HDU-5833: Zhu and 772002

高斯消元

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,$(1≤ai≤10^{18})$.

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

题意： t组样例，每组样例输入一个n，表示有n个数ai，要从这n个数中随便选几个出来，让他们的乘积成为一个完全平方数，问有多少种方案。方案数模1e9+7。

分析：首先，对于一个完全平方数来说，它的每个相同素因子，一定有偶数个，比如36，素因子2，2，3，3，2和3都是两个。那么我们将每一个输入的x进行素因子分解。一个列数为n的矩阵，行表示素数，从上到下分别为素数2，3，5，7……出现的次数。如果某个素因子出现奇数次，那么记为1，否则为0。如第一个数素因子分解之后，素因子3出现奇数次，其他都为偶数次，那么mp[1][0]=1，其他mp[i][0]为0。
由于完全平方数相同素因子为偶数个，也就是说，将这个矩阵进行高斯消元之后，增广矩阵最后一列为全0。我们求出自由元的个数fre，那么是否$2^{fre}$就为最终的答案？不……还要刨去所有数都不选的情况，因此得到$2^{fre}-1$……

• 代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=350;
const int mod=1e9+7;
int prime[350],cnt;
int mp[350][350];
void getPrime()
{
    prime[cnt++]=2;
    for(int i=3;i<=2000;i+=2)
    {
        int flag=1;
        for(int j=2;j*j<=i;j++)
        {
            if(i%j==0)
            {
                flag=0;
                break;
            }
        }
        if(flag) prime[cnt++]=i;
    }
}
int resolution(int col,LL x)
{
    int m=0;
    for(int i=0;i<cnt;i++)
    {
        while(x%prime[i]==0)
        {
            mp[i][col]^=1;
            x/=prime[i];
            m=max(m,i+1);
        }
    }
    return m;
}
int gauss(int A[][maxn],int n,int m)
{
    int res=0,r=0;
    for(int i=0;i<m&&r<m;i++)
    {
        for(int j=r;j<n;j++)
        {
            if(A[j][i]==1)
            {
                for(int k=i;k<=m;k++)
                    swap(A[j][k],A[r][k]);
                break;
            }
        }
        if(A[r][i]==0)
        {
            res++;
            continue;
        }
        for(int j=0;j<n;j++)
        {
            if(A[j][i]==1&&j!=r)
            {
                for(int k=i;k<=m;k++)
                    A[j][k]^=A[r][k];
            }
        }
        r++;
    }
    return res;
}
int main()
{
    int t,o=0;
    getPrime();
    scanf("%d",&t);
    while(t--)
    {
        memset(mp,0,sizeof(mp));
        LL x;
        int n,y=0,fre=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&x);
            y=max(y,resolution(i,x));
        }
        fre+=gauss(mp,y,n);
        LL ans=1;
        for(int i=1;i<=fre;i++)
            ans=ans*2%mod;
        ans--;
        printf("Case #%d:\n%lld\n",++o,ans);
    }
    return 0;
}