山东省第四届省赛Problem F:Alice and Bob（找规律）

B - Problem F:Alice and BobTime Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%lld & %lluSubmit Status use MathJax to parse formulasDescriptionAlice 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?
InputThe 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
OutputFor each question of each test case, please output the answer module 2012.
Sample Input

122 1234

Sample Output

20

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

题目意思就是给你n个数，q个查询，求2的q次放的系数是多少
假如n = 4.n个数分别a0,a1,a2,a3
把多项式拆开后：
让我们来列一个表格

P	1	2	3	4	5	6	7
	2^0	2^1	2^1+2^0	2^2	2^2+2^0	2^2+2^1	2^2+2^1+2^0
	a0	a1	a0*a1	a2	a2*a0	a1*a2	a2*a1*a0

答案很明显了，查询P的ans *= a[P的二进制数]
且当p的最高位二进制数大于n时，ans = 0;
代码如下#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 110
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define For(i,n) for(i=0;i<n;i++)
#define mod 2012
typedef long long ll;
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int main()
{
ll n,a[maxn],t,q,p;
ll i,j;
ll num;
ll ans;
cin >>t;
while(t--){
cin >> n;
int leng;
mem(a,0);
For(i,n)
cin >> a[i];
cin >>q;
ll k;
while(q--){
ans = 1;
i = 0;
num = 0;
cin >> p;
num = 1;
while(p){
if(p&1 && i<n)
ans = ans * a[i]%mod;
else if(i >= n){
ans = 0;
break;
}
p>>=1;
i++;
}
cout << ans%mod << endl;
}
}
return 0;
}