UVa10892

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

题意：

       输入正整数n，统计有多少a<=b满足lcm(a,b)=n。输出n以及满足条件的整数对数。

分析：

       只需要枚举并存储数n的所有因子于数组arr，然后两两组合arr中的数，如果两数的最小公倍数是n则满足条件的整数对加1。

#include <cstdio>
#include <cmath>
#define ll long long
const int MAX_N = 1000;
ll arr[MAX_N + 1]; // 存储所有的因子，从小到大
ll gcd(ll a,ll b){return b == 0 ? a : gcd(b,a % b);}
ll lcm(ll a,ll b){return a * b / gcd(a,b);}
// 找出所有的因子
int find_all(ll n){
    int m = sqrt(n + 0.5);
    int cnt = 0;
    for(int i = 1 ; i <= m ; i++)if(n % i == 0)
        arr[cnt++] = i,arr[cnt++] = n / i;
    if(cnt >= 2 && arr[cnt - 1] == arr[cnt - 2]) cnt--;
    return cnt; // 不同因子的个数
}
ll n;
int main(){
    while(scanf("%lld",&n) && n){
        int cnt = find_all(n),ans = 0;
        for(int i = 0 ; i < cnt ; i++)
            for(int j = i ; j < cnt ; j++)
            if(lcm(arr[i],arr[j]) == n) ans++;
        printf("%lld %d
",n,ans);
    }
    return 0;
}