SDUT 2608：Alice and Bob

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

122 1234

示例输出

20

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 <iostream>#include <math.h>#include <algorithm>using namespace std;int a[55];int kk[55];long long sum,summ;void solve(long long p,int n){    if(p>summ)    {        sum=0;        return;    }    sum=1;    for(int i=n-1; i>=0; i--)    {        if(p>=kk[i])sum=(sum*a[i])%2012,p-=kk[i];        if(p==0)break;    }    if(p!=0)sum=0;}int main(){    int N;    cin>>N;    while(N--)    {        int n;        cin>>n;        for(int i=0; i<n; i++)        {            cin>>a[i];            kk[i]=pow(2,i);            summ+=kk[i];        }        int k;        cin>>k;        while(k--)        {            long long p;            cin>>p;            solve(p,n);            cout<<sum<<endl;        }    }    return 0;}