素数

Description





Webster defines prime as:

prime (prim) n.[ME, fr. MF, fem. of prin first, Lprimus; akin to Lprior] 1 :first in time: original 2 a :
having no factor except itself and one

3 is a number

b :
having no common factor except one

12 and 25 are relatively

3
a : first in rank, authority or significance :principal b : having the highest quality or value

television
time

[from Webster's New Collegiate Dictionary]

The most relevant definition for this problem is 2a: An integer g>1 is said to be prime if and only if its only positive divisors are itself and one (otherwise it is said to be composite). For example, the number 21 is composite;
the number 23 is prime. Note that the decompositon of a positive number g into its prime factors, i.e.,



is unique if we assert that fi > 1 for all i and

for i<j.

One interesting class of prime numbers are the so-called Mersenne primes which are of the form 2p- 1. Euler proved that 231 - 1 is prime in 1772 -- all without the aid
of a computer.

Input

The input will consist of a sequence of numbers. Each line of input will contain one number g in the range -2 31 < g <2 31, but different of -1 and 1. The end of input will be indicated by an input line having a value
of zero.

Output

For each line of input, your program should print a line of output consisting of the input number and its prime factors. For an input number

,
where each fi is a prime number greater than unity (with

for i< j), the format
of the output line should be



When g < 0, if

, the format
of the output line should be

Sample Input

-190
-191
-192
-193
-194
195
196
197
198
199
200
0

Sample Output

-190 = -1 x 2 x 5 x 19
-191 = -1 x 191
-192 = -1 x 2 x 2 x 2 x 2 x 2 x 2 x 3
-193 = -1 x 193
-194 = -1 x 2 x 97
195 = 3 x 5 x 13
196 = 2 x 2 x 7 x 7
197 = 197
198 = 2 x 3 x 3 x 11
199 = 199
200 = 2 x 2 x 2 x 5 x 5

Miguel Revilla

2000-05-19

#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;

int main()
{
long long int n;
while (scanf("%lld",&n) && n)
{
if (n <0)
{
n = -n;
cout << "-"<< n<< " = -1 x ";
}
else
{
cout << n << " = ";
}
long long int k = sqrt((double) n);
for (long long int i=2; i<=k; i++)
{
if (n %i==0)
{
n = n/i;
cout << i<< " x ";
i = 1;
k = sqrt((double) n);
}
}
cout << n<< endl;
}
return 0;
}