uva583 Prime Factors（很水很水。。只是纪念一下脑搭的一早晨）

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本文深入探讨了算法和数据结构的基础理论，包括排序、动态规划、哈希算法等核心概念，旨在帮助读者理解算法设计与优化的重要性，并提供解决实际问题的方法论。

B - Prime Factors

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status Practice UVA 583

Description

Prime Factors

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 $\langle$ 3 is a number $\rangle$ b : having no common factor except one $\langle$ 12 and 25 are relatively $\rangle$ 3 a : first in rank, authority or significance : principal b : having the highest quality or value $\langle$ television time $\rangle$ [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.,

$\begin{displaymath}g = f_1 \times f_2 \times \dots \times f_n\end{displaymath}$

is unique if we assert that f_i > 1 for all i and $f_i \le f_j$ for i<j.

One interesting class of prime numbers are the so-called Mersenne primes which are of the form 2^p- 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 $g>0, g = f_1 \times f_2 \times\dots \times f_n$ , where each f _i is a prime number greater than unity (with $f_i \le f_j$ for i< j), the format of the output line should be

$\begin{displaymath}g \mbox{\tt\ = } f_1 \mbox{\tt\ x } f_2 \mbox{\tt\ x } \dots \mbox{\tt\ x } f_n\end{displaymath}$

When g < 0, if $\mid g \mid = f_1 \times f_2 \times \dots \times f_n$ , the format of the output line should be

$\begin{displaymath}g \mbox{\tt\ = -1 x } f_1 \mbox{\tt\ x } f_2 \mbox{\tt\ x } \dots\mbox{\tt\ x } f_n\end{displaymath}$

Sample Input

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

题意：将正数表成 $\begin{displaymath}g \mbox{\tt\ = } f_1 \mbox{\tt\ x } f_2 \mbox{\tt\ x } \dots \mbox{\tt\ x } f_n\end{displaymath}$ 这种递增素数相乘的形式

同理，将负数表成 $\begin{displaymath}g \mbox{\tt\ = -1 x } f_1 \mbox{\tt\ x } f_2 \mbox{\tt\ x } \dots\mbox{\tt\ x } f_n\end{displaymath}$ 这种递增素数相乘的形式

不考虑输入为 1 或者 -1 的情况。

#include 
  
   
#include 
   
    

int pri[47000],vis[47000],num=0;

void getPrime(int n)
{	
	for(int i=2;i
    
     n)
			{
				break;
			}
			vis[i*pri[j]]=1;
			if(i%pri[j]==0)
			{
				break;
			}
		}
	}

}

int main()
{
	long long n;
	getPrime(46340);// 46340是比sqrt(2^31)大一点的一个数。。
	while(scanf("%lld",&n)!=EOF && n) //%lld和%I64d的问题。。调试的时候要用%I64d，提交的时候一定要改成%lld
	{
		printf("%lld =",n);
		int flag=1;
		if(n<0)
		{
			printf(" %d",-1);
			n=n*-1;
			flag=0;
		}
		for(int i=0;i