UVA11542 Square(高斯消元 异或方程组)

Posted 2020-08-12

 

建立方程组消元，结果为2 ^(自由变元的个数) - 1

采用高斯消元求矩阵的秩

 

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 108, INF = 0x3F3F3F3F;
const double eps = 1e-8;
int a[N][N];

template<typename T>
int gauss_jordan(T A[N][N], int n, int m){
    int i, c;
    for(i = 0, c = 0; i < n && c < m; i++, c++){
        int r = i;
        for(int j = i + 1; j < n; j++){
            if(A[j][c]){
                r = j;
                break;
            }
        }
        if(A[r][c] == 0){
            i--;
            continue;
        }
        if(r != i){
            for(int j = 0; j <= m; j++){
                swap(A[r][j], A[i][j]);
            }
        }
        for(int k = 0; k < n; k++){
            if(k != i && A[k][c]){
                for(int j = m; j >= c; j--){
                    A[k][j] ^= A[i][j];
                }
            }
        }
    }
    return i;
}

const int MAXN = 508;
int prime[MAXN];
bool vis[MAXN];
int getPrime(int n){//求1~n的素数
	int tot=0;
	memset(vis,0,sizeof(vis));
	for(int i=2;i<=n;i++){
		if(!vis[i]){
            prime[tot++]=i;
		}
		for(int j=0;j<tot&&i*prime[j]<=n;j++){
			vis[i*prime[j]]=true;
			if(i%prime[j]==0){//让每个合数仅被其最小的质数筛去
                break;
			}
		}
	}
	return tot;
}

int main(){
    int cnt = getPrime(500);
    int t;
    cin>>t;
    while(t--){
        memset(a, 0, sizeof(a));
        int n;
        cin>>n;
        for(int j = 0; j < n; j++){
            LL x;
            cin>>x;
            for(int i = 0; i < cnt && prime[i]<= x; i++){
                while(x % prime[i] == 0){
                    a[i][j] ^= 1;
                    x /= prime[i];
                }
            }
        }
        LL ans = n - gauss_jordan(a, cnt, n);
        //cout<<ans<<"  ans\n";
        cout<<((1ll << ans) - 1)<<‘\n‘;
    }

    return 0;
}