10892 - LCM Cardinality
Time limit: 3.000 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 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：
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<cmath>
using namespace std;
int main()
{
long long n, nn, ans, i, count;
while (scanf("%lld", &n), n)
{
nn = n;
ans = 1;
for (i = 2; i * i <= n; i += 2)///不用求素数，因为范围很小(注意n在不断减小)
{
if (n % i == 0)
{
count = 0;
while (n % i == 0)
{
n /= i;
++count;
}
ans *= (count << 1) + 1;
}
if (i == 2)
--i;///小技巧
}
if (n > 1)
ans += ans << 1;
ans = (ans >> 1) + 1;
printf("%lld %lld\n", nn, ans);
}
return 0;
}