山东省第四届省赛Problem F:Alice and Bob（找规律）

B - Problem F:Alice and Bob

Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%lld & %llu

Submit Status

 use MathJax to parse formulas

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

122 1234

Sample Output

20

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

题目意思就是给你n个数，q个查询，求2的q次放的系数是多少

假如n = 4.n个数分别a0,a1,a2,a3

把多项式拆开后：

让我们来列一个表格

P	1	2	3	4	5	6	7
	2^0	2^1	2^1+2^0	2^2	2^2+2^0	2^2+2^1	2^2+2^1+2^0
	a0	a1	a0*a1	a2	a2*a0	a1*a2	a2*a1*a0

答案很明显了，查询P的ans *= a[P的二进制数]

且当p的最高位二进制数大于n时，ans = 0;

代码如下

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 110
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define For(i,n) for(i=0;i<n;i++)
#define mod 2012
typedef long long ll;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
 
int main()
{
    ll n,a[maxn],t,q,p;
    ll i,j;
    ll num;
    ll ans;
    cin >>t;
    while(t--){
        cin >> n;
        int leng;
        mem(a,0);
        For(i,n)
            cin >> a[i];
        cin >>q;
        ll k;
        while(q--){
            ans = 1;
            i = 0;
            num = 0;
            cin >> p;
            num = 1;
            while(p){
                if(p&1 && i<n)
                    ans = ans * a[i]%mod;
                else if(i >= n){
                    ans = 0;
                    break;
                }
                p>>=1;
                i++;
            }
            cout << ans%mod << endl;
        }
    }
    return 0;
}