sdutoj 2608 Alice and Bob

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2608

Alice and Bob

Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^

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

示例程序

分析：

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

给出一个多项式：(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1)，输入P，求X^p 前边的系数。

将p转换成一个二进制的数，然后分别乘上系数。

AC代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 int a[55],b[55];
 4 int main()
 5 {
 6     int t;
 7     scanf("%d",&t);
 8     while(t--)
 9     {
10         int n,p,i,j;
11         scanf("%d",&n);
12         memset(a,0,sizeof(a));
13         for(i=0;i<n;i++)
14         scanf("%d",&a[i]);
15         scanf("%d",&p);
16         int count,ans;
17         long long c;
18         while(p--)
19         {
20             count=1,ans=0;
21             scanf("%lld",&c);
22             if(c==0)
23             {
24                 printf("1\n");
25                 continue;
26             }
27                     
28             while(c)
29             {
30                 b[ans++]=c%2;
31                 c/=2;
32             }
33             for(i=0;i<ans;i++)
34             if(b[i])
35             {
36                 count=count*a[i]%2012;
37             }
38             printf("%d\n",count);
39         }
40     }    
41     return 0;
42 }

View Code