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 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

题意：给出a和b的最小公倍数N，找出符合条件的a、b有多少对。

分析：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)

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. 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） ，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。

#include <cstdio>
#include <cmath>

int n;

void get_ans() {
    int tmp = n;
    int m = (int)sqrt(n + 0.5);
    long long ans = 1;
    for(int i = 2; i <= m; i += 2) {
        if(n % i == 0) {
            int cnt = 0;
            while(n % i == 0) {
                n /= i;
                cnt++;
            }
            ans *= (cnt << 1) + 1;
        }
        if(i == 2) i--;
    }
    if(n > 1) ans += (ans<<1);
    ans = (ans >> 1) + 1;
    printf("%d %lld\n", tmp, ans);
}

int main() {
    while(~scanf("%d", &n) && n) {
        get_ans();
    }
    return 0;
}