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題意：給出一個長度為$N$的數列$a_i$，再給出$M$個目標數字$d_i$，問目標數字中有多少個等於$a_i$或$a_i+a_j$($i,j \in [1,N]$，不要求$i,j$不相等)。$1 \leq N , M , a_i , d_i \leq 2 \times 10^5$

首先，等於$a_i$相當於等於$a_i+0$，那麼把$0$丟進$a_i$中就變成了詢問目標數字中有多少個等於$a_i$中兩個數字之和。

考慮到$a_i$與$a_j$會對$a_i+a_j$產生貢獻，與卷積類似，故使用$FFT$解決。與生成函式類似（好像就是生成函式），將$A_{a_i}$與$B_{a_i}$$(i \in [1 , N])$設為$1$，其餘設為$0$，做$FFT$。設$FFT$結果陣列為$C$，那麼如果$C_{d_i}>0$，意味著$d_i$是能被$a_i$中某兩個元素表示的，答案++即可。

#include<bits/stdc++.h>
#define ld long double
#define eps 1e-2
//This code is written by Itst
using namespace std;

inline int read(){
    int a = 0;
    bool f = 0;
    char c = getchar();
    while(c != EOF && !isdigit(c)){
        if(c == '-')
            f = 1;
        c = getchar();
    }
    
 
while(c != EOF && isdigit(c)){
        a = (a << 3) + (a << 1) + (c ^ '0');
        c = getchar();
    }
    return f ? -a : a;
}

const int MAXN = 530010;
struct comp{
    ld x , y;

    comp(ld _x = 0 , ld _y = 0){
        x = _x;
        y = _y;
    }

    comp operator +(comp a){
        
 
return comp(x + a.x , y + a.y);
    }

    comp operator -(comp a){
        return comp(x - a.x , y - a.y);
    }

    comp operator *(comp a){
        return comp(x * a.x - y * a.y , x * a.y + y * a.x);
    }
}A[MAXN] , B[MAXN];
int need , dir[MAXN];
const ld pi = acos(-1);

bool cmp(ld a , ld b){
    return a - eps < b && a + eps > b;
}

inline void FFT(comp* a , int type){
    for(int i = 1 ; i < need ; ++i)
        if(i < dir[i])
            swap(a[i] , a[dir[i]]);
    for(int i = 1 ; i < need ; i <<= 1){
        comp wn(cos(pi / i) , type * sin(pi / i));
        for(int j = 0 ; j < need ; j += i << 1){
            comp w(1 , 0);
            for(int k = 0 ; k < i ; ++k , w = w * wn){
                comp x = a[j + k] , y = a[i + j + k] * w;
                a[j + k] = x + y;
                a[i + j + k] = x - y;
            }
        }
    }
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("12879.in" , "r" , stdin);
    freopen("12879.out" , "w" , stdout);
#endif
    while(int N = read()){
        memset(&A , 0 , sizeof(A));
        memset(&B , 0 , sizeof(B));
        int maxN = 0 , cnt = 0;
        for(int i = 1 ; i <= N ; ++i){
            int a = read();
            A[a].x = B[a].x = 1;
            maxN = max(maxN , a);
        }
        A[0].x = B[0].x = 1;
        need = 1;
        while(need <= maxN << 1)
            need <<= 1;
        for(int i = 1 ; i < need ; ++i)
            dir[i] = (dir[i >> 1] >> 1) | (i & 1 ? need >> 1 : 0);
        FFT(A , 1);
        FFT(B , 1);
        for(int i = 0 ; i < need ; ++i)
            A[i] = A[i] * B[i];
        FFT(A , -1);
        for(int M = read() ; M ; --M)
            if(!cmp(0 , A[read()].x))
                ++cnt;
        printf("%d\n" , cnt);
    }
    return 0;
}