uva 10892 LCM Cardinality

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原题:
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∗10 9 ).
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
中文：
给你一个数n，让你找出能形成最小公倍数为n的两个数的个数。

#include <bits/stdc++.h>using namespace std;long p[150001];bool tag[150001];int get_prime(){    long cnt=0;    for(long i=2;i<150001;i++)    {        if(!tag[i])            p[cnt++]=i;        for(long j=0;j<cnt&&p[j]*i<150001;j++)        {            tag[i*p[j]]=1;            if(i%p[j]==0)                break;        }    }    return cnt;}long decomposition(long x){    long k=0,cnt,res=1;    while(x>1)    {        cnt=0;        while(x%p[k]==0)        {            x/=p[k];            cnt++;        }        res*=(cnt*2)+1;        k++;    }    return res;}int main(){    ios::sync_with_stdio(false);    cout<<get_prime();    long n;    while(cin>>n,n)    {        if(n==1)            cout<<n<<" "<<1<<endl;        else        {            long ans=decomposition(n);            cout<<n<<" "<<ans/2+1<<endl;        }    }    return 0;}

解答：
先在纸上写一写发现，如果两个数要是能形成最小公倍数为n，那么必须要满足如下的形式。
首先把n用唯一分解定理分解可以得到
n=pa11pa22…pann
如果要想有两个数a，b的公倍数的是n。那么a用唯一分解定理分解，b用唯一分解定理分解，a和b的相同素因子的系数取最大的那个必须要等于ai。
也就是最小公倍数的那个公式
n=pmax{x1,y1}1pmax{x2,y2}2…pmax{xn,yn}n
其中xi为a分解后素因子的系数，yi为b分解后的素因子的系数。max{xi,yi}=ai
那么如果xi 取ai 的时候bi可以取得0到ai−1的任意一个数，同理bi。
不过两个数(a,b)和(b,a)会重复(n,n)也会重复。所以要除以2加1。
先筛个素数表，然后用唯一分解定理找出所有质因子的个数。最后计算即可

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