LCM Cardinality 暴力

LCM Cardinality

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

Submit Status

Description

 

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 LCMof (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 theLCM cardinality of that number N. In this problem your job is to find out the LCM cardinality of a number.

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.

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.

Sample Input                             Output for Sample Input

2 12 24 101101291 0

2 2 12 8 24 11 101101291 5

 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #include<math.h>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 int gcd(int x,int y)
 {
     if(y==)return x;
     return gcd(y,x%y);
 }
 int lcm(int x,int y)
 {
     return x/gcd(x,y)*y;
 }
 long long c[];
 int main()
 {
     int n,nu,i,j,size,ans;
     while(scanf("%d",&n),n)
     {
         ans=nu=;
         size=sqrt(n+0.5);
         for(i=; i<=size; i++)
         {
             if(n%i==)
             {
                 c[nu++]=i;
                 if(i*i!=n)
                     c[nu++]=n/i;
             }
         }
         for(i=; i<nu; i++)
         {
             for(j=i; j<nu; j++)
                 if(lcm(c[i],c[j])==n)ans++;
         }
         cout<<n<<" "<<ans<<endl;
     }
 }