【唯一分解定理】Pairs Forming LCM

最新推荐文章于 2022-02-09 16:45:00 发布

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最新推荐文章于 2022-02-09 16:45:00 发布

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本文链接：https://blog.youkuaiyun.com/qq_41548233/article/details/96589158

本文介绍了一种高效计算给定整数n下满足最小公倍数为n的数对(i, j)数量的方法。通过质因数分解并利用数学公式减少计算复杂度，避免了暴力求解可能导致的时间超时问题。

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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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const int N=1e7+5;
unsigned int prime[1000005],cnt;         
bool vis[N];

void is_prime()
{
    cnt=0;
    memset(vis,0,sizeof(vis));
    for(int i=2;i<N;i++)
    {
        if(!vis[i])
        {
            prime[cnt++]=i;
            for(int j=i+i;j<N;j+=i)
            {
                vis[j]=1;
            }
        }
    }
}

int main()
{
    is_prime();
    int t;
    cin>>t;
    for(int nowcase=1;nowcase<=t;nowcase++)
    {
        LL n;
        cin>>n;
        int ans=1;
        for(int i=0;i<cnt&&prime[i]*prime[i]<=n;i++)
        {
            if(n%prime[i]==0)
            {
                int e=0;
                while(n%prime[i]==0)
                {
                    n/=prime[i];
                    e++;
                }
                ans*=(2*e+1);
            }
        }
        if(n>1)
            ans*=(2*1+1);
        printf("Case %d: %d\n",nowcase,(ans+1)/2);
    }
}