NCPC2016-E- Exponial

题目描述

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 constructing really 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

a^b %c= a^(b%phi(c)+phi(c)) %c (b>=phi(c)） 
如果 phi(c)>b 直接 a^b%c

对这个题来说，当n>4可以直接用这个算了

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll fi(ll n)
{
    ll ans=n;
    for (int i=2;i*i<=n;i++)
    {
        if (n%i==0)
        {
            ans-=ans/i;
            while (n%i==0) n/=i;
        }
    }
    if (n>1) ans-=ans/n;
    return ans;
}
 
ll qpow(ll a, ll n, ll m) {
    a%=m;
    ll ret = 1;
    while(n)
    {
        if (n&1) ret=ret*a%m;
        a=a*a%m;
        n>>=1;
    }
    return ret;
}
ll f(ll n, ll m)
{
    if (m==1) return 0;
    if (n==1) return 1;
    if (n==2) return 2%m;
    if (n==3) return 9%m;
    if (n==4) return 262144%m;
    return qpow(n, f(n-1, fi(m)) % fi(m) + fi(m), m);
}
int main()
{
    ll n, m;
    while(cin >> n >> m)
    {
        cout << f(n, m) << endl;
    }
    return 0;
}