D - Pairs Forming LCM

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本文介绍了一种高效算法，用于计算特定整数n的所有可能的(i, j)对的数量，这些对满足最小公倍数等于n且i≤j。通过对质数分解和数学优化，避免了直接实现的超时问题。

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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

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

#include <cstring>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#define MAXN 100000005 
using namespace std;
bool isprime[MAXN];
int prime[MAXN/10], tot; //防止MLE
typedef long long LL;
void getPrime() {
    memset(isprime, true, sizeof(isprime));
    isprime[0] = isprime[1] = false;
    tot = 0;
    for(int i = 2; i < MAXN; i++) {
        if(isprime[i]) prime[tot++] = i;
        for(int j = 0; j < tot && prime[j] <= MAXN / i; j++) {
            isprime[i * prime[j]]=false;
            if(i % prime[j] == 0) break;
        }
    }
}

int main()
{

	int T;
	LL n;
	scanf("%d",&T);
	getPrime();
	int cas=1;
    while(T--)
	{
		scanf("%lld",&n);
		int ans=1;
		LL temp=n;
        for(LL i = 0; i < tot && prime[i]<=n; i++)
        {
            LL p=0;
            while(temp)
            {
                if(temp%prime[i]==0)
                    p++;
                else break;
				temp=temp/prime[i];

            }
            if(p>0) ans=ans*(2*p+1);
        }
        if (temp > 1) ans *= 3;
        printf("Case %d: %lld\n",cas++,(ans+1)/2);
	}
	return 0;
}