SDUT2608（Alice and Bob）

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

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

 

#include<stdio.h>#include<string.h>int main(){    long long p,tem,tt;    int n,q,t,i,k,a[55],sum;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        memset(a,0,sizeof(a));        for(i=0;i<n;i++)        scanf("%d",&a[i]);        scanf("%d",&q);        while(q--)        {            scanf("%lld",&p);            sum=1;            while(p>0)            {                k=0;tem=p;tt=1;                while(tem)                {                   if(tem>1)tt*=2;                    k++;tem/=2;                }                sum=(sum*a[k-1])%2012;                p-=tt;            }            printf("%d\n",sum%2012);        }    }}