Sdut 2608-alice and Bob (math problem)

Alice and Bob Time limit:1000ms Memory limit:65536k have questions? Dot here ^_^ Title Description

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 a integer P, please tell the coefficient of the x^p.

Can you help Bob answer these questions?

Enter the first line of the input is a number T, which means the number of the 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 was a number Q, and then following Q numbers P.

1 <= T <= 20

1 <= N <= 50

0 <= AI <= 100

Q <= 1000

0 <= P <= 1234567898765432

Output for each question of all test case, please output the answer Module 2012. Sample input

122 1234

Sample output

20

Tip 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 Source 2013 ACM College Student Program Design competition, Shandong Province
Test instructions: That is to ask you to find the coefficients of the two-item x^p.

Idea: Just find a simple formula to open can see the law, and the binary has a bit of a relationship.

#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;}

Sdut 2608-alice and Bob (math problem)