HDU 5833 Zhu and 772002（高斯消元）——2016中国大学生程序设计竞赛 - 网络选拔赛

传送门

Zhu and 772002

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 48 Accepted Submission(s): 16


[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

题目大意：

给出 n 个整数，从中选出 1 个或者多个，使得选出的整数乘积是完全平方数。一共有多少种选

法？对 MOD 取模。

解题思路：

赛后群里说 原题，已然蒙逼，刘汝佳白书 P160 , 一样一样的！！！，唉，亏我们还在推了那么

久。。。

其实我是做过一个类似的题所以没用多长时间，那个题目链接传送门

其实，那个题跟这个题目是差不多的，看那个题目就可以了，这个题目就是在那个题目的基础上先

处理一下就行了。。。

My Code：

/**
2016 - 08 - 14 下午
Author: ITAK

Motto:

今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 1e9+5;
const LL MAXN = 2e3+5;
const LL MOD = 1e9+7;
const double eps = 1e-7;
const double PI = acos(-1);
using namespace std;
/**+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/

LL equ, var;///equ个方程 var个变量
LL a[MAXN][MAXN];///增广矩阵
LL x[MAXN];///解的数目
bool free_x[MAXN];///判断是不是自由变元
LL free_num;///自由变元的个数
inline LL GCD(LL m, LL n)
{
if(n == 0)
return m;
return GCD(n, m%n);
}
inline LL LCM(LL a, LL b)
{
return a/GCD(a,b)*b;
}

LL Gauss()
{
LL Max_r;///当前列绝对值最大的存在的行
///col：处理当前的列
LL row=0;
for(LL col=0; row<equ&&col<var; row++,col++)
{
Max_r = row;
for(LL i=row+1; i<equ; i++)
if(abs(a[i][col]) > abs(a[Max_r][col]))
Max_r = i;

if(Max_r != row)
for(LL i=0; i<var+1; i++)
swap(a[row][i], a[Max_r][i]);

if(a[row][col] == 0)
{
row--;
continue;
}
for(LL i=row+1; i<equ; i++)
{
if(a[i][col])
{
for(LL j=col; j<var; j++)
{
a[i][j] ^= a[row][j];
}
}
}
}
return row;
}
const LL MAX = 2e3+5;
LL p[MAX];
bool prime[MAX];
LL k;
void isprime()
{
k = 0;
memset(prime, false, sizeof(prime));
for(LL i=2; i<MAX; i++)
{
if(!prime[i])
{
p[k++] = i;
for(LL j=i*i; j<MAX; j+=i)
prime[j] = true;
}
}
}
LL quick_mod(LL a, LL b)
{
LL ans = 1;
while(b)
{
if(b & 1)
ans = (ans*a)%MOD;
b>>=1;
a = (a*a)%MOD;
}
return ans;
}
int main()
{
isprime();
equ = k;
LL T, n;
scanf("%I64d",&T);
for(LL cas=1; cas<=T; cas++)
{
cin>>var;
memset(a, 0, sizeof(a));
for(LL i=0; i<var; i++)
{
LL x;
LL sum;
scanf("%I64d",&x);
for(LL j=0; j<equ; j++)
{
sum = 0;
if(x%p[j] == 0)
{
LL mm = x;
while(mm%p[j]==0)
{
sum++;
mm /= p[j];
}
}
///构造系数矩阵
if(sum & 1)
a[j][i] = 1;
else
a[j][i] = 0;
}
}
LL ans = var - Gauss();
LL ret = quick_mod(2LL, ans);
ret--;
ret = (ret%MOD+MOD)%MOD;
printf("Case #%I64d:\n%I64d\n",cas,ret);
}
return 0;
}