Alice and BobTime 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 apolynomial 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 theexpansion 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-1followed 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
122 1234
示例输出
20
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
题意：就是让你求二项式x^p的系数。
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;}
思想如钻子，必须集中在一点钻下去才有力量
