1059 Prime Factors （25 分）

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×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 = p​1​​^k​1​​*p​2​​^k​2​​*…*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

程式碼

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e5+10; 
typedef long long ll;
ll pri[maxn];
vector<ll>vec; 
void init()
{
	pri[1] = 1;
	vec.push_back(1); 
	for(ll i = 2; i < 500000; i++)
    {
    	if(pri[i] == 0 )
		{
			for(ll j = i + i; j < 500000; j += i)
            	pri[j] = 1;
            vec.push_back(i);
		}	
	}    
	return;
}    
int main()
{
	init();
	ll n;
	scanf("%lld",&n);
	ll size = vec.size();
	printf("%lld=",n);
	for(ll i = 0 ; i < size; i ++)
	{
		if(vec[i] == n)
		{
			printf("%lld\n",n);
			return 0;
		}	
	}
	ll k = n;
	int flag = 0;
	bool state = false;
	for(ll i = 1; i < size && k >= 2; i ++)
	{
		ll cnt = 0;
		while(k % vec[i] == 0)
		{
			cnt ++;
			k = k / vec[i];
		}
		if(cnt >= 1)
		{
			if(state) printf("*");
			printf("%lld",vec[i]);
			if(cnt >= 2)
				printf("^%lld",cnt);
			state = true;
		}
	}
	printf("\n");
    return 0;
}