Alice and Bob（2013年山东省第四届ACM大学生程序设计竞赛）

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

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

来源

2013年山东省第四届ACM大学生程序设计竞赛

分析

受到公式的影响，刚开始当成母函数做的，母函数都写好了，发现开不了最大的数组。郁闷啊！！！后来用一片面包换来的情报，与二进制有关。
恍然大悟，大难题秒变水题。

示例程序

 #include <iostream>

 #include <stdio.h>

 #include <stdlib.h>

 #include <algorithm>

 #include <cstring>

 using namespace std;

 int main()

 {

     int T;

     scanf("%d",&T);//测试组数

     while(T--)

     {

         int a[];//存储系数

         memset(a,,sizeof(a));//全部初始化为零

         int n,q;

         scanf("%d",&n);

         for(int i=;i<n;i++)

             scanf("%d",&a[i]);

         scanf("%d",&q);//查询次数

         while(q--)

         {

             long long p;

             scanf("%lld",&p);

             long long sum=,i=;

             while(p)//化为二进制

             {

                 if(p%==)

                     sum*=a[i];

                 if(sum>)

                     sum%=;

                 i++;

                 p/=;

             }

             printf("%lld\n",sum);

         }

     }

     return ;

 }