HDU 5833 Zhu and 772002 （高斯消元）

Zhu and 772002

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 706    Accepted Submission(s): 240

[align=left]Problem Description[/align]
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
numbers a1,a2,...,an.
The value of the prime factors of each number does not exceed
2000,
you can choose at least one number and multiply them, then you can get a number
b.

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.
[align=left]Input[/align]
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).
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]

2
3
3 3 4
3
2 2 2

[align=left]Sample Output[/align]

Case #1:
3
Case #2:
3

[align=left]Author[/align]
UESTC
[align=left]Source[/align]
2016中国大学生程序设计竞赛
- 网络选拔赛
[align=left]Recommend[/align]
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题意：给你n个数，选择一些数字乘积为平方数的选择方案数。

题解：将原题化成异或方程组，并用高斯消元求解矩阵的秩.

每一个数字分解质因数。比如4,6,10,15。

4 = 2^2 * 3^0 * 5^0 . (^在这里是指数)

6 = 2^1 * 3^1 * 5^0.
10 = 2^1 * 3^0 *  5^1.
15 = 2^0 * 3^1 * 5^1.
令Xi表示选择第i个数字，那么，如果p=2^(2x1+x2+x3) * 3^(x2+x4) * 5^(x3+x4) 是平方数，那么每个质因数上的指数为偶数，x1系数为2已经是偶数不考虑。可以转换为异或为0判断偶数，即奇数置为1，偶数置为0，然后n个数字m个质因数的增广矩阵消元（gauss消元）看有几个自由变量(取0或1无所谓)，答案是2^r - 1(全部都不取方案不算)（快速幂ok）。

另外：可以看一下卿学姐写的代码，觉得用了bitset很好，代码简便。点我打开链接

代码: 

#pragma comment(linker, "/STACK:102400000,102400000")
//#include<bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<map>
#include<cmath>
#include<queue>
#include<set>
#include<stack>
#include <utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define mst(a) memset(a, 0, sizeof(a))
#define M_P(x,y) make_pair(x,y)
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
const int lowbit(int x) { return x&-x; }
const double eps = 1e-8;
const int INF = 1e9+7;
const ll inf =(1LL<<62) ;
const int MOD = 1e9 + 7;
const int mod = 1000000007;
const int N = 2010;
const int M=100010;
template <class T1, class T2>inline void getmax(T1 &a, T2 b) {if (b>a)a = b;}
template <class T1, class T2>inline void getmin(T1 &a, T2 b) {if (b<a)a = b;}
int read()
{
int v = 0, f = 1;
char c =getchar();
while( c < 48 || 57 < c ){
if(c=='-') f = -1;
c = getchar();
}
while(48 <= c && c <= 57)
v = v*10+c-48, c = getchar();
return v*f;
}
typedef int Matrix

;
int prime
, vis
;
Matrix A;
int init(int m)
{
memset(vis, 0, sizeof(vis));
int ans = 0;
for (int i = 2; i < m; i++){
if (!vis[i]) {
prime[ans++] = i;
for (int j = i * i; j < m; j += i)
vis[j] = 1;
}
}
return ans;
}
int gauss(Matrix A, int m, int n) {
int i = 0, j = 0 ;
int 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;
}
ll q_mod(ll a,ll b,ll m)
{
ll ans = 1;
while(b)
{
if(b&1)
ans = (ans * a) % m;
b>>=1;
a = a * a % m;
}
return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
int m = init(2010);
int t;
int cas= 1;
t=read();
while (t--)
{
int n;
int maxx= 0;
ll x;
n=read();
memset(A, 0, sizeof(A));
for (int i = 0; i < n; i++)
{
scanf("%lld", &x);
for (int j = 0; j < m; j++)
while (x % prime[j] == 0)
{
maxx = max(maxx, j);
x /= prime[j];
A[j][i] ^= 1;
}
}
int r = gauss(A, maxx+1, n);
int ans = q_mod(2, (ll)(n-r), mod);
printf("Case #%d:\n%d\n", cas++ , ans-1);
}
return 0;
}