sdutoj 2608 Alice and Bob

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2608

Alice and Bob

Time 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 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

1
2
2 1
2
3
4

示例输出

2
0

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

示例程序

分析：

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

给出一个多项式：(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1)，输入P，求X^p 前边的系数。

将p转换成一个二进制的数，然后分别乘上系数。

AC代码：

 #include<stdio.h>
 #include<string.h>
 int a[],b[];
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         int n,p,i,j;
         scanf("%d",&n);
         memset(a,,sizeof(a));
         for(i=;i<n;i++)
         scanf("%d",&a[i]);
         scanf("%d",&p);
         int count,ans;
         long long c;
         while(p--)
         {
             count=,ans=;
             scanf("%lld",&c);
             if(c==)
             {
                 printf("1\n");
                 continue;
             }

             while(c)
             {
                 b[ans++]=c%;
                 c/=;
             }
             for(i=;i<ans;i++)
             if(b[i])
             {
                 count=count*a[i]%;
             }
             printf("%d\n",count);
         }
     }
     return ;
 }

官方标程：

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int mod = ;
 int a[], n;
 int Q;
 #define ll long long
 ll w[];
 #define fuck while(1)
 int main() {
     int t;
     freopen("data.in", "r", stdin);
 //    freopen("data.out", "w", stdout);
     scanf("%d", &t);
     w[] = ;
     for(int i = ; i < ; i++) w[i] = (w[i - ] << );
     while(t--) {
         scanf("%d", &n);
         if(n <=  || n > ) fuck;
         memset(a, , sizeof(a));
         for(int i = ; i < n; i++) {
             scanf("%d", &a[i]);
             if(a[i] <  || a[i] > ) fuck;
         }
         scanf("%d", &Q);
         while(Q--) {
             ll x;
             scanf("%I64d", &x);
             if(x > 1234567898765432ll) fuck;
             int ret = ;
             for(int i = ; x; i++, x /= ) {
                 if((x & )) {
                     ret = ret * a[i] % mod;
                 }
             }
             printf("%d\n", ret);
         }
     }
     return ;
 }

数据生成程序：

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <algorithm>
 #include <time.h>
 using namespace std;
 #define ll long long
 #define mod 1234567898765432ll
 ll r() {
     ll a = rand();
     a *= rand();
     a %= ;
     if(a < ) a += ;
     return a;
 }
 ll maxx;
 ll rr() {
     int xx = rand() % ;
     if(xx == ) return ;
     return r() * r() % maxx;
 }
 int main() {
     int t = ;
     freopen("data.in", "w", stdout);
     cout<<t<<endl;
     srand(time(NULL));
     while(t--) {
         int n = rand() %  + ;
         cout<<n<<endl;
         for(int i = ; i < n; i++) {
             int a = rand() % ;
             if(i) cout<<" ";
             cout<<a;
         }
         maxx = (1ll << (n));
         maxx = maxx + maxx / ;
         cout<<endl;
         int Q = rand() %  + ;
         ll woca = (1ll << n);
         Q = Q % woca + ;
         cout<<Q<<endl;
         while(Q--) {
             cout<<rr() % mod<<endl;
         }
     }
     return ;
 }