Codeforces Round #448 (Div. 2) C. Square Subsets

http://codeforces.com/contest/895/problem/C

题意：n个数里，随意选k个数，他们的乘积是完全平方数，问有多少种选择，白书上的原题 另外hdu 5833。
T的线性基是T的一个子集A={a1,a2,a3,...,an}。

A中元素互相xor所形成的异或集合，等价于原数集T的元素互相xor形成的异或集合。

可以理解为将原数集进行了压缩。做一个每位存储的是质因子的个数%2的线性基。对n个数遍历存入线性基中，最终无法被存入线性基中的数可以视作与线性基中的某几个数做异或操作后变成一个完全平方数。设有t个数无法被存入线性基中，那么答案就为2^t−1 。 

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=~0U>>1;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+10;
const int maxx=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

//void readString(string &s)
//{
//	static char str[maxn];
//	scanf("%s", str);
//	s = str;
//}

int n;
int a[maxn];
int d[70],pri[70],tot=0;
void init()
{
for(int i = 2; i < 70; ++i)
{
if(!d[i])
pri[++tot] = d[i] = i;
for(int j = 0, k; (k = i * pri[j]) < 70; ++j)
{
d[k] = pri[j];
if(d[i] == pri[j])
break;
}
}
}

int _pow(LL a,LL n)
{
LL ret=1;
while(n)
{
if(n&1)  ret=ret*a%mod;
a=a*a%mod;
n>>=1;
}
return ret;
}

vector<int>xor_guass;
void solve()
{
s_1(n);
FOR(1,n,i) s_1(a[i]);
FOR(1,n,i)
{
int t=0;
FOR(1,tot,j)
{
if(a[i]%pri[j]==0)
{
W(a[i]%pri[j]==0)
{
a[i]/=pri[j];
t^=(1<<j);
}
}
}
for(int xx:xor_guass)
t=min(t,t^xx);
if(t) xor_guass.pb(t);
}
int ans=_pow(2,n-xor_guass.size())-1;
print(ans);

}
int main()
{
//freopen( "1.in" , "r" , stdin );
//freopen( "1.out" , "w" , stdout );
int t=1;
init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}