## SDUT2608（Alice and Bob）

2018年02月22日 ⁄ 综合 ⁄ 共 1094字 ⁄ 字号 评论关闭

## 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.

```1
2
2 1
2
3
4```

```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,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\n",sum%2012);
}
}
}
```

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