[i].r=0;
    Conv(va,vb,len);

    len=1;
    while(len<r-l)len<<=1;
    for(int i=0; i<len; i++)
    {
        ans[l+mid+1+i]+=(LL)(va[i].r+0.5);
    }
    for(int i=mid+1; i<=r; i++)
    {
        va[i-mid-1].r=x[i];
        va[i-mid-1].i=0;
    }
    for(int i=r-mid; i<len; i++)va[i].i=va[i].r=0;
    for(int i=l; i<=mid; i++)
    {
        vb[mid-i].r=y[i];
        vb[mid-i].i=0;
    }
    for(int i=mid-l+1; i<len; i++)vb[i].i=vb[i].r=0;
    Conv(va,vb,len);
    for(int i=1; i<=r-l; i++)
    {
        ans[i]+=(LL)(va[i-1].r+0.5);
    }
}
int main()
{
    FIN;
    int T;
    cin>>T;
    while(T--)
    {
        mem(x,0);
        mem(y,0);
        mem(ans,0);
        scanf("%d%d%d",&n,&m,&q);
        int maxn=0;
        for(int i=1; i<=n; i++)
        {
            int w;
            scanf("%d",&w);
            x[w]++;
            maxn=max(maxn,w);
        }
        for(int i=1; i<=m; i++)
        {
            int w;
            scanf("%d",&w);
            y[w]++;
            maxn=max(maxn,w);
        }
        cdq(0,maxn);
        for(int i=1; i<=q; i++)
        {
            int x;
            scanf("%d",&x);
            printf("%lld\n",ans[x]);
        }
    }
    return 0;
}