UVA10892 最小公倍数素因子分解
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1833
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 LCMcardinality 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 Cis 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 |
思路：（引用别人的解释）
1. 设n=lcm(a,b)=(p1^r1)*(p2^r2)*(p3^r3)…(pm^rm)
又设a=(p1^a1)*(p2^a2)*(p3^a3)…(pm^am),b=(p1^b1)*(p2^b2)*(p3^b3)…(pm^bm)
则由lcm的定义有ri=max{ai,bi}
所以对于每个ri，ai和bi中至少有一个要取ri
2. 对于ai取ri的情况，bi可以取[0,ri-1]的任意整数，这有ri种情况；bi取ri的情况同样是ri种。最后加上ai和bi都取ri的情况，共有(2*ri+1)种情况
3. 最后，由于这么考虑把(a,b)和(b,a)算重复了，但(n,n)的情况只算了一遍，所以最后要ans=(ans+1)/2=ans/2+1（因为ans是奇数）
4. 优化：只考虑√n范围内的质数，但这样会存在漏掉一个大质数的情况（比如n=2*101等），这个大质数的幂次只能为1（即少算了一个*(2*1+1)），所以在这种情况发生时要补上ans*=3，写成位运算就是ans+=ans<<1了。
#include <iostream>
#include <cstdio>
#include <string.h>
#include <cmath>
using namespace std;
const int maxx=100005;
int fac[maxx];
int num[maxx],prime[maxx],p[maxx];
int cnt,k;
void isprime()
{
memset(prime,true,sizeof(prime));
k=0;
for(int i=2;i<maxx;i++)
{
if(prime[i])
{
p[k++]=i;
for(int j=i+i;j<maxx;j+=i)
prime[i]=false;
}
}
}
void gao(long long n)
{
cnt=0;
isprime();
long long lim=(int)sqrt(n);
for(int i=0;i<k;i++)
if(n%p[i]==0)
{
int a=0;
while(n%p[i]==0)
{
n/=p[i];
a++;
}
fac[cnt]=p[i];
num[cnt++]=a;
}
if(n>1)
{
fac[cnt]=n;
num[cnt++]=1;
}
}
int main()
{
long long n;
while(cin >> n)
{
if(n==0)
break;
gao(n);
long long ans=1;
for(int i=0;i<cnt;i++)
ans*=num[i]*2+1;
ans=(ans+1)/2;
cout << n << " " << ans<< endl;
}
return 0;
}