2016中国大学生程序设计竞赛 - 网络选拔赛 1002 Zhu and 772002 hdu5833

最新推荐文章于 2020-09-22 19:09:02 发布

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#高斯消元

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本文介绍了一个数学问题，即如何计算由多个数值通过不同组合相乘得到的乘积为完全平方数的不同方式的数量，并提供了一种解决方案，即通过质因数分解和高斯消元的方法来解决该问题。

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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

Sample Output

Case #1:
3
Case #2:
3

听说是白书原题。

直接按照质因子列异或方程组高斯消元即可

ans=2^自由元个数-1

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
long long mod=1000000007;
int prime[2001];
bool check[2001]; 
int tot;
inline void prepare()
{
     memset(check,false,sizeof(check));
     int i,j;
     for(i=2;i<=2000;i++)
     {
          if(!check[i])
          {
               tot++;
               prime[tot]=i;
          }
          for(j=1;j<=tot;j++)
          {
               if(i*prime[j]>2000)
                    break;
               check[i*prime[j]]=true;
          }
     }
}
int a[400][400],x[400],free_x[400];  
int equ,var,free_num;
long long b[301];
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;
    return var-k;
}
int main()
{
	prepare();
	int T,k=0;
	scanf("%d",&T);
	while(T>0)
	{
		T--;
		k++;
		int n;
		scanf("%d",&n);
		equ=tot;
		var=n;  
		memset(a,0,sizeof(a));  
  	    memset(x,0,sizeof(x));
  	    int i,j;
  	    for(i=1;i<=n;i++)
  	    	scanf("%I64d",&b[i]);
  	    for(i=1;i<=n;i++)
  	    {
  	    	long long t=b[i];
  	    	for(j=1;j<=tot;j++)
  	    	{
  	    		int sx=0;
  	    		while(t%prime[j]==0)
  	    		{
  	    			t/=prime[j];
  	    			sx++;
  	    		}
  	    		if(sx%2==1)
  	    			a[j-1][i-1]=1;
  	    	}
  	    }
  		int p=Gauss();
  		long long ans=1;
  		for(i=1;i<=p;i++)
  			ans=ans*(long long)2%mod;
		ans=(ans+mod-1)%mod;
  	    printf("Case #%d:\n%I64d\n",k,ans);
	}
}