Alice and Bob（2013年山东省第四届ACM大学生程序设计竞赛）

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

来源

2013年山东省第四届ACM大学生程序设计竞赛

分析

受到公式的影响，刚开始当成母函数做的，母函数都写好了，发现开不了最大的数组。郁闷啊！！！后来用一片面包换来的情报，与二进制有关。
恍然大悟，大难题秒变水题。

 

 

示例程序

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <cstring>
using namespace std;

int main()
{
    int T;
    scanf("%d",&T);//测试组数
    while(T--)
    {
        int a[55];//存储系数
        memset(a,0,sizeof(a));//全部初始化为零
        int n,q;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&q);//查询次数
        while(q--)
        {
            long long p;
            scanf("%lld",&p);
            long long sum=1,i=0;
            while(p)//化为二进制
            {
                if(p%2==1)
                    sum*=a[i];
                if(sum>2012)
                    sum%=2012;
                i++;
                p/=2;
            }
            printf("%lld
",sum);
        }
    }
    return 0;
}