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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;
}
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原文地址:http://www.cnblogs.com/IMGavin/p/5930604.html