Exponial（拓展欧拉定理）

Illustration of exponial(3) (not to scale), Picture by C.M. de Talleyrand-Périgord via Wikimedia Commons Everybody loves big numbers (if you do not, you might want to stop reading at this point). There are many ways of constructingreally big numbers known to humankind, for instance:

In this problem we look at their lesser-known love-child the exponial , which is an operation defined for all positive integers n as

For example, exponial(1) = 1 and   which is already pretty big. Note that exponentiation is right-associative:   .
Since the exponials are really big, they can be a bit unwieldy to work with. Therefore we would like you to write a program which computes exponial(n) mod m (the remainder of exponial(n) when dividing by m).

输入

The input consists of two integers n (1 ≤ n ≤ 109 ) and m (1 ≤ m ≤ 109 ).

输出

Output a single integer, the value of exponial(n) mod m.

样例输入

2 42

样例输出

2

拓展欧拉定理：

则想到用其降幂，递归即可。
但是要特判n=1,n=2,n=3,n=4,因为此时a^n可能小于phi(m)。

#include <bits/stdc++.h>
#define maxn 50005
using namespace std;
typedef long long ll;
ll prime[maxn];
bool vis[maxn];
int cnt;
void getprime()
{
    for(int i=2;i<maxn;i++)
    {
        if(!vis[i])
        {
            prime[cnt++]=i;
        }
        for(int j=0;j<cnt&&prime[j]*i<maxn;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0) break;
        }
    }
}
ll phi(ll n)
{
    ll res=n;
    for(int i=0;i<cnt;i++)
    {
        if(n%prime[i]==0) res=res/prime[i]*(prime[i]-1);
        while(n%prime[i]==0) n/=prime[i];
    }
    if(n>1) res=res/n*(n-1);
    return res;
}
ll qpow(ll k,ll n,ll mod)
{
    ll res=1;
    while(n)
    {
        if(n&1) res=res*k%mod;
        k=k*k%mod;
        n>>=1;
    }
    return res;
}
ll cal(ll n,ll m)
{
    if(m==1) return 0;
    if(n==1) return 1ll;
    if(n==2) return 2ll%m;
    if(n==3) return 9ll%m;
    if(n==4) return 262144ll%m;
    ll tmp=phi(m);
    return qpow(n,cal((n-1),tmp)+tmp,m);
}
int main()
{
    ll n,m;
    getprime();
    scanf("%lld%lld",&n,&m);
    ll ans=cal(n,m);
    printf("%lld\n",ans);
    return 0;
}