UVA 10892 LCM Cardinality 数学

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：给你m，求出多少对 数的LCM（a,b）=m

题解：必然是M的因子，我们对M求因子，我们知道因子特别少，我们就两层循环暴力去找对子就好了

//meek///#include<bits/stdc++.h>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include<iostream>
#include<bitset>
#include<vector>
using namespace std ;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
typedef long long ll;

const int N = 10005;
const int M = 1000001;
const int inf = 0x3f3f3f3f;
const int MOD = 1000000007;
const double eps = 0.000001;


int main() {
    ll n;vector<ll >G;
    while(~scanf("%lld",&n)) {
        if(n == 0) break;
        if(n == 1) {
            cout<<1<<" "<<1<<endl;continue;
        }
        ll ans = 0;
        ll aim = n;
        G.clear();
        G.pb(n);G.pb(1);
        for(ll i=2;i*i<=n;i++) {
            if(n%i==0) {
                  G.pb(i);if(n/i!=i) G.pb(n/i);
            }
        }
        sort(G.begin(),G.end());
        for(int i=0;i<G.size();i++)
        for(int j=i;j<G.size();j++) {
            ll a = G[i];
            ll b = G[j];
            if(a*b/__gcd(a,b) == aim)
                ans++;//,cout<<a<<" "<<b<<endl;
        }
        cout<<aim<<" "<<ans<<endl;
    }
    return 0;
}