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[55],b[55];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,p,i,j;
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        for(i=0;i<n;i++)
        scanf("%d",&a[i]);
        scanf("%d",&p);
        int count,ans;
        long long c;
        while(p--)
        {
            count=1,ans=0;
            scanf("%lld",&c);
            if(c==0)
            {
                printf("1
");
                continue;
            }
                    
            while(c)
            {
                b[ans++]=c%2;
                c/=2;
            }
            for(i=0;i<ans;i++)
            if(b[i])
            {
                count=count*a[i]%2012;
            }
            printf("%d
",count);
        }
    }    
    return 0;
}

 

官方标程：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
const int mod = 2012;
int a[70], n;
int Q;
#define ll long long
ll w[60];
#define fuck while(1)
int main() {
    int t;
    freopen("data.in", "r", stdin);
//    freopen("data.out", "w", stdout);
    scanf("%d", &t);
    w[0] = 1;
    for(int i = 1; i < 60; i++) w[i] = (w[i - 1] << 1);
    while(t--) {
        scanf("%d", &n);
        if(n <= 0 || n > 50) fuck;
        memset(a, 0, sizeof(a));
        for(int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            if(a[i] < 0 || a[i] > 100) fuck;
        }
        scanf("%d", &Q);
        while(Q--) {
            ll x;
            scanf("%I64d", &x);
            if(x > 1234567898765432ll) fuck;
            int ret = 1;
            for(int i = 0; x; i++, x /= 2) {
                if((x & 1)) {
                    ret = ret * a[i] % mod;
                }
            }
            printf("%d
", ret);
        }
    }
    return 0;
}

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 %= 100000000;
    if(a < 0) a += 100000000;
    return a;
}
ll maxx;
ll rr() {
    int xx = rand() % 100;
    if(xx == 0) return 0;
    return r() * r() % maxx;
}
int main() {
    int t = 20;
    freopen("data.in", "w", stdout);
    cout<<t<<endl;
    srand(time(NULL));
    while(t--) {
        int n = rand() % 50 + 1;
        cout<<n<<endl;
        for(int i = 0; i < n; i++) {
            int a = rand() % 100;
            if(i) cout<<" ";
            cout<<a;
        }
        maxx = (1ll << (n));
        maxx = maxx + maxx / 10000;
        cout<<endl;
        int Q = rand() % 1000 + 1;
        ll woca = (1ll << n);
        Q = Q % woca + 1;
        cout<<Q<<endl;
        while(Q--) {
            cout<<rr() % mod<<endl;
        }
    }
    return 0;
}