Square 
Input: Standard Input

Output: Standard Output

Given n integers you can generate 2ⁿ-1 non-empty subsets from them. Determine for how many of these subsets the product of all the integers in that is a perfect square. For example for the set {4,6,10,15} there are 3 such subsets. {4}, {6,10,15} and {4,6,10,15}. A perfect square is an integer whose square root is an integer. For example 1, 4, 9,16,…. .

Input

Input contains multiple test cases. First line of the input contains T(1≤T≤30) the number of test cases. Each test case consists of 2 lines. First line contains n(1≤n≤100) and second line contains n space separated integers.  All these integers are between 1 and 10^15.  None of these integers is divisible by a prime greater than 500.

Output

For each test case output is a single line containing one integer denoting the number of non-empty subsets whose integer product is a perfect square. The input will be such that the result will always fit into signed 64 bit integer.

Sample Input                              Output for Sample Input

Problemsetter: Abdullah al Mahmud

Special Thanks to: Manzurur Rahman Khan

题目大意：给n个整数，从中选1个或多个，他们的积是一个完全平方数。这样的情况有多少种？

解题思路：选s个整数出来(1<=s<=n),它们的积可以表示成a1^p1*a2.P2....an^pm,要想它是一个完全平方数，那pi必须是偶数。把每个数分解质因数，所有质因数的指数都模2，最后得出一个n*m的矩阵，进行消元，求出矩阵的秩r。那么就有n-r行每个元素都是0。从中选一行或多行（即求一个集合的真子集个数）都符合题目要求。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5
 6 typedef long long LL;
 7 const int maxn=500;
 8 const int maxm=105;
 9 typedef int Matrix[maxm][maxm];
10 int prime[maxn],num;
11 bool flag[maxn];
12 Matrix A;
13
14 int max(int a,int b){ return a>b?a:b;}
15 void swap(int& a,int& b){int t=a;a=b;b=t;}
16
17 void get_primes()
18 {
19     int i,j;num=0;
20     memset(flag,1,sizeof(flag));
21     for(i=2;i<maxn;i++)
22     {
23         if(flag[i])prime[num++]=i;
24         for(j=0;j<num&&i*prime[j]<maxn;j++)
25         {
26             flag[i*prime[j]]=false;
27             if(i%prime[j]==0) break;
28         }
29     }
30 }
31
32 int rank(int n,int m)
33 {
34     int i=0,j=0,k,r,u;
35     while(i<n&&j<m)
36     {
37         r=i;
38         for(k=i;k<n;k++)
39             if(A[k][j]){r=k;break;}
40         if(A[r][j])
41         {
42                if(r!=i) for(k=0;k<=m;k++) swap(A[r][k],A[i][k]);
43             for(u=i+1;u<n;u++) if(A[u][j])
44             for(k=i;k<=m;k++) A[u][k]^=A[i][k];
45                 i++;
46         }
47         j++;
48     }
49     return i;
50 }
51
52 LL Pow(LL a,LL b)
53 {
54     LL ret=1;
55     while(b)
56     {
57         if(b&1) ret*=a;
58         a*=a;b>>=1;
59     }
60     return ret;
61 }
62
63 int main()
64 {
65     get_primes();
66     int i,j,t,n,maxp;
67     LL p;
68     scanf("%d",&t);
69     while(t--)
70     {
71         scanf("%d",&n);
72         memset(A,0,sizeof(A));
73         maxp=0;
74         for(i=0;i<n;i++)
75         {
76             scanf("%lld",&p);
77             for(j=0;j<num;j++)
78             {
79                 if(p==1) break;
80                 while(p%prime[j]==0)
81                 {
82                     maxp=max(maxp,j);
83                     p/=prime[j];
84                     A[i][j]^=1;
85                 }
86             }
87         }
88         int r=rank(n,maxp+1);
89         printf("%lld\n",Pow(2,n-r)-1);
90     }
91     return 0;
92 }

uva 11542 高斯消元