H

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)'.

大体题意：问1-n有多少对数使得lcm=n;

解题思路：

先来看个知识点：

素因子分解：n = p1 ^ e1 * p2 ^ e2 *..........*pn ^ en

for i in range(1,n):

ei 从0取到ei的所有组合

必能包含所有n的因子。

现在取n的两个因子a,b

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)

题解：

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=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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 
 8 const int maxn=1e7+10;
 9 const int maxnn=1e6;
10 int T;
11 long long a;
12 bool ip[maxn];
13 unsigned int p[maxnn],num;//输入数据太大会导致超时
14 
15 void is_prime()
16 {
17     memset(ip,1,sizeof(ip));
18     ip[0]=ip[1]=0;
19     for(long long i=2;i<=maxn;i++)
20     {
21         if(ip[i])
22         {
23             p[num++]=i;
24             for(long long j=i+i;j<=maxn;j+=i)
25             {
26                 ip[j]=0;
27             }
28         }
29     }
30 }
31 
32 int main()
33 {
34     ios::sync_with_stdio(false);
35     cin>>T;
36     is_prime();
37     for(int i=1;i<=T;i++)
38     {
39         cin>>a;
40         long long ans=1;
41         for(int i=0;i<num&&p[i]*p[i]<=a;i++)
42         {
43             if(a%p[i]==0)
44             {
45                 int cnt=0;
46                 while(a%p[i]==0)
47                 {
48                     a/=p[i];
49                     cnt++;
50                 }
51                 ans*=(2*cnt+1);
52             }
53         }
54         if(a>1) ans*=(2*1+1);
55 
56         cout<<"Case"<<" "<<i<<":"<<" "<<(ans+1)/2<<endl;
57     }
58     return 0;
59 }