SDUT2608（Alice and Bob）

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

 

#include<stdio.h>

#include<string.h>
int main()
{
    long long p,tem,tt;
    int n,q,t,i,k,a[55],sum;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        for(i=0;i<n;i++)
        scanf("%d",&a[i]);
        scanf("%d",&q);
        while(q--)
        {
            scanf("%lld",&p);
            sum=1;
            while(p>0)
            {
                k=0;tem=p;tt=1;
                while(tem)
                {
                   if(tem>1)tt*=2;
                    k++;tem/=2;
                }
                sum=(sum*a[k-1])%2012;
                p-=tt;
            }
            printf("%d
",sum%2012);
        }
    }
}