1059. Prime Factors (25)

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#PAT 1059 #整数分解 #数论 #素数

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本文介绍了一个用于将正整数分解为其质因数的算法，并提供了完整的C++代码实现。该算法能有效地处理长整型范围内的数值，适用于数学竞赛及编程实践。

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1059. Prime Factors (25)

时间限制

50 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

题意：将一个整数分解为若干素数的乘积，注意N==1的情况即可。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;

typedef long long int ll;

vector< pair<ll,int> > my_vector;

int main(){

	ll num;
	cin >> num;
	cout << num << "=";
	if(num==1){
		cout << 1 << endl;
		return 0;
	}
	ll tmp = sqrt(num)+1;
	int count;
	for(ll i=2; i<=tmp; i++){
		if(num%i==0){
			count = 0;
			do{
				count++;
				num /= i;
			}while(num%i==0);
			my_vector.push_back({i,count});
		}
	}
	if(num!=1)
		my_vector.push_back({num,1});
	int size = my_vector.size();
	for(int i=0; i<size; i++){
		cout << my_vector[i].first;
		if(my_vector[i].second>1)
            cout << "^" << my_vector[i].second;
        if(i+1!=size)
            cout << "*";
	}
    cout <<endl;
	return 0;
}