Longge's problem ( gcd的积性）

Posted 2022-01-24 zhangbuang

Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N.

"Oh, I know, I know!" Longge shouts! But do you know? Please solve it.

Input

Input contain several test case.
A number N per line.

Output

For each N, output ,∑gcd(i, N) 1<=i <=N, a line

Sample Input

2
6

Sample Output

3
15

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这题推完柿子 之后，直接根号枚举因子，然后算phi 也能过....
但是这题想考的是gcd是一个积性函数    

gcd(i*j,n)=gcd(i,n)*gcd(j,n)

好了现在我们需要来学习真正的姿势了，我也是刚学的，利用gcd是积性函数的性质，根据前文说的，我们有这样的结论：n>1时 n=p1^a1*p2^a2*...*ps^as，p为n的质因子，那么f(n)是积性函数的充要条件是f(1)=1，及f(n) = f(p1^a1)*f(p2^a2)*...f(pr^ar)，所以只要求f(pi^ai)就好。现在来看具体做法。

f(pi^ai) =  Φ（pi^ai）+pi*Φ（pi^(ai-1)）+pi^2*Φ（pi^(ai-2)）+...+pi^(ai-1)* Φ（pi）+ pi^ai *Φ（1）

根据性质1，我们可以做出如下化简

f(pi^ai)=[pi^(ai-1)*(pi-1) ] +  [pi*pi^(ai-2)*(pi-1)]  +  [pi^2*pi^(ai-3)*(pi-1)]  +  [pi^3*pi^(ai-4)*(pi-1)]....[pi^(ai-1)*pi^(ai-ai)*(pi-1)]+pi^ai   ①

然后对①提取公因子（pi-1）

f(pi^ai)=(pi-1)[pi^(ai-1) ] +  [pi*pi^(ai-2)]  +  [pi^2*pi^(ai-3)]  +  [pi^3*pi^(ai-4)]....[pi^(ai-1)*pi^(ai-ai)]+[pi^ai/(pi-1)]  ②

紧接着我们发现出了最后一项每个[]每个方括号内乘积都等于pi^(ai-1),所以对②提取公因子pi^(ai-1)

f(pi^ai)=(pi-1)*pi^(ai-1)*ai+[pi/(pi-1)] ③

然后把(pi-1)/pi放进括号里得

f(pi^ai)=pi^(ai)*1+ai*(pi-1)/pi ④

这个 ④是单个f(pi^ai)的公式，我们提取所有的pi^(ai)相乘实际上就是n了，所以我们可以得到f(n)的公式：f(n)=n*∏（1+ai*(pi-1)/pi）

然后我们看代码吧！

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 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #ifdef WIN32
 5 #define LLD "%I64d"
 6 #else
 7 #define LLD "%lld"
 8 #endif
 9 #define ll long long
10 using namespace std;
11 ll n;
12 inline int phi(int x)
13     int ans=x;
14     for(int i=2;1ll*i*i<=1ll*x;i++)
15         if(x%i==0) 
16             ans=ans/i*(i-1);
17             while(x%i==0) x/=i;
18         
19     
20     if(x>1) ans=ans/x*(x-1);
21     return ans;
22 
23 int main()
24 //     freopen("poj2480.in","r",stdin);
25      while(scanf(LLD,&n)!=EOF)
26          ll ans=0;
27          for(int i=1;1ll*i*i<=1ll*n;i++)
28             if(n%i==0)
29                 int x=phi(n/i);ans+=1ll*x*i;
30                 if(i*i!=n) 
31                     int y=phi(i);ans+=1ll*y*(n/i);
32                      
33             
34          
35         printf(LLD"\n",ans);
36     
37     return 0;
38