hdu5833 Zhu and 772002

题解：2000以内的质因数为303个。

以质因数为方程个数，a[i]的取或不取为未知数，解xor的高斯方程

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long LL;
const int M=1000000007;
const int N=300+10;
int h,p[2002];
int n;
int RP(int a,int b)
{
int Ans=1;
for (;b;b>>=1){
if (b&1)Ans=1LL*Ans*a%M;
a=1LL*a*a%M;
}
return Ans;
}
void prime()
{
memset(p,0,sizeof p);
h=0;
for (int i=2;i<=2000;i++)if (!p[i]){
p[h]=i;h++;
for (int j=i;j<=2000;j+=i)p[j]=1;
}
}
typedef int Matrix

;
int Rank(Matrix A,int m,int n){
int i=0,j=0,k,r,u;
while (i<m && j<n){
r=i;
for (k=i;k<m;k++)
if (A[k][j]){r=k;break;}
if (A[r][j]){
if (r!=i)for (k=0;k<=n;k++)swap(A[r][k],A[i][k]);
for (u=i+1;u<m;u++)if (A[u][j])
for (k=i;k<=n;k++)A[u][k]^=A[i][k];
i++;
}
j++;
}
return i;
}
Matrix A;
int main()
{
prime();//printf("%d\n",h);
int Case;scanf("%d",&Case);
for (int C=1;C<=Case;C++){
printf("Case #%d:\n",C);
scanf("%d",&n);
memset(A,0,sizeof A);
int maxp=0;
for (int i=0;i<n;i++){
LL x;cin>>x;
for (int j=0;j<h;j++)while (x%p[j]==0){
maxp=max(j,maxp);x/=p[j];A[j][i]^=1;
}
}
int r=Rank(A,maxp+1,n);
if (r>=n)printf("%d\n",0);
else printf("%d\n",RP(2,n-r)-1);
}
return 0;
}