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[],b[] ;
         cin>>t ;
         memset(a,,sizeof(a));
         for(int i =  ; i <= t- ; i++)
         {
             cin>>a[i];
         }
         int p;
         cin>>p;
         for(int i =  ; i <= p ; i++)
         {
             long long q ;
             cin>>q ;
             int x = ;
             if(q == )
             {
                 cout<<""<<endl;
                 continue ;
             }
             while(q)
             {
                 b[x++] = q% ;
                 q = q/ ;
             }
             int sum =  ;
             for(int j =  ; j <= x- ; j++)
             {
                 if(b[j])
                 {
                     sum = sum*a[j]%;
                 }
             }
             cout<<sum<<endl ;
         }
     }
     return ;
 }