Square Number & Cube Number

Square Number:

Description

In mathematics, a square number is an integer that is the square of an integer. In other words, it is the product of some integer with itself. For example, 9 is a square number, since it can be written as 3 * 3.

Given an array of distinct integers (a1, a2, ..., an), you need to find the number of pairs (ai, aj) that satisfy (ai * aj) is a square number.

Input

The first line of the input contains an integer T (1 ≤ T ≤ 20) which means the number of test cases.

Then T lines follow, each line starts with a number N (1 ≤ N ≤ 100000), then N integers followed (all the integers are between 1 and 1000000).

Output

For each test case, you should output the answer of each case.

Sample Input

1
5
1 2 3 4 12

Sample Output

2

Cube Number:

Description

In mathematics, a cube number is an integer that is the cube of an integer. In other words, it is the product of some integer with itself twice. For example, 27 is a cube number, since it can be written as 3 * 3 * 3.

Given an array of distinct integers (a1, a2, ..., an), you need to find the number of pairs (ai, aj) that satisfy (ai * aj) is a cube number.

Input

The first line of the input contains an integer T (1 ≤ T ≤ 20) which means the number of test cases.

Then T lines follow, each line starts with a number N (1 ≤ N ≤ 100000), then N integers followed (all the integers are between 1 and 1000000).

Output

For each test case, you should output the answer of each case.

Sample Input

1
5
1 2 3 4 9

Sample Output

2

题意：

给你一列数，问两个数相乘组成（平方数&立方数）的种数有多少

题解：

对于平方数来说，每个平方数都能分解成若干质数的平方，所以枚举所有的素数，如果出现偶次幂直接忽略，若是奇数次幂，打表统计；

对于立方数来说，每个平方数都能分解成若干质数的立方，枚举所有的素数，若出现三次幂忽略，然后剩下的有两种情况：

例如剩下的一个数可以分解成三个质数a*b^2(a,b均为质数),那么他只能和a^2*b匹配；

代码（square number）：

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <deque>
#include <stack>
#include <iomanip>
#include <cstdlib>
using namespace std;
#define is_lower(c) (c>='a' && c<='z')
#define is_upper(c) (c>='A' && c<='Z')
#define is_alpha(c) (is_lower(c) || is_upper(c))
#define is_digit(c) (c>='0' && c<='9')
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define IO ios::sync_with_stdio(0);\
    cin.tie();\
    cout.tie();
#define For(i,a,b) for(int i = a; i <= b; i++)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const ll inf=0x3f3f3f3f;
;
const ll inf_ll=(ll)1e18;
const ll mod=1000000007LL;
;
];
int prime[maxn],prime1[maxn];
];
];
;
void getprime()
{
    memset(vis, false, sizeof(vis));
    int N = sqrt(maxn);
    ; i <= N; ++i)
    {
        if ( !vis[i] )
        {
            prime[++num] = i;
            prime1[num] = i*i;
        }
        ; j <= num && i * prime[j] <= N ;  j++)
        {
            vis[ i  *  prime[j] ]  =  true;
            ) break;
        }
    }
}
int main()
{
    int T;
    cin>>T;
    getprime();
    while(T--)
    {
        int x;
        memset(cnt,,sizeof(cnt));
        cin>>x;
        For(i,,x)
        {
            int xx;
            cin>>xx;
            ; xx>=prime1[j]&&j<=num; j++)
            {
                )
                    xx/=prime1[j];
            }
            cnt[xx]++;
        }
        ll ans = ;
         For(i,,maxn-)
        if(cnt[i])
            ans+= cnt[i]*(cnt[i]-)/;
        cout<<ans<<endl;
    }
    ;
}

Cube Number:

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <deque>
#include <stack>
#include <iomanip>
#include <cstdlib>
using namespace std;
#define is_lower(c) (c>='a' && c<='z')
#define is_upper(c) (c>='A' && c<='Z')
#define is_alpha(c) (is_lower(c) || is_upper(c))
#define is_digit(c) (c>='0' && c<='9')
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define IO ios::sync_with_stdio(0);\
    cin.tie();\
    cout.tie();
#define For(i,a,b) for(int i = a; i <= b; i++)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const ll inf=0x3f3f3f3f;
;
const ll inf_ll=(ll)1e18;
const ll mod=1000000007LL;
;
];
ll prime[maxn],prime_2[maxn],prime_3[maxn];
];
ll cnt[maxn+];
;
void getprime()
{
    memset(vis, false, sizeof(vis));
    int N = maxn;
    ; i <= N; ++i)
    {
        if ( !vis[i] )
        {
            prime[++num] = i;
            prime_2[num] = i * i;
            prime_3[num] = i * i * i;
        }
        ; j <= num && i * prime[j] <= N ;  j++)
        {
            vis[ i  *  prime[j] ]  =  true;
            ) break;
        }
    }
}
int main()
{
    int T;
    cin>>T;
    getprime();
    while(T--) {
        int x;
        ;
        cin >> x;
        memset(cnt, , sizeof(cnt));
        ,xx; i <= x && cin>>xx; i++) {
            ; j <= num && xx >= prime_3[j]; j++)
                )
                    )
                        xx /= prime_3[j];
            cnt[xx]++;
            ) {
                res += cnt[xx] - ;
                continue;
            }
            ;
            ; j <=num && xx >= prime_2[j]; j++)
                )
                    ) {
                        xx /= prime_2[j];
                        tem *= prime_2[j];
                    }
            ) { // 大于1000的素数的平方一定不存在；
                xx = sqrt(tem) * xx * xx; // 和另一半匹配，一定不大于maxn 加条件判断；
                if(xx < maxn)
                    res += cnt[xx];
            }
        }
        cout << res << endl;
    }
    ;
}