hdu5833 Zhu and 772002 （高斯消元的简单应用）

hdu5833 Zhu and 772002 （高斯消元的简单应用）：http://acm.hdu.edu.cn/showproblem.php?pid=5833

题面描述：

Zhu and 772002

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 520    Accepted Submission(s): 174

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  a1,a2,...,an . 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≤n≤300) ，next line there are  n  numbers  a1,a2,...,an,(1≤ai≤1018) .

 

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
  

 

题目大意：

输入n个数，范围在10的18次方以内，将其质因数分解之后他们的质因子都在2000以内，从中选出1-多个，使得所选数的乘积为完全平方数，计算一共有多少种选法。

题目分析：

不超过2000的素因子，可以考虑每个数的唯一分解式，用01向量表示一个数，再用n个01变量xi来表示我们的选择，其中xi=1表示要选第i个数，xi=0表示不选它，则可以对每个素数的幂列出一个模2的方程。

代码实现如下：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>

using namespace std;

const long long mod=1000000007;

long long equ,var;
long long a[330][330];
long long x[330];
long long free_x[330];
long long free_num;

int Gauss()
{
    int max_r, col, k;
    free_num = 0;
    for(k = 0, col = 0; k < equ && col < var; k++, col++)
    {
        max_r = k;
        for(int i = k+1 ; i < equ; i++)
            if(abs(a[i][col]) > abs(a[max_r][col]))
                max_r = i;
        if(a[max_r][col] == 0)
        {
            k--;
            free_x[free_num++] = col;
            continue;
        }
        if(max_r != k)
        {
            for(int j = col; j < var+1; j++)
                swap(a[k][j],a[max_r][j]);
        }
        for(int i = k+1; i < equ; i++)
            if(a[i][col] != 0)
                for(int j = col; j < var+1; j++)
                    a[i][j] ^= a[k][j];
    }
    for(int i = k; i < equ; i++)
        if(a[i][col] != 0)
            return -1;
    if(k < var)return var-k;
    for(int i = var-1; i >= 0; i--)
    {
        x[i] = a[i][var];
        for(int j = i+1; j < var; j++)
            x[i] ^= (a[i][j] && x[j]);
    }
    return 0;
}

const int MAXN = 2200;
long long prime[MAXN+1];
void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i = 2; i <= MAXN; i++)
    {
        if(!prime[i])prime[++prime[0]] = i;
        for(int j = 1; j <= prime[0] && prime[j] <= MAXN/i; j++)
        {
            prime[prime[j]*i] = 1;
            if(i%prime[j] == 0)break;
        }
    }
}

long long quick_mod(long long a,long long b)
{
    long long ans=1;
    while(b)
    {
        if(b&1)
        {
            ans=(ans*a)%mod;
            b--;
        }
        b/=2;
        a=a*a%mod;
    }
    return ans;
}

long long data[330];
char str1[330],str2[330];

int main()
{
    getPrime();
    int T,m;
    int casenum=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&m);
        {
            for(int i = 0; i < m; i++)
                scanf("%lld",&data[i]);
            int t=303;
            equ = t;
            var = m;
            for(int i = 0; i < t; i++)
                for(int j = 0; j < m; j++)
                {
                    int cnt = 0;
                    while(data[j]%prime[i+1] == 0)
                    {
                        cnt++;
                        data[j] /= prime[i+1];
                    }
                    a[i][j] = (cnt&1);
                }
            for(int i = 0; i < t; i++)
                a[i][m] = 0;
            int ret = Gauss();
            long long ansans=quick_mod(2,free_num);
            ansans=(ansans-1)%mod;
            printf("Case #%d:\n",casenum++);
            printf("%lld\n",ansans);
        }
    }
    return 0;
}