Uva 10892 - LCM Cardinality 解题报告（因式分解）

Problem F
LCM Cardinality
Input: Standard Input

Output: Standard Output

Time Limit: 2 Seconds

 

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                             Output for Sample Input

2

12

24

101101291

0

2 2

12 8

24 11

101101291 5

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Problem setter: Shahriar Manzoor

Special Thanks: Derek Kisman

    解题报告： 问最大公约数为n的数有多少对。首先，可以将n因式分解，写出p1^e1 * p2^e2 ... * pn*en。一对数中，必然至少有一个可以整除pi^ei，另外一个就有（1+ei）种情况。统计所有的情况，去重即可。代码如下：

#include <cstdio>#include <algorithm>#include <cstring>using namespace std;void work(int n){    int nn=n;    int ans=1;    for(int i=2;i*i<=n;i++) if(n%i==0)    {        int t=0;        while(n%i==0) n/=i,t++;        ans*=(t*2+1);    }    if(n>1)        ans*=3;    ans-=1;    ans/=2;    ans+=1;    printf("%d %d\n", nn, ans);}int main(){    int n;    while(~scanf("%d",&n) && n)        work(n);}