标签:
题目链接:
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1006 Accepted Submission(s): 348
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=2010;
const double eps=1e-12;
int vis[maxn],cnt=0,m,a[maxn][maxn],n;
LL prime[maxn];
inline void get_prime()
{
for(int i=2;i<maxn;i++)
{
if(!vis[i])
{
for(int j=2*i;j<maxn;j+=i)vis[j]=1;
prime[++cnt]=(LL)i;
}
}
}
int gauss()
{
int k,i=1,j=1;
while(i<=m&&j<=n)
{
k=i;
while(!a[k][j]&&k<=m)k++;
if(a[k][j])
{
if(k!=i)for(int x=0;x<=n;x++)swap(a[k][x],a[i][x]);//cout<<"@@@"<<endl;
for(int x=i+1;x<=m;x++)
{
if(a[x][j])
for(int u=0;u<=n;u++)a[x][u]^=a[i][u];
}
i++;
}
j++;
}
return n-i;
}
int main()
{
get_prime();
int t,Case=0;
LL x;
read(t);
while(t--)
{
mst(a,0);
m=0;
read(n);
For(i,1,n)
{
read(x);
for(int j=1;j<=cnt;j++)
{
while(x%prime[j]==0){a[j][i]^=1;x/=prime[j];m=max(m,j);}
}
}
n++;
int num=gauss();
LL ans=1;
for(int i=1;i<=num;i++)ans=2*ans%mod;
ans=(ans-1+mod)%mod;
printf("Case #%d:\n%lld\n",++Case,ans);
}
return 0;
}
标签:
原文地址:http://www.cnblogs.com/zhangchengc919/p/5774564.html