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
1 2 2 1 2 3 4
2 0
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 #include <cstring> 6 using namespace std; 7 8 int main() 9 { 10 int T; 11 scanf("%d",&T);//测试组数 12 while(T--) 13 { 14 int a[55];//存储系数 15 memset(a,0,sizeof(a));//全部初始化为零 16 int n,q; 17 scanf("%d",&n); 18 for(int i=0;i<n;i++) 19 scanf("%d",&a[i]); 20 scanf("%d",&q);//查询次数 21 while(q--) 22 { 23 long long p; 24 scanf("%lld",&p); 25 long long sum=1,i=0; 26 while(p)//化为二进制 27 { 28 if(p%2==1) 29 sum*=a[i]; 30 if(sum>2012) 31 sum%=2012; 32 i++; 33 p/=2; 34 } 35 printf("%lld\n",sum); 36 } 37 } 38 return 0; 39 }
Alice and Bob（2013年山东省第四届ACM大学生程序设计竞赛）,布布扣,bubuko.com
Alice and Bob（2013年山东省第四届ACM大学生程序设计竞赛）
原文：http://www.cnblogs.com/vivider/p/3707901.html