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

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

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 ;
 }