pata1059（一不小心就超时，再一不小心就漏掉测试点！！！）

本文提供了一个质因数分解的算法实现，该算法能够接收一个正整数N作为输入，并将其分解为所有质因数的形式，即N=p1^k1 *p2^k2 *…*pm^km，其中pi为N的质因数，ki为pi的指数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1059. Prime Factors (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HE, Qinming

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p₁^k₁ * p₂^k₂ *…*p_m^k_m.

Input Specification:

Each input file contains one test case which gives a positive integer N in the range of long int.

Output Specification:

Factor N in the format N = p₁^k₁ * p₂^k₂ *…*p_m^k_m, where p_i's are prime factors of N in increasing order, and the exponent k_i is the number of p_i -- hence when there is only one p_i, k_i is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

#include<iostream>
#include<cmath>
#include<vector>
using namespace std;
typedef long long ll;
bool isprime(ll a){
	ll _end=(ll)sqrt(a);
	for(ll i=2;i<=_end;i++){
		if(a%i==0){
			return false;
		}
	}
	return true;
}
int main()
{
	ll n;
	cin>>n;
	vector<ll> v1;
	ll _end1=(ll)sqrt(n);
	for(ll i=2;i<=_end1;i++){
		if(isprime(i)&&(n%i==0)){
			v1.push_back(i);
		}
	}
	cout<<n<<"=";
	if(n==1){
		cout<<n;
	}else if(isprime(n)){
		cout<<n;
	}else{
		for(ll i=0;i<v1.size()&&n!=0;i++){
		ll count=0;
		while(n%v1[i]==0){
			count++;
			n/=v1[i];
		}
		if(count==1){
			cout<<v1[i];
		}else{
			cout<<v1[i]<<"^"<<count;
		}
		if(i<v1.size()-1){
			cout<<"*";
		}
	}
	}
	
	return 0;
}