Pairs Forming LCM

题目：

B - Pairs Forming LCM

Time Limit:2000MS     Memory Limit:32768KB    

Description

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给定一个数字，n，求1~n之间可以找到的最小公倍数为n的对数

解题思路：

素因子分解：n=p1^x1*p2^x2**************pn^xn;

a=p1^y1*p2^y2*p3^y3*****************pn^yn;

b=p1^c1*p2^c2*p3^c3*****************pn^cn;

gcd(a,b)=p1 ^ min(a1,b1) * p2 ^ min(a2,b2) *..........*pn ^ min(an,bn)

lcm(a,b)=p1 ^ max(a1,b1) * p2 ^ max(a2,b2) *..........*pn ^ max(an,bn)

 

所以，当lcm(a,b)==n时，max(a1,b1)==e1,max(a2,b2)==e2,…max(ak,bk)==ek

 

当ai == ei时，bi可取 [0, ei] 中的所有数  有 ei+1 种情况，bi==ei时同理。

 

那么就有2(ei+1)种取法,但是当ai = bi = ei 时有重复，所以取法数为2(ei+1)-1=2*ei+1。
除了 (n, n) 所有的情况都出现了两次  那么满足a<=b的有 (2*ei + 1)) / 2 + 1 个

 

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
const int mx=10000002;
ll pos[664583];
ll top=0,n,t,k;
bool vis[mx];
void get_prime()
{
    for(int i=2;i<mx;i++)
    {
        if(vis[i]) continue;
        pos[top++]=i;
        for(int j=2;j*i<mx;j++)
        {
            vis[i*j]=1;
        }
    }
}
ll solve(ll n)
{
    ll ans=1;
    for(int i=0;i<top && pos[i]*pos[i]<=n;i++)
    {
        if(n%pos[i]==0)
        {
            int cnt=0;
            while(n%pos[i]==0)
            {
                n/=pos[i];
                cnt++;
            }
            ans*=(2*cnt+1);
        }
    }
    if(n>1) ans*=(2*1+1);
    return (ans+1)/2;
} 
int main()
{
    get_prime();
    scanf("%lld",&t);
    k=t;
    while(t--)
    {
        scanf("%lld",&n);
        printf("Case %lld: %lld\n",k-t,solve(n));
    }
}