UVA 10892 LCM Cardinality

Question:
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
题意大意：给你一个正整数n，让你找出LCM（a,b）=n的对数（a<=b）.
思路:这题比较简单，只需要找出给出数n的因子即可，将其推入一个数组，再对数组进行操作，找出p[i]*p[j]/gcd(p[i],p[j])==n的对数
(http://acm.hust.edu.cn/vjudge/contest/121559#problem/D)

#include <iostream>#include <cstdio>#include <vector>#include <cmath>using namespace std;typedef long long LL;vector <LL> p;LL gcd(LL a,LL b){    if(b==0)        return a;    else return  gcd(b,a%b);}int main(){    LL n,num;    while (1)    {        num=0;        scanf("%lld",&n);        if(n==0)            break;        for(LL i=1;i<=sqrt(n);i++)        {            if(i==sqrt(n))                p.push_back(i);  //一个因子只需要推入一次即可，此操作避免重复重复推入一个因子            else if(n%i==0)            {                p.push_back(i);                p.push_back(n/i);            }        }        for(int i=0;i<p.size();i++)        {            for(int j=i;j<p.size();j++)            {                if(p[i]*p[j]/gcd(p[i],p[j])==n)                        num++;            }        }        printf("%lld %lld\n",n,num);        p.clear();    }    return 0;}

这题不难，只要你敢想敢做就行，不要被题吓到，有时就是这么简单