H - Pairs Forming LCM（lcm唯一分解定理+大力素数筛）_h

本文链接：https://blog.youkuaiyun.com/qq_44134712/article/details/99087159

本文探讨了一段代码的功能，该代码用于计算满足特定条件的整数对数量，即两个整数的最小公倍数等于给定的整数n。通过对质因数分解的理解，文章提出了一种更高效的算法实现，避免了超时问题，并提供了详细的数学解释和优化后的代码。

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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

由于
a=p1 ^ a1 * p2 ^ a2 *…*pn ^ an

b=p1 ^ b1 * p2 ^ b2 *…*pn ^ bn

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)

分解质因子，先对n素因子分解，n = p1 ^ e1 * p2 ^ e2 *…*pk ^ ek，

lcm(a,b)=p1 ^ max(a1,b1) * p2 ^ max(a2,b2) *…*pk ^ max(ak,bk)

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=2ei+1。
除了 (n, n) 所有的情况都出现了两次那么满足a<=b的有 (2ei + 1)) / 2 + 1个
代码：


//Full of love and hope for life

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <queue>
#define inf 0x3f3f3f3f
//https://paste.ubuntu.com/
//https://www.cnblogs.com/zzrturist/    //博客园
//https://blog.youkuaiyun.com/qq_44134712     //csdn

using namespace std;

typedef long long ll;
const ll maxn=1e7+10;

/*
ll lcm(ll a,ll b){
    return a/__gcd(a,b)*b;
}
*/

bool prime[maxn];
vector<ll>n;

void init(){
    prime[1]=1;
    for(ll i=2;i<=maxn;i++){
        if(prime[i]==0){
                n.push_back(i);
            for(ll j=i+i;j<=maxn;j+=i){
                prime[j]=1;
            }
        }
    }
}

int main(){
    init();
    ll a,k=0,b;
    cin >> a;
    while(a--){
        k++;
        cin >> b;
        ll x=b;
        ll ans=1;
        for(ll i=0;i<n.size()&&n[i]*n[i]<=x;i++){
            if(x%n[i]==0){
                ll cnt=0;
                while(x%n[i]==0){
                    x/=n[i];
                    cnt++;
                }
                ans*=(2*cnt+1);
            }
        }
        if(x>1){
            ans*=(2*1+1);
        }
        cout << "Case " << k << ": ";
        cout << (ans+1)/2 << endl;
    }
    return 0;
}