CF895C Square Subsets [线性基]

线性基的题…

考虑平方数只和拆解质因子的个数的奇偶性有关系

比如说你 \(4\) 和 \(16\) 的贡献都是一样的。因为

\(4 = 2^2 , 16 = 2^4\)

\(2\) 和 \(4\) 奇偶性相同

然后考虑如何线性基，不难想到，二进制可以表示奇偶性，

所以异或和每一位是0的时候就是一个平方数了。

我们考虑把 线性基的元素设为 \(|S|\) 个

那么你手头只剩下 \(n-|S|\) 个数字还可以被线性基表示的。

如果可以表示，那么说明了这些 \(2^{n-|S|}-1\) 个子集异或和都可以和线性基拼凑成0

所以目标答案就是 \(2^{n-|S|}-1\)

// by Isaunoya
#include <bits/stdc++.h>
using namespace std;

#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
#define int long long

const int _ = 1 << 21;
struct I {
	char fin[_], *p1 = fin, *p2 = fin;
	inline char gc() {
		return (p1 == p2) && (p2 = (p1 = fin) + fread(fin, 1, _, stdin), p1 == p2) ? EOF : *p1++;
	}
	inline I& operator>>(int& x) {
		bool sign = 1;
		char c = 0;
		while (c < 48) ((c = gc()) == 45) && (sign = 0);
		x = (c & 15);
		while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
		x = sign ? x : -x;
		return *this;
	}
	inline I& operator>>(double& x) {
		bool sign = 1;
		char c = 0;
		while (c < 48) ((c = gc()) == 45) && (sign = 0);
		x = (c - 48);
		while ((c = gc()) > 47) x = x * 10 + (c - 48);
		if (c == '.') {
			double d = 1.0;
			while ((c = gc()) > 47) d = d * 0.1, x = x + (d * (c - 48));
		}
		x = sign ? x : -x;
		return *this;
	}
	inline I& operator>>(char& x) {
		do
			x = gc();
		while (isspace(x));
		return *this;
	}
	inline I& operator>>(string& s) {
		s = "";
		char c = gc();
		while (isspace(c)) c = gc();
		while (!isspace(c) && c != EOF) s += c, c = gc();
		return *this;
	}
} in;
struct O {
	char st[100], fout[_];
	signed stk = 0, top = 0;
	inline void flush() {
		fwrite(fout, 1, top, stdout), fflush(stdout), top = 0;
	}
	inline O& operator<<(int x) {
		if (top > (1 << 20)) flush();
		if (x < 0) fout[top++] = 45, x = -x;
		do
			st[++stk] = x % 10 ^ 48, x /= 10;
		while (x);
		while (stk) fout[top++] = st[stk--];
		return *this;
	}
	inline O& operator<<(char x) {
		fout[top++] = x;
		return *this;
	}
	inline O& operator<<(string s) {
		if (top > (1 << 20)) flush();
		for (char x : s) fout[top++] = x;
		return *this;
	}
} out;
#define pb emplace_back
#define fir first
#define sec second

template < class T > inline void cmax(T & x , const T & y) {
	(x < y) && (x = y) ;
}
template < class T > inline void cmin(T & x , const T & y) {
	(x > y) && (x = y) ;
}

int ans = 0 ;
int p[30] ;
int pri[30] = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 } ;
void ins(int x) {
	for(int i = 19 ; ~ i ; i --) {
		if(x & (1 << i)) {
			if(! p[i]) {
				p[i] = x ;
			}
			x ^= p[i] ;
		}
	}
}
const int mod = 1e9 + 7 ;
int qpow(int x , int y) {
	int ans = 1 ;
	for( ; y ; y >>= 1 , x = x * x % mod)
		if(y & 1) ans = ans * x % mod ;
	return ans ;
}
signed main() {
#ifdef _WIN64
	freopen("testdata.in" , "r" , stdin) ;
#endif
	int n ;
	in >> n;
	rep(i , 1 , n) {
		int x ;
		in >> x ;
		int res = 0 ;
		rep(j , 0 , 18) {
			if(x % pri[j] == 0) {
				int now = 0 ;
				while(x % pri[j] == 0) {
					x /= pri[j] ;
					now ^= 1 ;
				}
				res ^= (now << j) ;
			}
		}
		ins(res) ;
	}
	for(int i = 19 ; ~ i ; i --) if(p[i]) n -- ;
	out << ( qpow(2 , n) - 1 ) % mod << '\n' ;
	return out.flush(), 0;
}