SDUT 2608：Alice and Bob

Alice and Bob

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

    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?

输入

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

输出

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

示例输入

122 1234

示例输出

20

提示

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

来源

 2013年山东省第四届ACM大学生程序设计竞赛

#include <iostream>
#include <math.h>
#include <algorithm>
using namespace std;
int a[55];
int kk[55];
long long sum,summ;
void solve(long long p,int n)
{
    if(p>summ)
    {
        sum=0;
        return;
    }
    sum=1;
    for(int i=n-1; i>=0; i--)
    {
        if(p>=kk[i])sum=(sum*a[i])%2012,p-=kk[i];
        if(p==0)break;
    }
    if(p!=0)sum=0;
}
int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int n;
        cin>>n;
        for(int i=0; i<n; i++)
        {
            cin>>a[i];
            kk[i]=pow(2,i);
            summ+=kk[i];
        }
        int k;
        cin>>k;
        while(k--)
        {
            long long p;
            cin>>p;
            solve(p,n);
            cout<<sum<<endl;
        }
    }
    return 0;
}