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 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).
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欧拉降幂定理：当b>phi(p)时，有a^b%p=a^(b%phi(p)+phi(p))%p思路：当n>=6时，欧拉降幂定理一定适用，因为f(5)>1e9,也就是一定有欧拉降幂定理的b>phi(p)这个条件，所以f(n)%p=n^f(n-1)%p=n^(f(n-1)%phi(p)+phi(p))%p;再递归地求f(n-1)%phi(p)当n<=5时，f(n)%p=n^f(n-1)%p,因为不一定有f(n-1)>phi(p)成立，所以不能用欧拉降幂定理求，直接手动求出f(n)%p即可；从1e9递归到5很慢，但当p=1时，可以直接返回f(n)%p=0而不用递归到下一层;AC代码：
#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;
}
原文地址：https://www.cnblogs.com/lllxq/p/9748184.html