UVa11889 Minimum Sum LCM 分解质因数_11889分解质因数-优快云博客

本文链接：https://blog.youkuaiyun.com/Megumin/article/details/51891490

本文介绍了一种计算正整数N的最小公倍数(LCM)基数的方法，即有多少不同的整数对的LCM等于N。文章提供了一个具体的实现算法，并通过样例输入输出展示了如何计算不同数值的LCM基数。

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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 LCMcardinality 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 Cis its cardinality. These two numbers are separated by a single space.

Sample Input

2
12
24
101101291

Output for Sample Input

2 2
12 8

24 11

101101291 5

对于n的每个素因子，均作为一个乘数时，和最小，须注意特殊情况如n=1以及n为素数幂。

#include <iostream>
#include <cstdio>
#include <map>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
#define LL long long
using namespace std;
#define maxn 100005
LL prime[maxn],c;
LL v[maxn];
void p()
{
    LL i,j,n=maxn,m;
    c=0;
    m=(LL)sqrt(n+0.5);
    memset(v,0,sizeof(v));
    for(i=2;i<=m;i++)
        if(!v[i]){
            for(j=i*i;j<=n;j+=i)
                v[j]=1;
        }
    for(j=2;j<=n;j++){
        if(!v[j]){
            prime[c++]=j;
        }
    }
}
int main()
{
    LL n,ans,temp;
    int flag,ca=1;
    p();
    while(~scanf("%lld",&n),n){
        ans=flag=0;
        if(n==1) ans=2;
        else{
            for(int i=0;n>1&&i<c;i++){
                if(n%prime[i]==0){
                    flag++;
                    temp=prime[i];
                    n/=prime[i];
                    while(n%prime[i]==0){
                        n/=prime[i];
                        temp*=prime[i];
                    }
                    ans+=temp;
                }
            }
            if(flag==1) ans++;
            else if(flag==0){
                ans=n+1;
            }
        }
        printf("Case %d: %lld\n",ca++,ans);
    }
    return 0;
}