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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?
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
122 1234
20
思路：只要随便找个式子化简开就能看出规律，和二进制有点关系。
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
const int mod =2012;
LL a[1010];
int main()
{
LL T;
LL n,q;
LL p;
LL ans,j,i;
scanf("%lld",&T);
while(T--) {
scanf("%lld",&n);
memset(a,0,sizeof(a));
for(i=0; i<n; i++)
scanf("%lld",&a[i]);
scanf("%lld",&q);
while(q--) {
ans=1;
j=0;
scanf("%lld",&p);
while(p) {
if(p%2==1) {
ans=(ans*a[j])%mod;
}
p=p/2;
j++;
}
printf("%lld\n",ans);
}
}
return 0;
}
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原文地址：http://blog.csdn.net/u013486414/article/details/44682637
