6.CF1881D 质因数分解与重新分配2024-06-05

题目

You are given an array $a$ consisting of $n$ positive integers. You can perform the following operation on it:

Choose a pair of elements $a_{i}$ and $a_{j}$ ( $1 \leq i, j \leq n$ and $i \neq j$ );
Choose one of the divisors of the integer $a_{i}$ , i.e., an integer $x$ such that $a_{i} mod x = 0$ ;
Replace $a_{i}$ with $\frac{a_{i}}{x}$ and $a_{j}$ with $a_{j} \cdot x$ .

Determine whether it is possible to make all elements in the array the same by applying the operation a certain number of times (possibly zero).

For example, let's consider the array $a$ = [ $100, 2, 50, 10, 1$ ] with $5$ elements. Perform two operations on it:

Choose $a_{3} = 50$ and $a_{2} = 2$ , $x = 5$ . Replace $a_{3}$ with $\frac{a_{3}}{x} = \frac{50}{5} = 10$ , and $a_{2}$ with $a_{2} \cdot x = 2 \cdot 5 = 10$ . The resulting array is $a$ = [ $100, 10, 10, 10, 1$ ];
Choose $a_{1} = 100$ and $a_{5} = 1$ , $x = 10$ . Replace $a_{1}$ with $\frac{a_{1}}{x} = \frac{100}{10} = 10$ , and $a_{5}$ with $a_{5} \cdot x = 1 \cdot 10 = 10$ . The resulting array is $a$ = [ $10, 10, 10, 10, 10$ ].

After performing these operations, all elements in the array $a$ become equal to $10$ .
https://codeforces.com/contest/1881/problem/D

Input

The first line of the input contains a single integer $t$ ( $1 \leq t \leq 2000$ ) — the number of test cases.

Then follows the description of each test case.

The first line of each test case contains a single integer $n$ ( $1 \leq n \leq 10^{4}$ ) — the number of elements in the array $a$ .

The second line of each test case contains exactly $n$ integers $a_{i}$ ( $1 \leq a_{i} \leq 10^{6}$ ) — the elements of the array $a$ .

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^{4}$ .

7
5
100 2 50 10 1
3
1 1 1
4
8 2 4 2
4
30 50 27 20
2
75 40
2
4 4
3
2 3 1

Output

For each test case, output a single line:

"YES" if it is possible to make all elements in the array equal by applying the operation a certain (possibly zero) number of times;
"NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

YES
YES
NO
YES
NO
YES
NO

题解

解题思路

显然这是一道数论题，我们观察到，无论怎么进行操作，总数的积不变；

那么操作的本质就是质因数的分解和重新分配，只需让相同质因数的个数是 $n$ 的倍数，就可以平均分配。

Code

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false); cin.tie(nullptr);
using namespace std;
typedef long long LL;

signed main(){
	IOS;
	int _;cin>>_;
	while(_--){
		int n;cin>>n;

		map<int,int> mp;
        //质因数分解
		auto f=[&](int x){
			for(int i=2;i<=sqrt(x);i++){
				while(x%i==0){
					mp[i]++;
					x/=i;
				}
			}
			if(x>1) mp[x]++;
		};

		for(int i=1;i<=n;i++){
			int t;cin>>t;
			f(t);
		}
		bool fl=0;
		for(auto i:mp){
			auto t=i.second;
			if(t%n!=0){
				cout<<"NO"<<endl;
				fl=1;break;
			}
		}
		if(!fl) cout<<"YES"<<endl;
	}
	return 0;
}