UVA 10892 - LCM Cardinality

Problem F LCM Cardinality Input: Standard Input

Output: Standard Output

Time Limit: 2 Seconds

A pair of numbers has a unique LCM but a single number can be the LCM of more than one possible pairs. For example 12 is the LCM of (1, 12), (2, 12), (3,4) etc. For a given positive integer N, the number of different integer pairs with LCM is equal to N can be called the LCM cardinality of that number N. In this problem your job is to find out the LCM cardinality of a number.

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Input

The input file contains at most 101 lines of inputs. Each line contains an integer N (0<N<=2*10⁹). Input is terminated by a line containing a single zero. This line should not be processed.

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Output

For each line of input except the last one produce one line of output. This line contains two integers N and C. Here N is the input number and C is its cardinality. These two numbers are separated by a single space.

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Sample Input                             Output for Sample Input

2 12 24 101101291 0

2  2 12  8 24  11 101101291  5

#include <iostream>
#include <stdio.h>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
#include <map>
#include <stack>
#include <math.h>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ;
const int M= ;
bool isprime[M+] ;
int prime[M] ,id;
void make_prime(){
    id= ;
    memset(isprime,,sizeof(isprime)) ;
    for(int i=;i<=M;i++){
        if(!isprime[i])
            prime[++id]=i ;
        for(int j=;j<=id&&prime[j]*i<=M;j++){
            isprime[i*prime[j]]= ;
            if(i%prime[j]==)
               break ;
        }
    }
}
LL gao(LL x){
   LL sum ;
   LL ans= ;
   for(int i=;i<=id&&prime[i]*prime[i]<=x;i++){
       if(x%prime[i]==){
            sum= ;
            while(x%prime[i]==){
                sum++ ;
                x/=prime[i] ;
            }
            ans=ans*(sum+sum+) ;
       }
       if(x==)
           break  ;
   }
   if(x!=)
       ans*= ;
   return (ans+)>> ;
}
int main(){
   LL x ;
   make_prime() ;
   while(cin>>x&&x){
       cout<<x<<" "<<gao(x)<<endl ;
   }
   return  ;
}