A1059. Prime Factors (25)

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 <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
using namespace std;


const int maxn=100000;
int prime[maxn],pnum=0;
bool p[maxn]={0};
void Find_Prime(){
    for(int i=2;i<maxn;i++)
    {
        if(p[i]==false)
        {
            prime[pnum++]=i;
            for(int j=i+i;j<maxn;j+=i)
            {
                p[j]=true;
            }
        }
    }
} 

struct factor{
    int x,cnt;
}fac[10];
int main(){
   Find_Prime();
   int n,num=0;
   scanf("%d",&n);
   if(n==1){ 
   printf("1=1");
   return 0;
   }else{
       printf("%d=",n);
   }
   
   
   int sqr=(int)sqrt(1.0*n);
   int count=0;
   for(int i=2;i<=sqr;)
   {
       if(n%i==0)
       {
           fac[num].x=i;
           fac[num].cnt=0;
           while(n%i==0)
           {
               fac[num].cnt++;
               n/=i;
           }
           num++;    
       }
       
       count++;
       i=prime[count];
   }
   if(n>sqr)
   {
       fac[num].x=n;
       fac[num++].cnt=1;
   } 
   for(int i=0;i<num;i++)
   {
       if(i>0)printf("*");
       printf("%d",fac[i].x);
       if(fac[i].cnt>1)
       {
           printf("^%d",fac[i].cnt);
       }
   } 
    return 0;
}