2013年山东省第四届ACM大学生程序设计竞赛 Alice and Bob

 

Alice and Bob

Time Limit: 1000ms   Memory limit: 65536K 

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

1
2
2 1
2
3
4

示例输出

2
0

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

这个题的话一开始挺没有思路的，后来看到题解是说 将p转换成一个二进制的数，然后分别乘上系数。

一开始挺难理解的，T给我讲了，才明白了，其实可以在纸上自己先算算再自己写出二进制来，推一下，就可以理解了

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std ;
int main()
{
    int n ;
    cin>>n ;
    while(n--)
    {
        int t ;
        int a[190],b[191] ;
        cin>>t ;
        memset(a,0,sizeof(a));
        for(int i = 0 ; i <= t-1 ; i++)
        {
            cin>>a[i];
        }
        int p;
        cin>>p;
        for(int i = 1 ; i <= p ; i++)
        {
            long long q ;
            cin>>q ;
            int x = 0;
            if(q == 0)
            {
                cout<<"1"<<endl;
                continue ;
            }
            while(q)
            {
                b[x++] = q%2 ;
                q = q/2 ;
            }
            int sum = 1 ;
            for(int j = 0 ; j <= x-1 ; j++)
            {
                if(b[j])
                {
                    sum = sum*a[j]%2012;
                }
            }
            cout<<sum<<endl ;
        }
    }
    return 0;
}