SDUT 2608 Alice and Bob （巧妙的二进制）

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

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

 

 

看到2,，自然想到二进制。。。

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int main(){

    //freopen("input.txt","r",stdin);

    int t,n,a[60];
    int q;
    long long p;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&q);
        while(q--){
            cin>>p;
            int loc;
            int ans=1,cnt=0;
            while(p){
                loc=p&1;
                //printf("%d",loc);
                if(cnt>=n){
                    ans=0;
                    break;
                }
                if(loc==1)
                    ans=ans*a[cnt]%2012;
                cnt++;
                p>>=1;
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}