1059. Prime Factors (25)-PAT

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本文介绍了一种用于分解任意正整数为质因子的方法，并提供了相应的代码实现。

1059. Prime Factors (25)

时间限制

50 ms

内存限制

32000 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

推荐指数：※

来源：http://pat.zju.edu.cn/contests/pat-a-practise/1059

这道题目主要是分解质因子。

1.考虑1的情况。

#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
bool is_prime(long n){
	long i,tmp;
	tmp=sqrt(n)+1;
	for(i=3;i<=tmp;i++){
		if(n%i==0)
			return false;
	}
	return true;
}
int main()
{
	long n,i,tmp,k;
	cin>>n;
	cout<<n<<"=";
	if(n==1){
		cout<<1;
		return 0;
	}
	tmp=n;
	for(i=2;i<=tmp;i++){
		k=0;
		if(is_prime(i)==true){
			while(n%i==0){
				k++;
				n=n/i;
			}
		}
		if(k>1){
			cout<<i<<"^"<<k;
		}
		else if(k==1){
			cout<<i;
		}
		if(n!=1&k>=1)
			cout<<"*";
		else if(n==1)
			break;
	}
	return 0;
}