[2013山东ACM省赛] 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

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

解题思路：

求多项式相乘展开式中x的某一指数的系数。

(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1)给定这个式子，观察x的指数，2^0   2^1  2^2  。。很容易联想到二进制。

后来发现有规律，展开式中没有指数相同的两项，也就是说不能合并公因式。而某一x指数的系数化为二进制以后就可以找到规律了。

比如 求指数为13的系数，把13化为二进制    1  1   0  1    从右到左分别对应  a0   a1   a 2   a3  ，那么所求系数就是  a0  *   a2   *  a 3

再举例说明：

代码：

#include <iostream>
#include <stack>
#include <string.h>
using namespace std;
int a[51];
int t,n,q;
long long p;  

int main()
{
    cin>>t;while(t--)
    {
        cin>>n;
        for(int i=0;i<n;i++)
            cin>>a[i];
        cin>>q;
        while(q--)
        {
            stack<int>s;
            cin>>p;
            int result=1;
            int cnt=-1; //这里使用了-1，为了方便，因为a数组是从0开始的
            int yu;
            while(p) //把数化为二进制存到栈中
            {
                cnt++;
                yu=p%2;
                s.push(yu);
                p/=2;
            }
            if(cnt>n-1) //当数的二进制位数大于n时，不存在直接输出0
            {
                cout<<0<<endl;
                continue;
            }
            else
            {
                while(!s.empty())
                {
                    if(s.top()==1)
                    {
                        result*=a[cnt];//取数相乘
                        if(result>2012)
                        result%=2012;
                    }
                    s.pop();
                    cnt--;
                }
            }
            cout<<result<<endl;
        }
    }
    return 0;
}  

[2013山东ACM省赛] Alice and Bob