C++11(clang++ 3.9) 解法, 执行用时: 55ms, 内存消耗: 380K, 提交时间: 2019-06-01 15:00:09

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
LL n,p,R;
LL qpow(LL a,LL x){
	LL res = 1;
	while(x){
		if(x&1) res = res * a % p;
		a = a * a % p;
		x >>= 1;
	}
	return res;
}
int main(){
	cin >> n >> R >> p;
	LL pre = 1;
	for(int i = 2;i <= n; ++i){
		pre += 2;
		pre = pre*qpow(pre+1,p-2)%p; 
	}
	cout << pre * R % p;
	return 0;
}

C++14(g++5.4) 解法, 执行用时: 69ms, 内存消耗: 480K, 提交时间: 2019-06-06 22:32:31

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll n,r,p;
ll quickpow(ll a,ll b){
	ll r=1;
	a%=p;
	while(b){
		if(b&1) r=(r*a)%p;
		a=(a*a)%p;
		b>>=1;
	}
	return r;
}
int main(){
	cin>>n>>r>>p;
	ll ans=1;
	for(int i=1;i<n;i++){
		ans=(ans+2)%p*quickpow(ans+3,p-2)%p;
	}
	ans=ans*r%p;
	cout<<ans<<endl;
	return 0;
}