Description

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 di?erent 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 ?nd out the LCM cardinality of a number.

Input

The input ?le contains at most 101 lines of inputs. Each line contains an integer N \((0 \lt N \le 2*10^9)\) . 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

Resume

對於給定的數N，求出以N為最小公倍數的有序數對個數。

Analysis

剛看到題目，苦思冥想，以為要什麽神奇的數論方法來解決，甚至想到了標準分解，無奈要求的質數太大。
後來在同學的啟迪下，發現一個數的因數並不多，大概在 \(\ln N\) 級別上（沒找到準確的理論），那就可以直接暴力枚舉解決嘍。
時間復雜度 \(O(N) = \sqrt {N} {(\ln N)}^2\) 左右（迷。

Code

//////////////////////////////////////////////////////////////////////
//Target: UVa10892 - LCM Cardinality
//@Author: Pisceskkk
//Date: 2019-2-16
//////////////////////////////////////////////////////////////////////

#include<cstdio>
#include<algorithm>
using namespace std;
int n,cnt,f[10000],ans;
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}
int lcm(int a,int b){return 1ll*a*b/gcd(a,b);}        //1LL 防止爆int
int main(){
    while(~scanf("%d",&n) && n){
        ans = cnt = 0;
        for(int i=1;i*i<=n;i++){
            if(n%i == 0){
                f[++cnt]=i;
                if(i*i != n)f[++cnt]=n/i;
            }
        }
        sort(f+1,f+1+cnt);
        for(int i=1;i<=cnt;i++){
            for(int j=1;j<=i;j++){
                if(lcm(f[j],f[i]) == n)ans++;
            }
        }
        printf("%d %d\n",n,ans);
    }
    return 0;
}
 

UVa 10892 - LCM Cardinality