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

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<iostream>

 4 using namespace std ;

 5 int main()

 6 {

 7     int n ;

 8     cin>>n ;

 9     while(n--)

10     {

11         int t ;

12         int a[190],b[191] ;

13         cin>>t ;

14         memset(a,0,sizeof(a));

15         for(int i = 0 ; i <= t-1 ; i++)

16         {

17             cin>>a[i];

18         }

19         int p;

20         cin>>p;

21         for(int i = 1 ; i <= p ; i++)

22         {

23             long long q ;

24             cin>>q ;

25             int x = 0;

26             if(q == 0)

27             {

28                 cout<<"1"<<endl;

29                 continue ;

30             }

31             while(q)

32             {

33                 b[x++] = q%2 ;

34                 q = q/2 ;

35             }

36             int sum = 1 ;

37             for(int j = 0 ; j <= x-1 ; j++)

38             {

39                 if(b[j])

40                 {

41                     sum = sum*a[j]%2012;

42                 }

43             }

44             cout<<sum<<endl ;

45         }

46     }

47     return 0;

48 }

View Code