LightOJ

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

求小于等于n的数中最小公倍数等于N的组合而且 i<=j  ;

#include<stack>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<map>
#include<iostream>
#include<string.h>
#include<algorithm>
#define maxn 1000005
#define maxm 10000005
#define MAXN 100005
#define MAXM 10005
#define mem(a,b) memset(a,b,sizeof(a))
#define ll unsigned long long
#define inf 0x3f3f3f3f
using namespace std;
bool vis[maxm+5];
ll tot;
ll p[maxn];
void get_prime(){
     tot=0;
    mem(vis,true);
    for(ll i=2;i<=maxm;i++){
        if(vis[i]){
            p[tot++]=i;
            for(ll j=i*i;j<=maxm;j+=i)vis[j]=false;
        }
    }
}
ll f(ll n){
    if(n==1)return 1;
    ll ans=1;
    for(int i=0;i<tot;i++){
        if(n%p[i]==0){
            ll x=0;
           while(n%p[i]==0){x++;n/=p[i];}
            ans*=(2*x+1);
        }
    }
    if(n!=1)ans*=(2*1+1);
    return ans/2+1;
}
int main(){
    get_prime();
    int t,test=0;
    scanf("%d",&t);
    while(t--){
    ll n;
    scanf("%lld",&n);
   printf("Case %d: %lld
",++test,f(n));
    }
}