UVA-10892 LCM Cardinality

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.

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
2
12
24
101101291
0

Sample Output
2 2
12 8
24 11
101101291 5

分析：将n质因数分解，得到n=lcm(a,b)=(p1^r1)(p2^r2)(p3^r3)…(pm^rm)，pm小于n。其中，ri=max(ai,bi)。若ai=ri,则bi有ri+1种可能性，若bi=ri，则ai有ri+1种可能性，除去ai=bi=ri，总共有2*ri+1种情况。若n是素数，则ri=0。由于a,b可以互换，则ans=ans/2+1(考虑a=b)。

Source:

#include<stdio.h>int main(){    int n,m,c,p,count,ans;    while(scanf("%d",&n)!=EOF)    {        if(n==0)            return 0;        m=n;        ans=1;        for(p=2;p<n;p+=2)             //质因数分解        {            count=0;            while(n%p==0)            {                count++;                n/=p;            }            ans*=2*count+1;            if(p==2)                p--;        }        if(n>1)                       //考虑n是素数的情况            ans*=3;        ans=ans/2+1;                  //除去a=b的情况        printf("%d %d\n",m,ans);    }    return 0;}