传送门

Zhu and 772002

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 48    Accepted Submission(s): 16

Problem Description

Zhu and 772002 are both good at math. One day, Zhu wants to test the ability of 772002, so he asks 772002 to solve a math problem.

But 772002 has a appointment with his girl friend. So 772002 gives this problem to you.

There are $n$ numbers $a_1, a_2, ..., a_n$. The value of the prime factors of each number does not exceed $2000$, you can choose at least one number and multiply them, then you can get a number $b$.

How many different ways of choices can make $b$ is a perfect square number. The answer maybe too large, so you should output the answer modulo by $1000000007$.

 

Input

First line is a positive integer $T$ , represents there are $T$ test cases.

For each test case:

First line includes a number $n(1 \leq n \leq 300)$，next line there are $n$ numbers $a_1, a_2, ..., a_n, (1 \leq a_i \leq 10^{18})$.

 

Output

For the i-th test case , first output Case #i: in a single line.

Then output the answer of i-th test case modulo by $1000000007$.

 

Sample Input

2 

3 

3 3 4 

3 

2 2 2

 

Sample Output

Case #1: 

3 

Case #2: 

3

 

Author

UESTC

 

Source

2016中国大学生程序设计竞赛 - 网络选拔赛

题目大意：

给出 $n$ 个整数，从中选出 $1$ 个或者多个，使得选出的整数乘积是完全平方数。一共有多少种选

法？对 $MOD$ 取模。

解题思路：

赛后群里说 原题，已然蒙逼，刘汝佳白书 $P_{160}$ , 一样一样的！！！，唉，亏我们还在推了那么

久。。。

其实我是做过一个类似的题所以没用多长时间，那个题目链接传送门

其实，那个题跟这个题目是差不多的，看那个题目就可以了，这个题目就是在那个题目的基础上先

处理一下就行了。。。

$My$ $Code：$

/**
2016 - 08 - 14 下午
Author: ITAK

Motto:

今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 1e9+5;
const LL MAXN = 2e3+5;
const LL MOD = 1e9+7;
const double eps = 1e-7;
const double PI = acos(-1);
using namespace std;
/**+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/

LL equ, var;///equ个方程 var个变量
LL a[MAXN][MAXN];///增广矩阵
LL x[MAXN];///解的数目
bool free_x[MAXN];///判断是不是自由变元
LL free_num;///自由变元的个数
inline LL GCD(LL m, LL n)
{
    if(n == 0)
        return m;
    return GCD(n, m%n);
}
inline LL LCM(LL a, LL b)
{
    return a/GCD(a,b)*b;
}

LL Gauss()
{
    LL Max_r;///当前列绝对值最大的存在的行
    ///col：处理当前的列
    LL row=0;
    for(LL col=0; row<equ&&col<var; row++,col++)
    {
        Max_r = row;
        for(LL i=row+1; i<equ; i++)
            if(abs(a[i][col]) > abs(a[Max_r][col]))
                Max_r = i;

        if(Max_r != row)
            for(LL i=0; i<var+1; i++)
                swap(a[row][i], a[Max_r][i]);

        if(a[row][col] == 0)
        {
            row--;
            continue;
        }
        for(LL i=row+1; i<equ; i++)
        {
            if(a[i][col])
            {
                for(LL j=col; j<var; j++)
                {
                    a[i][j] ^= a[row][j];
                }
            }
        }
    }
    return row;
}
const LL MAX = 2e3+5;
LL p[MAX];
bool prime[MAX];
LL k;
void isprime()
{
    k = 0;
    memset(prime, false, sizeof(prime));
    for(LL i=2; i<MAX; i++)
    {
        if(!prime[i])
        {
            p[k++] = i;
            for(LL j=i*i; j<MAX; j+=i)
                prime[j] = true;
        }
    }
}
LL quick_mod(LL a, LL b)
{
    LL ans = 1;
    while(b)
    {
        if(b & 1)
            ans = (ans*a)%MOD;
        b>>=1;
        a = (a*a)%MOD;
    }
    return ans;
}
int main()
{
    isprime();
    equ = k;
    LL T, n;
    scanf("%I64d",&T);
    for(LL cas=1; cas<=T; cas++)
    {
        cin>>var;
        memset(a, 0, sizeof(a));
        for(LL i=0; i<var; i++)
        {
            LL x;
            LL sum;
            scanf("%I64d",&x);
            for(LL j=0; j<equ; j++)
            {
                sum = 0;
                if(x%p[j] == 0)
                {
                    LL mm = x;
                    while(mm%p[j]==0)
                    {
                        sum++;
                        mm /= p[j];
                    }
                }
                ///构造系数矩阵
                if(sum & 1)
                    a[j][i] = 1;
                else
                    a[j][i] = 0;
            }
        }
        LL ans = var - Gauss();
        LL ret = quick_mod(2LL, ans);
        ret--;
        ret = (ret%MOD+MOD)%MOD;
        printf("Case #%I64d:\n%I64d\n",cas,ret);
    }
    return 0;
}