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。
1 #include <cstdio>
2 #include <cmath>
3 #define ll long long
4 const int MAX_N = 1000;
5 ll arr[MAX_N + 1]; // 存储所有的因子，从小到大
6 ll gcd(ll a,ll b){return b == 0 ? a : gcd(b,a % b);}
7 ll lcm(ll a,ll b){return a * b / gcd(a,b);}
8 // 找出所有的因子
9 int find_all(ll n){
10 int m = sqrt(n + 0.5);
11 int cnt = 0;
12 for(int i = 1 ; i <= m ; i++)if(n % i == 0)
13 arr[cnt++] = i,arr[cnt++] = n / i;
14 if(cnt >= 2 && arr[cnt - 1] == arr[cnt - 2]) cnt--;
15 return cnt; // 不同因子的个数
16 }
17 ll n;
18 int main(){
19 while(scanf("%lld",&n) && n){
20 int cnt = find_all(n),ans = 0;
21 for(int i = 0 ; i < cnt ; i++)
22 for(int j = i ; j < cnt ; j++)
23 if(lcm(arr[i],arr[j]) == n) ans++;
24 printf("%lld %d\n",n,ans);
25 }
26 return 0;
27 }