Pairs Forming LCM （LCM+ 唯一分解定理）题解

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

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

题意：

在a,b中(a,b<=n)(1 ≤ n ≤ 10¹⁴),有多少组(a,b) (a<b)满足lcm(a,b)==n;

思路：

这里要学个新东西：快问问神奇海螺

首先，我们已经知道了唯一分解定理：n = p1 ^ e1 * p2 ^ e2 *..........*pn ^ en

那么，n的任意两个因子a，b肯定能表示为：

a=p1 ^ a1 * p2 ^ a2 *..........*pn ^ an

b=p1 ^ b1 * p2 ^ b2 *..........*pn ^ bn

现在给出公式：

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)

现在我们就可以求解题目了。

要a，b的LCM是n，所以max（a1，b1）== e1，max（a2，b2）== e2以此类推到n，也就是说每一个ai，bi中至少有一个等于ei，求这种组合方式有多少。按上面的思路第pi组有2*（ei+1）种组合方式，但是还要再减去一种重复的 ai==bi==ei ，所以结果是 2*ei+1 种。

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<stack> 
#include<set>
#include<vector>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
const int N=1e7+5;
const int NN=1e6;
const int MOD=1000; 
using namespace std;
bool prime[N];
ll p[NN];    //这里只能1e6不然就MLE了
int pn;
void get_prime(){
	pn=0;
	memset(prime,false,sizeof(prime));
	prime[0]=prime[1]=true;
	for(ll i=2;i<N;i++){
		if(!prime[i]){
			p[pn++]=i;
			for(ll j=i*i;j<N;j+=i){
				prime[j]=true;
			}
		}
	}
}
ll deal(ll n){
	ll res=1;
	for(ll i=0;i<pn && p[i]*p[i]<=n;i++){
		if(n%p[i]==0){
			int tmp=0;
			while(n%p[i]==0){
				tmp++;
				n/=p[i];
			}
			res*=(2*tmp+1);
		}
	}
	if(n>1) res*=3;	 
	res=res/2+1;
	return res;
}
int main(){
	get_prime();
	int T,num=1;
	ll ans,n;
	scanf("%d",&T);
	while(T--){
		scanf("%lld",&n);
		ans=deal(n);
		printf("Case %d: %lld
",num++,ans);
	}
	return 0;
}