小x的奇遇-adventure

题意：

给定函数f(n),g(n),求 G k ( n ) G_k(n) Gk(n)
f ( n ) = { 1 , n = 1 ∑ i = 1 n − 1 [ g c d ( i , n − i ) = = 1 ] , n > 1 f(n)=\begin{cases} 1,n=1\\ \sum_{i=1}^{n-1}[gcd(i,n-i)==1],n>1 \end{cases} f(n)={1,n=1∑i=1n−1[gcd(i,n−i)==1],n>1
g ( n ) = ∑ d ∣ n f ( n d ) g(n)=\sum_{d|n}f(\frac{n}{d}) g(n)=d∣n∑f(dn)
G k ( n ) = { f ( g ( n ) ) , k = 1 g ( G k − 1 ( n ) ) , k > 1 ∧ k m o d 2 = 0 f ( G k − 1 ( n ) ) , k > 1 ∧ k m o d 2 = 1 G_k(n)=\begin{cases} f(g(n)),k=1\\ g(G_{k-1}(n)),k>1 \land k\ mod\ 2=0\\ f(G_{k-1}(n)),k>1 \land k\ mod\ 2=1 \end{cases} Gk(n)=⎩⎪⎨⎪⎧f(g(n)),k=1g(Gk−1(n)),k>1∧k mod 2=0f(Gk−1(n)),k>1∧k mod 2=1

Solution:

对于f(n)函数的处理：

g c d ( i , n − i ) = g c d ( i , n ) gcd(i,n-i)=gcd(i,n) gcd(i,n−i)=gcd(i,n)
当n>1时
f ( n ) = ∑ i = 1 n − 1 [ g c d ( i , n − i ) = = 1 ] = ∑ i = 1 n − 1 [ g c d ( i , n ) = = 1 ] = φ ( n ) f(n)=\sum_{i=1}^{n-1}[gcd(i,n-i)==1]\\ =\sum_{i=1}^{n-1}[gcd(i,n)==1]\\ =\varphi(n) f(n)=i=1∑n−1[gcd(i,n−i)==1]=i=1∑n−1[gcd(i,n)==1]=φ(n)
所以
f ( n ) = φ ( n ) f(n)=\varphi(n) f(n)=φ(n)

对于g(n)函数的处理：

g ( n ) = ∑ d ∣ n f ( n d ) n = 1 , g ( 1 ) = 1 n > 1 , g ( n ) = ∑ d ∣ n f ( n d ) = ∑ d ∣ n φ ( n d ) = n g(n)=\sum_{d|n}f(\frac{n}{d})\\ n=1,g(1)=1\\ n>1,g(n)=\sum_{d|n}f(\frac{n}{d}) \\ =\sum_{d|n}\varphi(\frac{n}{d}) \\ =n g(n)=d∣n∑f(dn)n=1,g(1)=1n>1,g(n)=d∣n∑f(dn)=d∣n∑φ(dn)=n
所以 g ( n ) = n g(n)=n g(n)=n

对于 G k ( n ) G_k(n) Gk(n)函数的处理：

G k ( n ) = { f ( g ( n ) ) , k = 1 g ( G k − 1 ( n ) ) , k > 1 ∧ k m o d 2 = 0 f ( G k − 1 ( n ) ) , k > 1 ∧ k m o d 2 = 1 = { φ ( n ) , k = 1 G k − 1 ( n ) , k > 1 ∧ k m o d 2 = 0 φ ( G k − 1 ( n ) ) , k > 1 ∧ k m o d 2 = 1 G_k(n)=\begin{cases} f(g(n)),k=1\\ g(G_{k-1}(n)),k>1 \land k\ mod\ 2=0\\ f(G_{k-1}(n)),k>1 \land k\ mod\ 2=1 \end{cases}\\ =\begin{cases} \varphi(n),k=1\\ G_{k-1}(n),k>1 \land k\ mod\ 2=0\\ \varphi(G_{k-1}(n)),k>1 \land k\ mod\ 2=1 \end{cases} Gk(n)=⎩⎪⎨⎪⎧f(g(n)),k=1g(Gk−1(n)),k>1∧k mod 2=0f(Gk−1(n)),k>1∧k mod 2=1=⎩⎪⎨⎪⎧φ(n),k=1Gk−1(n),k>1∧k mod 2=0φ(Gk−1(n)),k>1∧k mod 2=1
易得求 G k ( n ) G_k(n) Gk(n)，就是求 ⌊ k + 1 2 ⌋ \lfloor\frac{k+1}{2}\rfloor ⌊2k+1⌋次 φ \varphi φ
并且在求 l o g 2 n log_2n log2n次后，恒有 φ ( 1 ) = 1 \varphi(1)=1 φ(1)=1

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
unordered_map<ll,ll> mpsmu;
const int N=1e6;
const int mod=1e9+7;
int prime[N+5],notprime[N+5],cnt=0;
int phi[N+5];
void initPhi()
{phi[1]=1;notprime[1]=1;for(int i=2;i<=N;i++){if(!notprime[i]){prime[cnt++]=i;phi[i]=i-1;}for(int j=0;j<cnt&&1ll*i*prime[j]<=N;j++){notprime[i*prime[j]]=1;if(i%prime[j])phi[i*prime[j]]=phi[i]*(prime[j]-1);else{phi[i*prime[j]]=phi[i]*prime[j];break;}}}
}
int t;
ll n,k;
int main()
{initPhi();scanf("%d",&t);while(t--){scanf("%lld%lld",&n,&k);k=(k+1)/2;while(k--){if(n==1)break;if(n<=N)n=phi[n];else{ll x=n;for(int i=0;i<cnt;i++){int c=0;while(x%prime[i]==0)x/=prime[i],c++;if(c>0)n=n/prime[i]*(prime[i]-1);if(x<=N&&!notprime[x]){n=n/x*(x-1);x=1;break;}if(x==1)break;}if(x>1)n=n/x*(x-1);}}n=n%mod;printf("%lld\n",n);}return 0;
}