Codeforces Round #429 (Div. 2) E. On the Bench

E. On the Bench

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A year ago on the bench in public park Leha found an array of n numbers. Leha believes that permutation p is right if for all 1 ≤ i < n condition, that a_pi·a_pi + 1 is not perfect square, holds. Leha wants to find number of right permutations modulo 10⁹ + 7.

Input

First line of input data contains single integer n (1 ≤ n ≤ 300) — length of the array.

Next line contains n integers a₁, a₂, ... , a_n (1 ≤ a_i ≤ 10⁹) — found array.

Output

Output single integer — number of right permutations modulo 10⁹ + 7.

Examples

Input

3
1 2 4

Output

2

Input

7
5 2 4 2 4 1 1

Output

144

Note

For first example:

[1, 2, 4] — right permutation, because 2 and 8 are not perfect squares.

[1, 4, 2] — wrong permutation, because 4 is square of 2.

[2, 1, 4] — wrong permutation, because 4 is square of 2.

[2, 4, 1] — wrong permutation, because 4 is square of 2.

[4, 1, 2] — wrong permutation, because 4 is square of 2.

[4, 2, 1] — right permutation, because 8 and 2 are not perfect squares.

题意 ：求相邻的元素相乘不为平方数的方案数（a[0]=a[1]=1, 视 a[0] 与 a[1]  不同)

思路 ：

　　　　每个数可以表示为  p1^a1 * p2^a2 * .....

　　　　如果 两个数A,B相乘为平方数 则   a1%2 = a1' %2  ,  a2%2 = a2'%2 .....

　　　　即 对应质因子的幂次 奇偶性相同 这样就可以划分出T组

　　　　然后题目就转化为 T种物品 相同种类物品不能放在相邻 求方案数

　　　　这题就变成原题 ：https://csacademy.com/contest/archive/task/distinct_neighbours/statement/

　　　　　　　　　　　　http://acm.hdu.edu.cn/showproblem.php?pid=6116

　　　　做法为dp

　　　　dp [ i ] [ j ]  表示  插入第 i 组的物品 出现了 左右为相同物品的空隙个数为 j   的方案数

　　　　那 dp [ T ]  [ 0 ]  就是最终答案了

　　　　附： cs官方题解 （原题的题解）

　　　　　　First we group all the distinct values in the array. Then we can solve the problem using dynamic programming:

　　　　　　Let dp[i][j] = the number of distinct arrays that can be obtained using the elements of the first i groups such that there are exactly j pairs of consecutive positions having the same value. The answer can be found in dp[distinctValues][0].

　　　　　　Now let's say the sum of frequences of the first i values is X. This means the arrrays we can build using the elements from these i groups have size X, so we can insert the elments of group i + 1 in X + 1 gaps: before the first element, after the last, and between any two consecutive. We can fix the number k of gaps where we want to insert at least one element from group i + 1, but we also need to fix the number l of these k gaps which will be between elements that previously had the same value. State dp[i][j] will update the state dp[i + 1][j - l + frequence[i + 1] - k].

　　　　　　The number of ways of choosing k gaps such that exactly l are between consecutive elements having the same value can be computed easily using combination formulas. We　 are left with finding out the number of ways of inserting frequence[i + 1] elements in k gaps. This is also a classical problem with a simple answer: Comb(frequence[i + 1] - 1, k - 1).

　　　　具体转移方程见代码 ：

　　　　

　　　　

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
#include <assert.h>
#include <math.h>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
#include <functional>

#define mp make_pair
#define pb push_back
#define mes(a,b) memset(a,b,sizeof(a))
#define mes0(a) memset(a,0,sizeof(a))
#define lson l,mid,pos<<1
#define rson mid+1,r,pos<<1|1
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define fi first
#define se second
#define sss(a) a::iterator
#define all(a) a.begin(),a.end()

using namespace std;

typedef double DB;
typedef long long LL;
typedef pair<int,int> pii;
typedef pair<long long ,int> pli;
typedef pair<int,long long > pil;
typedef pair<string,int> psi;
typedef pair<long long ,long long > pll;

const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const int maxn = +;
const int mod = 1e9+;
LL dp[][];
LL C[][];
LL fact[];
int cnt[];
int a[];
int vis[];
int sz;
int check(LL x)
{
    LL l=,r=1e9;
    LL now=l;
    while (l<=r){
        LL mid=(l+r)>>;
        if (mid*mid<=x)now=mid,l=mid+;
        else r=mid-;
    }
    return now*now==x;
}
inline LL M(LL x)
{
    return x%mod;
}
void init()
{
    C[][]=;
    fact[]=;
    for (int i=;i<=;i++){
        C[i][]=;
        for (int j=;j<=i;j++){
            C[i][j]=M(C[i-][j]+C[i-][j-]);
        }
    }
    for (int i=;i<=;i++)fact[i]=M(fact[i-]*i);
}
void slove()
{
    dp[][cnt[]-]=;
    int lim=cnt[];
    for (int i=;i<sz;i++){
        for (int j=;j<lim;j++){  /// dp[i-1][j]
            for (int k=;k<cnt[i];k++){/// group[i]分成k+1组 ,cnt[i]-1-k个空隙
                for (int m=;m<=min(j,k+);m++){ /// 选了m个左右相同的空隙插入
                    dp[i][j+cnt[i]--k-m]=M(dp[i][j+cnt[i]--k-m]+dp[i-][j]*C[cnt[i]-][k]%mod*C[j][m]%mod*C[lim--j+][k+-m]);
                }
            }
        }
        lim+=cnt[i];
    }
    LL ans=dp[sz-][];
    for (int i=;i<sz;i++){
        ans*=fact[cnt[i]];
        ans%=mod;
    }
    cout<<ans<<endl;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    scanf("%d",&n);
    for (int i=;i<n;i++){
        scanf("%d",a+i);
    }
    for (int i=;i<n;i++){
        if (vis[i]==){
            for (int j=i;j<n;j++){
                if (check(1LL*a[i]*a[j])){
                    cnt[sz]++;
                    vis[j]=;
                }
            }
            sz++;
        }
    }
    init();
    slove();
    return ;
}