HDU5833(2016CCPC網賽)——Zhu and 772002(異或方程組,素數分解)
阿新 • • 發佈:2019-01-26
Zhu and 772002
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 510 Accepted Submission(s): 170
Problem Description Zhu and 772002 are both good at math. One day, Zhu wants to test the ability of 772002, so he asks 772002 to solve a math problem.
But 772002 has a appointment with his girl friend. So 772002 gives this problem to you.
There are n
How many different ways of choices can make b is a perfect square number. The answer maybe too large, so you should output the answer modulo by 1000000007
Input First line is a positive integer T , represents there are T test cases.
For each test case:
First line includes a number n(1≤n≤300),next line there are n numbers a1,a2,...,an,(1≤ai≤1018).
Output For the i-th test case , first output Case #i: in a single line.
Then output the answer of i-th test case modulo by 1000000007
Sample Input 2 3 3 3 4 3 2 2 2
Sample Output Case #1: 3 Case #2: 3 完全平方數就是每一個素因子的個數都是偶數
那麼我們把每個數進行質因數分解,奇數個記為1,偶數個記為0,最後就是求一個異或方程組的解的個數。
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
using namespace std;
const int MAXN =310;
const int INF =1000000007 ;
const int MOD =1000000007;
const double EPS=1e-8;
const double pi = acos(-1);
int prime[MAXN];
int A[MAXN][MAXN];
int num=0;
bool is_prime[2010]={true};
long long powmod(long long a,long long n)//快速冪取膜
{
long long res=1;
while(n){
if(n&1) res=res*a%MOD;
a=a*a%MOD;
n>>=1;
}
return res;
}
long long solve(int m,int n)//根據係數矩陣求自由變元數量
{
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++;
}
long long ans=powmod(2,n-i)-1;
return ans;
}
long long a[MAXN];
void getprime()//求出2000以內素數表
{
memset(is_prime,true,sizeof(is_prime));
is_prime[2]=true;
for(int i=2;i<=2000;i++){
if(is_prime[i]){
prime[num++]=i;
for(int j=2*i;j<=2000;j+=i){
is_prime[j]=false;
}
}
}
}
void get_key(int i)//素數分解
{
long long temp=a[i];
for(int k=0;k<num;k++){
while(temp%prime[k]==0){
A[k][i]^=1;
temp/=prime[k];
}
}
}
int main()
{
int t;
scanf("%d",&t);
getprime();
for(int cas =1;cas<=t;cas++){
int n;
scanf("%d",&n);
memset(A,0,sizeof(A));
printf("Case #%d:\n",cas);
for(int i=0;i<n;i++){
scanf("%I64d",a+i);
get_key(i);
}
printf("%I64d\n",solve(num,n));
}
}