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

233 3 432 2 2

 

Sample Output

Case #1:3Case #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;}