NCPC2016-E-Exponial

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題目描述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 de?ned 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).輸入The input consists of two integers n (1 ≤ n ≤ 109 ) and m (1 ≤ m ≤ 109 ).輸出Output a single integer, the value of exponial(n) mod m.範例輸入

2 42

範例輸出

2

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

對這個題來說，當n>4可以直接用這個算了

#include <bits/stdc++.h>#define ll long longusing 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;}

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NCPC2016-E-Exponial