PAT(甲级)1059

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

#include<iostream>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;
//////////////////////////////////////////////////////
//know the way to use recurssive
//
///////////////////////////////////////////////////////
bool isprime(int n){
	if(n == 1)
	    return false;
    if(n == 2)
        return true;
    else if(n & 0x1){
    	int limit = sqrt(n);
	    for(int i =3;i<=limit;i +=2)
	        if(n % i ==0)
	            return false;
	    return true;
	}else
	    return false;
}

vector <int> v;

void dfactor(int n){
	if(n <=1)
	    return ;
	if(isprime(n)){
		v.push_back(n);
		return;
	}else{
		for(int i=2;i<n;i++){
			if(n % i ==0){
				dfactor(i);
			    dfactor(n/i);
			    return ;
			}
		}
	}
}

int main()
{
	int n;
	cin >>n;
	if(n==1){
		cout <<"1=1" <<endl;
		return 0;
	}
	dfactor(n);
	int i = v.size();
	vector<int>::iterator iter = v.begin();
	sort(v.begin(),v.end());
	cout <<n <<'=';
	int last = *iter;
	cout <<last;
	i--;
	iter++;
	int count=1;
	while(i){
		if(last != *iter){
		    if(count != 1){
		    	cout <<'^' <<count;
		    	count =1;
			}    
		    cout <<'*' <<*iter;
		    last = *iter;			
	    }
		else{
			count++;
		}
		iter++;
		i--;
	}
	if(count != 1)
	    cout <<'^' <<count;
	cout <<endl;
	return 0;
}