题目大意：求2^(2^(2^(2^(2^...)))) mod p的值
SB出题人被各种乱艹系列……
其实是某天脑洞比较大突然想算算这东西= = 然后就发现了这个好玩的性质= =
其实+∞个2看着吓人其实没啥可怕的= =
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 10001000
using namespace std;
int phi[M],prime[1001001],tot;
bool not_prime[M];
int Phi(int x)
{
int i,re=x;
for(i=2;i*i<=x;i++)
if(x%i==0)
{
re/=i;re*=i-1;
while(x%i==0)
x/=i;
}
if(x^1) re/=x,re*=x-1;
return re;
}
int Quick_Power(long long x,int y,int p)
{
long long re=1;
while(y)
{
if(y&1) (re*=x)%=p;
(x*=x)%=p; y>>=1;
}
return re;
}
int Solve(int p)
{
if(p==1) return 0;
int temp=0;
while(~p&1) p>>=1,++temp;
int phi_p=Phi(p);
int re=Solve(phi_p);
(re+=phi_p-temp%phi_p)%=phi_p;
re=Quick_Power(2,re,p)%p;
return re<<temp;
}
int main()
{
int T,p;
for(cin>>T;T;T--)
{
scanf("%d",&p);
printf("%d\n",Solve(p));
}
return 0;
}
