Acwing game 57

4485. Than the size

Given a length of n Array of a1,a2,…,an And a length of n Array of b1,b2,…,bn.

Please calculate , Array a Whether the sum of all elements of the array is greater than or equal to the array b The sum of all the elements of .

Input format
The first line contains integers n.

The second line contains n It's an integer a1,a2,…,an.

The third line contains n It's an integer b1,b2,…,bn.

Output format
If the array a The sum of all elements of is greater than or equal to the array b The sum of all the elements of , The output Yes, Otherwise output No.

Data range
The first three test points meet 1≤n≤5.
All test points meet 1≤n≤50,0≤ai,bi≤1000.

sample input 1：
5
1 2 3 4 5
2 1 4 3 5
sample output 1：
Yes
sample input 2：
5
1 1 1 1 1
1 0 1 0 1
sample output 2：
Yes
sample input 3：
3
2 3 9
1 7 9
sample output 3：
No

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// Author: Changersh
// Problem:  Than the size 
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/4488/
// When: 2022-06-29 14:28:38
// 
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 


#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include<tr1/unordered_map>
#include <tr1/unordered_set>

using namespace std::tr1;
using namespace std;
typedef long long ll;
const int N = 5e4 + 50;
const int MOD = 1e9 + 7;

int n, a, b, sum1, sum2;
int main() {
    
	// scanf("%d", &T);while (T--)solve();
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
    
		scanf("%d", &a);
		sum1 += a;
	}
	for (int i = 0; i < n; i++) {
    
		scanf("%d", &b);
		sum2 += b;
	}
	if (sum1 >= sum2)printf("Yes\n");
	else printf("No\n");
	return 0;
}

4486. Digital operation

Given an integer n, You can do this number any number of times （ It can be 0 Time ） Change operation .

Each operation is one of the following two ：

The integer n Multiply by any positive integer x.
The integer n Replace with n√（ The precondition for this operation is n√ Integers ）.
Please calculate , By doing this ,n The lowest possible value that can be reached , And the minimum number of operations required to reach the minimum possible value .

Input format
An integer n.

Output format
a line , Two integers , Respectively n The lowest possible value that can be reached , And the minimum number of operations required to reach the minimum possible value .

Data range
All test points meet 1≤n≤106.

sample input 1：
20
sample output 1：
10 2
sample input 2：
5184
sample output 2：
6 4

number theory Decomposing the prime factor

Prime factorization theorem ： A number can be decomposed into its prime factors / The product of the same power
Two kinds of operations , One is multiplied by a number x, This does not change the number of types of qualitative factors
The second is to open the square , In fact, the number of times of each prime factor is divided by two , It will not change the number of types of qualitative factors
So the smallest possible value is The product of all prime factors .

Take a 2^m, More than all times , Because operation two , The root opening sign is actually the number of all prime factors divided by two .

And multiply at every step you need to multiply x In fact, it is equal to multiplying a total at the beginning x , This is a small number of steps , So put it in the first step

Actually, the number of times is m Value , Because you have to write a double root , open m Time .
Find the prime factor by trial division , Remember to divide all the prime factors , Calculate the index of the current quality factor ,2^m>= The largest index .
after , If n>1, explain n It is its own qualitative factor , The index is 1 .
The previous content means that all quality factor indexes are the same by default , But generally it is different , As long as there is a prime factor less than 2^m, On behalf of this n Smaller than what we constructed N, So you need to multiply by a total X, They count + 1,m++

// Author: Changersh
// Problem:  Digital operation 
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/4489/
// When: 2022-06-29 14:58:27
//
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <map>
#include <set>
#include <string>
#include <tr1/unordered_map>
#include <tr1/unordered_set>
#include <vector>

using namespace std::tr1;
using namespace std;
typedef long long ll;
const int N = 5e4 + 50;
const int MOD = 1e9 + 7;

//  Decomposing the prime factor 

int main() {
    
  int n;
  scanf("%d", &n);
  vector<int> s;  //  The number of existential factors 
  int res = 1;
  int m = 0; //  The maximum number of times should be less than or equal to  2 Integer power of 
  //  Try to divide into prime factors 
  for (int i = 2; i * i <= n; i++) {
    
    if (n % i == 0) {
    
    	int c = 0;  //  frequency 
    	while (n % i == 0) n /= i, c++;
    	res *= i; //  result 
    	s.push_back(c);
    	while (1 << m < c) m++;
    }
  }
  if (n > 1) {
     //  explain  n  Prime number 
  	res *= n;
  	s.push_back(1);
  	while (1 << m < 1) m++;
  }
  for (auto x : s) {
    
  	if (x < 1 << m) {
    
  		m++;
  		break;
  	}
  }
	printf("%d %d", res, m);
	
	
  return 0;
}