△2013山东省ACM竞赛-Alice and Bob

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)*.......*(a_n-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, .... a_n-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 <= a_i <= 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

题目大意：多项式相乘，写几项就发现规律了。

(a0x+1)(a1x^2+1)(a2x^4+1)

=a0a1a2x^7+a1a2x^6+a0a2x^5+a2x^4+a0a1x^3+a1x^2+a0x+1    列出来后发现正好该数化为二进制（不用逆置），如果为1，则相乘

7:a0a1a2   即1+2+4

6:a1a2      即2+4

5:a0a2      即1+4

4:a2         即4

3:a0a1     即1+2

2:a1        即2

1:a0        即1

0:1

可以看出，对应为：

a3   a2   a1   a0

8     4     2     1

#include <iostream>
#include<cstdio>
#include<string.h>
using namespace std;

#define maxn 105

int a[maxn],pos[maxn];

int main()
{
    int t,n,i,j,q,ans;
    long long p;//因为0 <= p <= 1234567898765432
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a));//这个很妙，比如n=3，则p>=8以后都要=0，就靠这一句
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        scanf("%d",&q);
        while(q--)
        {
            scanf("%lld",&p);
            if(p==0)
            {
                printf("1
");//别忘了
                continue;
            }
            memset(pos,0,sizeof(pos));
            i=0;
            while(p)
            {
                pos[i++]=p%2;//求p转化为二进制
                p/=2;
            }
            ans=1;
            for(j=0;j<i;j++)
            {
                if(pos[j])//如果那一位二进制=1
                {
                    ans=ans*a[j]%2012;//则就是那几位相乘，一边乘一遍取余2012
                }
            }
             printf("%d
",ans);
        }
    }
    return 0;
}