传送门

•题意

求$2^{2^{2^{2^{2^{2^{...^{2}}}}}}}$ (无穷个2) 对p取模的值

•思路

设答案为f(p)

$2^{2^{2^{2^{2^{2^{...^{2}}}}}}}\%p$

$=2^{(2^{2^{2^{2^{2^{...^{2}}}}}}\%\varphi(p)+ \varphi(p))}\%p$

$=2^{(2^{2^{2^{2^{2^{...^{2}}}}}}\%\varphi(p)+ \varphi(p))}\%p$

$=2^{(2^{(2^{2^{2^{2^{...^{2}}}}}\%\varphi(\varphi(p)+\varphi(\varphi(p))))}\%\varphi(p)+ \varphi(p))}\%p$

...

得到递推式     $2^{f(\varphi(p))+\varphi(p)}(mod\ p)$

利用欧拉降幂

$a^{b}=\begin{cases}a^{b\%\varphi(p)}  \ \ \ \ \ \ \ \ \ \  gcd(a,p)=1 \\ a^{b} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   gcd(a,p)\neq 1,b \leqslant \varphi(p)\\a^{b\%\varphi(p)+\varphi(p)}  \ \ gcd(a,p)\neq1,b\geqslant \varphi(p)  \\ \end{cases}$

由于2的幂数是无穷的，肯定$>p$,所以可以直接使用$a^{b\%\varphi(p)+\varphi(p)} $

•代码

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 ll qpow(ll a,ll b,ll mod)
 5 {
 6     ll res=1;
 7     while(b)
 8     {
 9         if(b&1)
10             res=res*a%mod;
11         a=a*a%mod;
12         b>>=1;
13     }
14     return res;
15 }
16
17 ll phi(ll x)
18 {
19     ll res=x;
20     for(int i=2;i*i<=x;i++)
21     {
22         if(x%i==0)
23         {
24             while(x%i==0)
25                 x/=i;
26             res=res-res/i;
27         }
28     }
29     if(x>1)
30         res=res-res/x;
31     return res;
32 }
33
34 ll solve(ll m)
35 {
36     if(m==1)
37         return 0;
38
39     ll p=phi(m);
40     return qpow(2,solve(p)+p,m);
41 }
42
43 int main()
44 {
45     int t;
46     cin>>t;
47     while(t--)
48     {
49         ll m;
50         cin>>m;
51         cout<<solve(m)<<endl;
52     }
53 }

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