UPC2222: Alice and Bob

原创

wx5915393277dca 2022-08-06 00:04:37 ©著作权

文章标签 #include ios i++ 文章分类 虚拟化云计算

©著作权归作者所有：来自51CTO博客作者wx5915393277dca的原创作品，请联系作者获取转载授权，否则将追究法律责任

2222: Alice and Bob

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 263

Solved: 77

[

Submit][

Status][

Web Board]

Description

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?

Input

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

Output

For each question of each test case, please output the answer module 2012.

Sample Input

122 1234

Sample Output

HINT

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

Source

2013年山东省第四届ACM大学生程序设计竞赛

#include<iostream>
#include<stack>
#include<cstdio>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,a[60];
        cin>>n;
        for(int i=0;i<n;i++) cin>>a[i];
        int q;
        cin>>q;
        
        while(q--)
        {
            stack<int>s;
            long long p;
            cin>>p;
            int x,cnt=-1;
            int result=1;
            
            
            while(p){
                cnt++;
                x=p%2;
                s.push(x);
                p/=2;
            }

            
            if(cnt > n-1){
                printf("0\n");
            }
            
            
            else{
                    
                while(!s.empty())
                {
                    if(s.top()==1)
                    {
                       result *= a[cnt];
                       if(result > 2012)
                       result %= 2012;
                    }
                    s.pop();
                    cnt--;
                }
                printf("%d\n",result);
            }
        }
    }
    return 0;
}