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：

 

 1 #include<stdio.h>

 2 #include<string.h>

 3 int a[55],b[55];

 4 int main()

 5 {

 6     int t;

 7     scanf("%d",&t);

 8     while(t--)

 9     {

10         int n,p,i,j;

11         scanf("%d",&n);

12         memset(a,0,sizeof(a));

13         for(i=0;i<n;i++)

14         scanf("%d",&a[i]);

15         scanf("%d",&p);

16         int count,ans;

17         long long c;

18         while(p--)

19         {

20             count=1,ans=0;

21             scanf("%lld",&c);

22             if(c==0)

23             {

24                 printf("1\n");

25                 continue;

26             }

27                     

28             while(c)

29             {

30                 b[ans++]=c%2;

31                 c/=2;

32             }

33             for(i=0;i<ans;i++)

34             if(b[i])

35             {

36                 count=count*a[i]%2012;

37             }

38             printf("%d\n",count);

39         }

40     }    

41     return 0;

42 }

View Code

 

 

 

官方标程：

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstdlib>

 4 #include <cstring>

 5 #include <algorithm>

 6 using namespace std;

 7 const int mod = 2012;

 8 int a[70], n;

 9 int Q;

10 #define ll long long

11 ll w[60];

12 #define fuck while(1)

13 int main() {

14     int t;

15     freopen("data.in", "r", stdin);

16 //    freopen("data.out", "w", stdout);

17     scanf("%d", &t);

18     w[0] = 1;

19     for(int i = 1; i < 60; i++) w[i] = (w[i - 1] << 1);

20     while(t--) {

21         scanf("%d", &n);

22         if(n <= 0 || n > 50) fuck;

23         memset(a, 0, sizeof(a));

24         for(int i = 0; i < n; i++) {

25             scanf("%d", &a[i]);

26             if(a[i] < 0 || a[i] > 100) fuck;

27         }

28         scanf("%d", &Q);

29         while(Q--) {

30             ll x;

31             scanf("%I64d", &x);

32             if(x > 1234567898765432ll) fuck;

33             int ret = 1;

34             for(int i = 0; x; i++, x /= 2) {

35                 if((x & 1)) {

36                     ret = ret * a[i] % mod;

37                 }

38             }

39             printf("%d\n", ret);

40         }

41     }

42     return 0;

43 }

View Code

 

 

数据生成程序：

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstdlib>

 4 #include <algorithm>

 5 #include <time.h>

 6 using namespace std;

 7 #define ll long long

 8 #define mod 1234567898765432ll

 9 ll r() {

10     ll a = rand();

11     a *= rand();

12     a %= 100000000;

13     if(a < 0) a += 100000000;

14     return a;

15 }

16 ll maxx;

17 ll rr() {

18     int xx = rand() % 100;

19     if(xx == 0) return 0;

20     return r() * r() % maxx;

21 }

22 int main() {

23     int t = 20;

24     freopen("data.in", "w", stdout);

25     cout<<t<<endl;

26     srand(time(NULL));

27     while(t--) {

28         int n = rand() % 50 + 1;

29         cout<<n<<endl;

30         for(int i = 0; i < n; i++) {

31             int a = rand() % 100;

32             if(i) cout<<" ";

33             cout<<a;

34         }

35         maxx = (1ll << (n));

36         maxx = maxx + maxx / 10000;

37         cout<<endl;

38         int Q = rand() % 1000 + 1;

39         ll woca = (1ll << n);

40         Q = Q % woca + 1;

41         cout<<Q<<endl;

42         while(Q--) {

43             cout<<rr() % mod<<endl;

44         }

45     }

46     return 0;

47 }

View Code