Euler generalized Euler descending descending + board title (ACM)

Euler descending: particularly large power can be used when rapid power to significantly reduce the time complexity, and when the power as large as 10 ^ when power is not very fast 10000 line, this time you need to use Euler descending it theorem as follows:

It proved to say good-bye.

Put a topic FZU 1759

Topic Source: https://vjudge.net/problem/FZU-1759

I.e., evaluation of a ^ b% c, b thief large range, so Euler title descending board.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<queue>
#define MAX_len 50100*4
using namespace std;
typedef long long ll;
char b[1000100];
ll a,mod;
ll euler(ll n)
{    
     ll i,j,res=n,temp=n;  
     for(i=2;i*i<=temp;i++)
    {  
         if(temp%i==0)
         {  
             res=res/i*(i-1);
             while(temp%i==0) 
                temp/=i;  
         }  
     }  
     if(temp>1) 
        res=res/temp*(temp-1);  
     return res;  
}  
ll quickpow(ll a,ll n)
{
	ll res=1;
	while(n)
	{
		if(n&1)
		{
			res=(res*a)%mod;
		}
		n>>=1;
		a=(a*a)%mod;
	}
	return res%mod;
}
int main()
{
	ll i,j;
	while(scanf("%lld %s %lld",&a,b,&mod)!=EOF)
	{
		ll len=strlen(b);
		ll t1=euler(mod);
		ll ans=0;
		for(i=0;i<len;i++)
		{
			ans=(ans*10+b[i]-'0')%t1;
		}
		ans+=t1;
		printf("%lld\n",quickpow(a,ans));
	 } 
	return 0;
} 

BZOJ 3884

Topic Source: https://www.lydsy.com/JudgeOnline/problem.php?id=3884

As the index is infinite, we should go to the modulo operation.

Extended Euler's theorem: a ^ b≡a ^ (b% φ (p) + φ (p)) (mod p) a is any integer, b, p is a positive integer, and b> φ (p) (a, p need not be prime to).

f(p)=(mod p)=(mod p)

Is equal to F (Phi (P)) = (P MOD)

We found whereby f (x) is recursive, but requires the termination condition, when phi (phi (... phi (p))) == 1 when any number is equal to the MOD end condition 1 0.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<queue>
#define MAX_len 50100*4
using namespace std;
typedef long long ll;
char b[1000100];
ll a,mod;
ll euler(ll n)
{    
     ll i,j,res=n,temp=n;  
     for(i=2;i*i<=temp;i++)
    {  
         if(temp%i==0)
         {  
             res=res/i*(i-1);
             while(temp%i==0) 
                temp/=i;  
         }  
     }  
     if(temp>1) 
        res=res/temp*(temp-1);  
     return res;  
}  
ll quickpow(ll a,ll n,ll mod)
{
    ll res=1;
    while(n)
    {
        if(n&1)
        {
            res=(res*a)%mod;
        }
        n>>=1;
        a=(a*a)%mod;
    }
    return res%mod;
}
ll solve(ll mod)
{
    if(mod==1)
    return 0;
    return quickpow(2,solve(euler(mod))+euler(mod),mod);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll p;
        scanf("%lld",&p);
        printf("%lld\n",solve(p));
     } 
    return 0;
}