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?
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
For each question of each test case, please output the answer module 2012.
1
2
2 1
2
3
4
2
0
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
题目大意：多项式相乘，写几项就发现规律了。
(a0x+1)(a1x^2+1)(a2x^4+1)
=a0a1a2x^7+a1a2x^6+a0a2x^5+a2x^4+a0a1x^3+a1x^2+a0x+1 列出来后发现正好该数化为二进制（不用逆置），如果为1，则相乘
7:a0a1a2 即1+2+4
6:a1a2 即2+4
5:a0a2 即1+4
4:a2 即4
3:a0a1 即1+2
2:a1 即2
1:a0 即1
0:1
可以看出，对应为：
a3 a2 a1 a0
8 4 2 1
1 #include <iostream> 2 #include<cstdio> 3 #include<string.h> 4 using namespace std; 5 6 #define maxn 105 7 8 int a[maxn],pos[maxn]; 9 10 int main() 11 { 12 int t,n,i,j,q,ans; 13 long long p;//因为0 <= p <= 1234567898765432 14 scanf("%d",&t); 15 while(t--) 16 { 17 scanf("%d",&n); 18 memset(a,0,sizeof(a));//这个很妙，比如n=3，则p>=8以后都要=0，就靠这一句 19 for(i=0;i<n;i++) 20 { 21 scanf("%d",&a[i]); 22 } 23 scanf("%d",&q); 24 while(q--) 25 { 26 scanf("%lld",&p); 27 if(p==0) 28 { 29 printf("1\n");//别忘了 30 continue; 31 } 32 memset(pos,0,sizeof(pos)); 33 i=0; 34 while(p) 35 { 36 pos[i++]=p%2;//求p转化为二进制 37 p/=2; 38 } 39 ans=1; 40 for(j=0;j<i;j++) 41 { 42 if(pos[j])//如果那一位二进制=1 43 { 44 ans=ans*a[j]%2012;//则就是那几位相乘，一边乘一遍取余2012 45 } 46 } 47 printf("%d\n",ans); 48 } 49 } 50 return 0; 51 }