正题
题目链接:https://www.luogu.org/problemnew/show/P4139
题目大意
求2222222...%p2^{2^{2^{2^{2^{2^{2^{...}}}}}}}\%p2222222...%p
解题思路
欧拉定理
ab={ab%φ(p)+φ(p)(b>φ(p))ab%φ(p)(b≤φ(p))a^b=\left\{\begin{matrix}a^{b\%\varphi(p)+\varphi(p)}(b>\varphi(p)) \\ a^{b\%\varphi(p)}(b\leq\ \varphi(p)) \end{matrix}\right.ab={ab%φ(p)+φ(p)(b>φ(p))ab%φ(p)(b≤ φ(p))
然后还有一个定理
φ(φ(φ(φ(φ(φ(φ(...)))))))\varphi(\varphi(\varphi(\varphi(\varphi(\varphi(\varphi(...)))))))φ(φ(φ(φ(φ(φ(φ(...)))))))这样子下去最多只有logloglog层就到0了。
所以我们得出一个算法,我们不停递归solve(p)solve(p)solve(p),然后令模数不断减小也就是solve(p)=2solve(φ(p))+φ(p)%psolve(p)=2^{solve(\varphi(p))+\varphi(p)}\%psolve(p)=2solve(φ(p))+φ(p)%p
直到p=0p=0p=0
codecodecode
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int P=1e7;
int T,p,phi[P+10];
void Get_phi()
{phi[1]=1;for(int i=2;i<=P;i++)if(!phi[i])for(int j=i;j<=P;j+=i){if(!phi[j]) phi[j]=j;phi[j]=phi[j]/i*(i-1);}
}
int power(int x,int b,int p)
{int ans=1;while(b){if(b&1) ans=(long long)ans*x%p;x=(long long)x*x%p;b>>=1; }return ans;
}
int solve(int x)
{if(x==1) return 0;return power(2,solve(phi[x])+phi[x],x);
}
int main()
{scanf("%d",&T);Get_phi();while(T--){scanf("%d",&p);printf("%d\n",solve(p));}
}