Pairs Forming LCM

Find the result of the following code:

long long pairsFormLCM( int n ) {
    long long res = 0;
    for( int i = 1; i <= n; i++ )
        for( int j = i; j <= n; j++ )
           if( lcm(i, j) == n ) res++; // lcm means least common multiple
    return res;
}

A straight forward implementation of the code may time out. If you analyze the code, you will find that the code actually counts the number of pairs (i, j) for which lcm(i, j) = n and (i ≤ j).

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 10¹⁴).

Output

For each case, print the case number and the value returned by the function 'pairsFormLCM(n)'.

Sample Input

15

2

3

4

6

8

10

12

15

18

20

21

24

25

27

29

Sample Output

Case 1: 2

Case 2: 2

Case 3: 3

Case 4: 5

Case 5: 4

Case 6: 5

Case 7: 8

Case 8: 5

Case 9: 8

Case 10: 8

Case 11: 5

Case 12: 11

Case 13: 3

Case 14: 4

Case 15: 2

挺狗的一个题目：本来以为是个规律题目，，一直打表找规律。

题解：给的就是一个lcm，问你有多少种组合

根据 LCM=p1^r1 * p2^r2 * p3^r3 .....

其中r1=max(a1,b1)  r2=max(a2,b2)....rn=max(an,bn)

因此当r1=a1时，，b1的取值为（0，a1-1）(先假设a1!=b1) 一共有r1种，然后如果r1=b1，同理也有r1种

因此一共有（2r1）种。如果a1=b1还会有一种，所以总共有2*r1+1种，所以先计算sum*=2*ans+1。(ans是每一步的质数的个数)

到目前为止我们已经求出来了一共有多少种选择，，即sum种。2*ans+1是奇数，一共有sum对，应题目要求我们需要i<=j

sum/2会保留整数，因此我们需要加一个 1

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
const int N=1E7+7;
const int NN=1E6+7;
bool p[N]={1,1,0};
int prime[NN];
int k=0; 
void pre(){
    memset(prime,0,sizeof(prime));
    for(int i=2;i<=N;i++){
        if(p[i]==0){
            prime[k++]=i;
            for(int j=i+i;j<=N;j+=i)
                p[j]=1;
        }
    }
    
}    
int main(){
    pre();
    int t;
    cin>>t;
    for(int ii=1; ii<=t; ii++){
        ll n;
        cin>>n;
        ll sum=1;
        for(int i=0;i<k&&prime[i] <=n;i++){
            int ans=0;
            if(n%prime[i]==0){
                while(n%prime[i]==0){
                    ans++;
                    n/=prime[i];
                }
                sum*=(2*ans+1);
            }            
        }
        if(n>1) sum*=2+1;
        printf("Case %d: %lld
",ii,sum/2+1);  
    }
    
    return 0;
}