HDU 5833 Zhu and 772002 高斯消元

题目链接

HDU 5833 Zhu and 772002 高斯消元

题解

完全平方数有因子的偶数次幂乘积构成
对于因数个数这就构成了%2意义下的方程组，对于因子列异或方程组
求自由元的自由组合方案数
因为不是很熟悉bitset加上hdu巨坑评测机，然后调了一下午？　

代码

#include<bits/stdc++.h> 
using namespace std; // 每天好心情从namespace开始 
const int maxn = 157; 
#define LL long long 
inline LL read() { 
	LL x = 0,f = 1;
	char c = getchar(); 
	while(c < '0' || c > '9'){if(c == '-')f = - 1;  c = getchar();}  
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar(); 
	return x * f; 
} 
int n; 
const int mod = 1000000007; 
long long A[100007]; 
int prime[100007],is[100007],num = -1; 
void getprime() { 
	for(int i = 2;i <= 2000;++ i) { 
		if(!is[i]) prime[++ num] = i;  
		for(int j = 0;j <= num && prime[j] * i <= 2000;++ j) { 
			is[i * prime[j]] = 1; 
			if(i % prime[j] == 0) break; 
		} 
	} 
} 
bitset<330>a[305]; 
int guass(int equ,int var) { 
	int r,c,t,j = 0; 
	for(r = c = 0;r < equ && c < var;++ r, ++c) { 
		for(t = r;t < equ;++ t) if(a[t][c]) break; 
		if(t == equ) { 
			 r --;continue; 
		} else swap(a[t],a[r]); 
		for(int i = r + 1; i < equ;++ i) if(a[i][c]) a[i] ^= a[r]; 
	} 
	return var - r; 
} 
int solve() { 
	for(int i = 0; i < 305; i++) a[i].reset(); 
	for(int i = 0;i < n;++ i)  
		for(int j = 0;j <= num;++ j) { 
			LL tmp = A[i]; 
			if(tmp % prime[j] == 0) { 
				bool flag = 0; 
				while(tmp % prime[j] == 0) { 
					tmp /= prime[j]; 
					flag ^= 1; 
				} 
				a[j][i] = flag; 
			} 
		} 
	int zyy = guass(num + 1,n); 
	LL ret = 1; 
	for(int i = 1;i <= zyy;++ i)ret <<= 1,ret %= mod; 
	return ret - 1; 
} 
int main() { 
	getprime(); 
	int t = read(); 
	int cas = 0;  
	while(t --) { 
		++ cas; 
		printf("Case #%d:\n",cas); 
		n = read(); 
		for(int i = 0;i < n;++ i) A[i] = read();  
		printf("%d\n",solve()); 
	} 
	return 0; 
}