bzoj3884: 上帝与集合的正确用法 扩展欧拉定理

题意：求(2^{2^{2^{2^{...}}}}\%p)
题解：可以发现用扩展欧拉定理不需要很多次就能使模数变成1，后面的就不用算了
(a^b\%c=a^{b\%phi c} gcd(b,c)==1)
(a^b\%c=a^{b\%phi c+phi c} gcd(b,c)!=1)

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=10000000+10,maxn=400000+10,inf=0x3f3f3f3f;

int prime[N],cnt,phi[N];
bool mark[N];
void init()
{
    phi[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!mark[i]){prime[++cnt]=i;phi[i]=i-1;}
        for(int j=1;j<=cnt&&i*prime[j]<N;j++)
        {
            mark[i*prime[j]]=1;
            phi[i*prime[j]]=phi[i]*phi[prime[j]];
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
        }
    }
}
int main()
{
    init();
    int T;scanf("%d",&T);
    while(T--)
    {
        vi v;
        ll p;scanf("%lld",&p);
        ll pp=p;
        while(p!=1)
        {
            v.pb(p);
           // printf("%lld
",p);
            p=phi[p];
        }
        ll now=1;
        for(int i=(int)v.size()-1;i>=0;i--)
        {
            now=qp(2,now,v[i])+(i?v[i]:0);
        }
        printf("%lld
",now%pp);
    }
    return 0;
}
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