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
把n,m唯一分解，n=(p1^a1)(p2^a2)……(pk^ak) m=(p1^b1)(p2^b2)……(pk^bk)
lcm=(p1^max(a1,b1))(p2^max(a2,b2))……(pk^max(ak,bk))
gcd=(p1^min(a1,b1))(p2^min(a2,b2))……(pk^min(ak,bk))
#include <bits/stdc++.h>
using namespace std;
const int N=1e7+50;
int prime[N/10];///这里开1e6 否则RE （玄学 为什么？？）
bool vis[N];
int tot=1;
void Er()
{
memset(vis,0,sizeof(vis));
vis[0]=vis[1]=1;
for(int i=2;i<N;i++)
{
if(!vis[i])
{
prime[tot++]=i;
for(int j=i+i;j<N;j+=i)
vis[j]=1;
}
}
}
long long solve(long long n)
{
long long ans=1;
for(int i=1;i<tot&&prime[i]*prime[i]<=n;i++)
{
if(n%prime[i]==0)
{
int tmp=0;
while(n%prime[i]==0)
{
n/=prime[i];
tmp++;
}
ans*=(2*tmp+1);
}
}
if(n>1)
ans*=3;
return ans;
}
int main()
{
Er();
int t,k=0;
long long n;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
long long ans=solve(n);
ans/=2;
ans++;
cout<<"Case "<<++k<<": "<<ans<<'\n';
}
return 0;
}
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