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 = ;

const int M = ;

const int inf = 0x3f3f3f3f;

const int MOD = ;

const double eps = 0.000001;

int main() {

    ll n;vector<ll >G;

    while(~scanf("%lld",&n)) {

        if(n == ) break;

        if(n == ) {

            cout<<<<" "<<<<endl;continue;

        }

        ll ans = ;

        ll aim = n;

        G.clear();

        G.pb(n);G.pb();

        for(ll i=;i*i<=n;i++) {

            if(n%i==) {

                  G.pb(i);if(n/i!=i) G.pb(n/i);

            }

        }

        sort(G.begin(),G.end());

        for(int i=;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 ;

}

代码

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UVA 10892 LCM Cardinality 数学

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