HDU 5833_hdu - 5833-优快云博客

本文链接：https://blog.youkuaiyun.com/spring371327/article/details/52719866

Zhu and 772002

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1521 Accepted Submission(s): 542

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

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;

typedef long long int ll;

const int MAXN=305;
const int mod = 2;//可改 
const int MOD = 1e9+7;
int a[MAXN][MAXN];//增广矩阵
int x[MAXN];//解集
bool free_x[MAXN];//标记是否是不确定的变元


/*
void Debug(int equ,int var)
{
    int i, j;
    for (i = 0; i < equ; i++)
    {
        for (j = 0; j < var + 1; j++)
        {
            cout << a[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
}
*/


inline int gcd(int a,int b)
{
    int t;
    while(b!=0)
    {
        t=b;
        b=a%b;
        a=t;
    }
    return a;
}
inline int lcm(int a,int b)
{
    return a/gcd(a,b)*b;//先除后乘防溢出
}

// 高斯消元法解方程组(Gauss-Jordan elimination).(-2表示有浮点数解，但无整数解，
//-1表示无解，0表示唯一解，大于0表示无穷解，并返回自由变元的个数)
//有equ个方程，var个变元。增广矩阵行数为equ,分别为0到equ-1,列数为var+1,分别为0到var.
int Gauss(int equ,int var)
{
    int i,j,k;
    int max_r;// 当前这列绝对值最大的行.
    int col;//当前处理的列
    int ta,tb;
    int LCM;
    int temp;
    int free_x_num;
    int free_index;

    for(int i=0;i<=var;i++)
    {
        x[i]=0;
        free_x[i]=true;
    }

    //转换为阶梯阵.
    col=0; // 当前处理的列
    for(k = 0;k < equ && col < var;k++,col++)
    {// 枚举当前处理的行.
// 找到该col列元素绝对值最大的那行与第k行交换.(为了在除法时减小误差)
        max_r=k;
        for(i=k+1;i<equ;i++)
        {
            if(abs(a[i][col])>abs(a[max_r][col])) max_r=i;
        }
        if(max_r!=k)
        {// 与第k行交换.
            for(j=k;j<var+1;j++) swap(a[k][j],a[max_r][j]);
        }
        if(a[k][col]==0)
        {// 说明该col列第k行以下全是0了，则处理当前行的下一列.
            k--;
            continue;
        }
        for(i=k+1;i<equ;i++)
        {// 枚举要删去的行.
            if(a[i][col]!=0)
            {
                LCM = lcm(abs(a[i][col]),abs(a[k][col]));
                ta = LCM/abs(a[i][col]);
                tb = LCM/abs(a[k][col]);
                if(a[i][col]*a[k][col]<0)tb=-tb;//异号的情况是相加
                for(j=col;j<var+1;j++)
                {
                    a[i][j] = (mod + (a[i][j]%mod)*ta - (a[k][j]%mod)*tb)%mod;
                }
            }
        }
    }

    //Debug(equ,var);

    // 1. 无解的情况: 化简的增广阵中存在(0, 0, ..., a)这样的行(a != 0).
    for (i = k; i < equ; i++)
    { // 对于无穷解来说，如果要判断哪些是自由变元，那么初等行变换中的交换就会影响，则要记录交换.
        if ( (a[i][col]%mod) != 0) return -1;
    }
    // 2. 无穷解的情况: 在var * (var + 1)的增广阵中出现(0, 0, ..., 0)这样的行，即说明没有形成严格的上三角阵.
    // 且出现的行数即为自由变元的个数.
    if (k < var)
    {
        // 首先，自由变元有var - k个，即不确定的变元至少有var - k个.
        for (i = k - 1; i >= 0; i--)
        {
            // 第i行一定不会是(0, 0, ..., 0)的情况，因为这样的行是在第k行到第equ行.
            // 同样，第i行一定不会是(0, 0, ..., a), a != 0的情况，这样的无解的.
            free_x_num = 0; // 用于判断该行中的不确定的变元的个数，如果超过1个，则无法求解，它们仍然为不确定的变元.
            for (j = 0; j < var; j++)
            {
                if (a[i][j] != 0 && free_x[j]) free_x_num++, free_index = j;
            }
            if (free_x_num > 1) continue; // 无法求解出确定的变元.
            // 说明就只有一个不确定的变元free_index，那么可以求解出该变元，且该变元是确定的.
            temp = a[i][var];
            for (j = 0; j < var; j++)
            {
                if (a[i][j] != 0 && j != free_index) temp -= a[i][j] * x[j];
            }
            x[free_index] = temp / a[i][free_index]; // 求出该变元.
            free_x[free_index] = 0; // 该变元是确定的.
        }
        return var - k; // 自由变元有var - k个.
    }
    // 3. 唯一解的情况: 在var * (var + 1)的增广阵中形成严格的上三角阵.
    // 计算出Xn-1, Xn-2 ... X0.
    for (i = var - 1; i >= 0; i--)
    {
        temp = a[i][var];
        for (j = i + 1; j < var; j++)
        {
            if (a[i][j] != 0) temp -= a[i][j] * x[j];
        }
        if (temp % a[i][i] != 0) return -2; // 说明有浮点数解，但无整数解.
        x[i] = temp / a[i][i];
    }
    return 0;
}


int prime[1000];
int spring[2005];
int count_prime = 0;
void Init(){
	
	for(int i=2; i<=2000; i++){
		if(spring[i]==0){
			prime[count_prime++] = i;
			for(int j=2*i; j<=2000; j+=i)
				spring[j] = 1;
		}
	}
} 

ll quick_pow(ll a,ll m){  
    ll ans = 1;  
    while( m ){  
        if( m%2 )  
            ans = (ans*a)%MOD;  
        a = (a*a)%MOD;  
        m /= 2;  
    }
	return ans; 
}



int main(){
    
    Init();
    int T;
    scanf("%d",&T);
    int N;
    int Case = 0;
    ll num;
    int Count=0;
    while(T--){
    	scanf("%d",&N);
    	for(int i=0; i<N; i++){
    		scanf("%I64d",&num);
    		for(int j=0; j<303; j++){
    			Count = 0;
    			while( num && num%prime[j]==0 ){
    				Count++;
    				num /= prime[j];
				}
				a[j][i] = Count%2;
			}
		}
		for(int i=0; i<303; i++)
			a[i][N] = 0;
		int ans = Gauss(303,N);
    	if(ans<=-1){
    		printf("Case #%d:\n",++Case);
    		printf("0\n");
		}
		else{
			printf("Case #%d:\n",++Case);
			ll ANS = quick_pow((ll)2,(ll)ans);
			ANS--;
			printf("%d\n",(int(ANS)+MOD)%MOD);
		}
	}
    return 0;
}