[數學][歐拉降冪定理]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
歐拉降冪定理：當b>phi(p)時，有a^b%p=a^(b%phi(p)+phi(p))%p
思路：當n>=6時，歐拉降冪定理一定適用，因為f(5)>1e9,也就是一定有歐拉降冪定理的b>phi(p)這個條件，所以f(n)%p=n^f(n-1)%p=n^(f(n-1)%phi(p)+phi(p))%p;再遞迴地求f(n-1)%phi(p)
當n<=5時，f(n)%p=n^f(n-1)%p,因為不一定有f(n-1)>phi(p)成立，所以不能用歐拉降冪定理求，直接手動求出f(n)%p即可；
從1e9遞迴到5很慢，但當p=1時，可以直接返回f(n)%p=0而不用遞迴到下一層;
AC代碼：

#include <cstdio>long long n,m;long long mod;long long phi(long long x){    long long res=x;    for(long long i=2; i*i<=x; ++i)    {        if(x%i==0)        {            res=res-res/i;            while(x%i==0)                x/=i;        }    }    if(x>1)        res=res-res/x;    return res;}long long qpow(long long a,long long n,long long mod){    long long res=1;    while(n)    {        if(n&1)        {            res*=a;            res%=mod;        }        n>>=1;        a=(a*a)%mod;    }    return res;}long long solve(long long n,long long m){    if(m==1) return 0;    if(n==1) return 1;    else if(n==2) return 2%m;    else if(n==3) return 9%m;    else if(n==4) return qpow(4,9,m);    long long tem=phi(m);    return qpow(n,solve(n-1,tem)+tem,m);}int main(){    while(~scanf("%lld%lld",&n,&m))    {        printf("%lld\n",solve(n,m));    }    return 0;}

[數學][歐拉降冪定理]Exponial