第四届 山东省ACM Alice and Bob

Alice and Bob

Time Limit: 1000MS Memory Limit: 65536KB

Submit Statistic

Problem 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.

Example Input

122 1234

Example 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

Author

 2013年山东省第四届ACM大学生程序设计竞赛

简单的二进制拆分！！！

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <vector>
#include <bitset>
#include <cstdio>
#include <string>
#include <numeric>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long  ll;
typedef unsigned long long ull;

int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};//up down left right
bool inmap(int x,int y,int n,int m)
{
    if(x<1||x>n||y<1||y>m)
        return false;
    return true;
}
int hashmap(int x,int y,int m)
{
    return (x-1)*m+y;
}

#define eps 1e-8
#define inf 0x7fffffff
#define debug puts("BUG")
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define root 1,26,1
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define maxn 105
#define MOD 2012

int a[maxn];
int gao(ll p)
{
    int ans=1;
    for(int i=1;p;++i,p>>=1)
        if(p%2)
            ans=(ans*a[i])%MOD;
    return ans;
}
int main()
{
    //read;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;++i)
            scanf("%d",a+i);
        scanf("%d",&m);
        ll p;
        while(m--)
        {
            scanf("%lld",&p);
            printf("%d
",gao(p));
        }
    }
    return 0;
}