山东13年省赛 Aliceand Bob

Problem F: Alice and Bob

Description

Alice and Bob like playing games very much.Today, they introduce a new game.

There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). Then Alice ask Bob Q questions. In the expansion of the Polynomial, Given an integer P, please tell the coefficient of the x^P.

Can you help Bob answer these questions?

Input

The first line of the input is a number T, which means the number of the test cases.

For each case, the first line contains a number n, then n numbers a0, a1, .... an-1 followed in the next line. In the third line is a number Q, and then following Q numbers P.

1 <= T <= 20

1 <= n <= 50

0 <= ai <= 100

Q <= 1000

0 <= P <= 1234567898765432

Output

For each question of each test case, please output the answer module 2012.

Sample Input

1
2
2 1
2
3
4

Sample Output

2
0

HINT

The expansion of the (2*x^(2^0) + 1) * (1*x^(2^1) + 1) is 1 + 2*x^1 + 1*x^2 + 2*x^3

解题思路：完全靠位运算即可，可记住这个规律。

求多项式相乘展开式中x的某一指数的系数。

(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1)给定这个式子，观察x的指数，2^0   2^1  2^2  。。很容易联想到二进制。

后来发现有规律，展开式中没有指数相同的两项，也就是说不能合并公因式。而某一x指数的系数化为二进制以后就可以找到规律了。

比如 求指数为13的系数，把13化为二进制    1  1   0  1    从右到左分别对应  a0   a1   a 2   a3  ，那么所求系数就是  a0  *   a2   *  a 3

再举例说明：

 #include <iostream>
 #include <stack>
 #include <string.h>
 using namespace std;
 int a[];
 int t,n,q;
 long long p;  

 int main()
 {
     cin>>t;while(t--)
     {
         cin>>n;
         for(int i=;i<n;i++)
             cin>>a[i];
         cin>>q;
         while(q--)
         {
             stack<int>s;
             cin>>p;
             int result=;
             int cnt=-; //这里使用了-1，为了方便，因为a数组是从0开始的
             int yu;
             while(p) //把数化为二进制存到栈中
             {
                 cnt++;
                 yu=p%;
                 s.push(yu);
                 p/=;
             }
             if(cnt>n-) //当数的二进制位数大于n时，不存在直接输出0
             {
                 cout<<<<endl;
                 continue;
             }
             else
             {
                 while(!s.empty())
                 {
                     if(s.top()==)
                     {
                         result*=a[cnt];//取数相乘
                         if(result>)
                         result%=;
                     }
                     s.pop();
                     cnt--;
                 }
             }
             cout<<result<<endl;
         }
     }
     return ;
 }