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[51];  
int t,n,q;  
long long p;  
  
int main()  
{  
    cin>>t;while(t--)  
    {  
        cin>>n;  
        for(int i=0;i<n;i++)  
            cin>>a[i];  
        cin>>q;  
        while(q--)  
        {  
            stack<int>s;  
            cin>>p;  
            int result=1;  
            int cnt=-1; //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%2;  
                s.push(yu);  
                p/=2;  
            }  
            if(cnt>n-1) //当数的二进制位数大于n时，不存在直接输出0 
            {  
                cout<<0<<endl;  
                continue;  
            }  
            else  
            {  
                while(!s.empty())  
                {  
                    if(s.top()==1)  
                    {  
                        result*=a[cnt];//取数相乘  
                        if(result>2012)  
                        result%=2012;  
                    }  
                    s.pop();  
                    cnt--;  
                }  
            }  
            cout<<result<<endl;  
        }  
    }  
    return 0;  
}