1.题目思路：1.只考虑“+” :生成函数经典模型（“+”以及最终查询的“和”有多个不同的值）

　　　　　　2.考虑“-”：架设生成函数中的次数可以为负数，那么就跟“+”一样处理就好，但FFT无法处理负数，所以我们考虑把最终查询的和+100002，而每一个次数都+50001，这就意味着负数变成了正数，可以使用FFT，而且不影响最终结果

　　　　　　3.处理条件：因为已经将其转化为一个生成函数（即多项式），那么我们不难发现次数小的一定比次数大的要小，但我们不能枚举每一对的大小关系，为了能用到FFT的时间优势，于是我们考虑分治，比mid小的显然比mid大的要小，如此，就把a_l~mid与b_mid+1~r 的贡献和 a_mid+1~r 与 b_l~mid的贡献相加，然后再分治就做完了

2.知识点：1.生成函数

　　　　　2.分治确定大小关系

　　　　　3.通过CDQ分治限制大小关系，从而解决一些有要求的题，满足一些条件

Code:

#include <bits/stdc++.h>
using namespace std;
const double FI = acos(-1);
struct edge
{ 
    double x;
    double y; 
}A[500010],B[500010];
int a[500010],b[500010],inv1[500010];
int n;
int T,x,m,Q;
int len1,len2;
long long ans[500010];
edge operator + (edge &a,edge &b)
{ 
    return (edge){a.x + b.x,a.y + b.y}; 
}
edge operator - (edge &a,edge &b)
{ 
    return (edge){a.x - b.x,a.y - b.y}; 
}
edge operator *(const edge &a,const edge &b)
{ 
    return (edge){a.x * b.x - a.y * b.y,a.x  * b.y + a.y * b.x}; 
}
void FFT(edge *a,int leng,int type)
{
    for (int i = 0; i < leng; i++) 
        if(i < inv1[i]) 
            swap(a[i],a[inv1[i]]);
    for (int i = 1; i < leng; i <<= 1 )
    {
        edge wn = (edge){cos(FI / i),type * sin(FI / i)};
        for (int p = i << 1, j = 0; j < leng; j += p)
        {
            edge w = (edge){1,0};
            for (int k = 0; k < i; k++,w = w * wn)
            {
                edge x = a[j + k],y = w * a[i + j + k]; 
                a[j + k] = x + y; 
                a[i + j + k] = x - y;
            }
        }
    }
    if (type == 1) return;
    for (int i = 0; i < leng; i++) a[i].x /= leng;
}

void CDQ(int l,int r)
{
    if (l == r) return;
    int mid = (l + r) >> 1;
    int n,L = 0;
    for (n = 1; n <= r - l + 1; n <<= 1) L++;
    for (int i = 0; i < n; i++) 
        inv1[i] = (inv1[i >> 1] >> 1) | ((i & 1) << (L - 1));
    for (int i = 0; i < n; i++) 
        A[i] = B[i] = (edge){0,0};
    for (int i =  l; i <= mid; i++) 
        A[i - l].x = a[i];
    for (int i = mid + 1; i <= r ; i++) B[i - mid - 1].x = b[i];
    FFT(A,n,1); FFT(B,n,1);
    for (int i = 0; i < n; i++) A[i] = A[i] * B[i];
    FFT(A,n,-1);
    for (int i = 0; i < n; i++) 
        ans[i + mid + 1 + l] += (long long)(A[i].x + 0.5);
    for (int i = 0; i < n; i++) 
        A[i] = B[i] = (edge){0,0};
    for (int i = mid + 1; i <= r; i++) 
        A[i - mid - 1].x = a[i];
    for (int i = l; i <= mid; i++) 
        B[mid - i].x = b[i];
    FFT(A,n,1); FFT(B,n,1);
    for (int i = 0; i <  n; i++) A[i] = A[i] * B[i];
    FFT(A,n,-1);
    for (int i = 0; i < n; i++) ans[i + 1] += (long long)(A[i].x + 0.5);
    CDQ(l,mid); CDQ(mid + 1,r);
}
int main()
{ 
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&Q); 
        memset(ans,0,sizeof(ans)); 
        memset(a,0,sizeof(a)); 
        memset(b,0,sizeof(b)); 
        len1 = len2 = 0;
        for (int i = 1; i <= n; i++) 
        {
            scanf("%d",&x);
            a[x]++;
            len1 = max(len1,x);
        }
        for (int i = 1; i <= m; i++) 
        {
            scanf("%d",&x);
            b[x]++;
            len2 = max(len2,x);
        }    
        for (int i = 1; i <= len1 && i <= len2; i++) 
            ans[0] += 1ll * a[i] * b[i];
        CDQ(0,max(len1,len2));
        for (int i = 1; i <= Q; i++) 
            scanf("%d",&x),printf("%lld
",ans[x]);
    }
    return 0;
}