BZOJ4836 二元运算（分治FFT）

　　设A(n)为a中n的个数，B(n)为b中n的个数。如果只考虑加法显然是一个卷积，减法翻转一下也显然是一个卷积。

　　问题在于两者都有。容易想到分开处理。那么可以考虑分治。即对于值域区间[l,r]，分别计算A[l,mid]和B[mid+1,r]的贡献及A[mid+1,r]和B[l,mid]的贡献，然后再递归处理[l,mid]和[mid+1,r]。一定程度上类似于cdq分治。

　　注意结果可能爆int，用NTT的话不太方便。

 

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 270000
const double PI=3.14159265358979324;
struct complex
{
    double x,y;
    complex operator +(const complex&a) const
    {
        return (complex){x+a.x,y+a.y};
    }
    complex operator -(const complex&a) const
    {
        return (complex){x-a.x,y-a.y};
    }
    complex operator *(const complex&a) const
    {
        return (complex){x*a.x-y*a.y,x*a.y+y*a.x};
    }
}c[N],d[N];
int T,n,m,q,a[N],b[N],r[N];
long long f[N];
void DFT(int n,complex *a,int p)
{
    for (int i=0;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
    for (int i=2;i<=n;i<<=1)
    {
        complex wn=(complex){cos(2*PI/i),p*sin(2*PI/i)};
        for (int j=0;j<n;j+=i)
        {
            complex w=(complex){1,0};
            for (int k=j;k<j+(i>>1);k++,w=w*wn)
            {
                complex x=a[k],y=w*a[k+(i>>1)];
                a[k]=x+y,a[k+(i>>1)]=x-y;
            }
        }
    }
}
void mul(int n,complex *a,complex *b)
{
    for (int i=0;i<n;i++) r[i]=(r[i>>1]>>1)|(i&1)*(n>>1);
    for (int i=0;i<n;i++) a[i].y=a[i].x-b[i].x,a[i].x=a[i].x+b[i].x;
    DFT(n,a,1);
    for (int i=0;i<n;i++) a[i]=a[i]*a[i];
    DFT(n,a,-1);
    for (int i=0;i<n;i++) a[i].x=a[i].x/n/4;
}
void solve(int l,int r)
{
    if (l==r) {f[0]+=1ll*a[l]*b[l];return;}
    int mid=l+r>>1;
    solve(l,mid);
    solve(mid+1,r);
    int t=1;while (t<r-l+1) t<<=1;
    for (int i=0;i<t;i++) c[i].x=c[i].y=d[i].x=d[i].y=0;
    for (int i=l;i<=mid;i++) c[i-l].x=a[i];
    for (int i=mid+1;i<=r;i++) d[i-mid-1].x=b[i];
    mul(t,c,d);
    for (int i=l+mid+1;i<=mid+r;i++) f[i]+=(long long)(c[i-l-mid-1].x+0.5);
    for (int i=0;i<t;i++) c[i].x=c[i].y=d[i].x=d[i].y=0;
    for (int i=mid+1;i<=r;i++) c[i-mid-1].x=a[i];
    for (int i=l;i<=mid;i++) d[mid-i].x=b[i];
    mul(t,c,d);
    for (int i=1;i<=r-l;i++) f[i]+=(long long)(c[i-1].x+0.5);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4836.in","r",stdin);
    freopen("bzoj4836.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    while (T--)
    {
        n=read(),m=read(),q=read();
        memset(f,0,sizeof(f));
        memset(a,0,sizeof(a));memset(b,0,sizeof(b));
        int x=0,y;
        for (int i=1;i<=n;i++) x=max(y=read(),x),a[y]++;
        for (int i=1;i<=m;i++) x=max(y=read(),x),b[y]++;
        solve(0,x);
        while (q--) printf(LL,f[read()]);
    }
    return 0;
}