快速幂

#include<bits/stdc++.h>
using namespace std;
const int N=5e4+5,INF=0x3f3f3f3f;
typedef long long LL;
LL x,n;
LL power(LL a,LL b){
	LL ans=1;
	while(b){
		if(b&1)ans*=a;
		b>>=1;
		a*=a;
	}
	return ans;
}
int main()
{
    cin>>x>>n;
    cout<<power(x,n);
}

最短

#include<bits/stdc++.h>//写cmath浪费位置
long long x,n,ans;int main(){std::cin>>x>>n;ans = pow(x,n);std::cout<<ans;}

#include <iostream>
#include <stack>
#include <cmath>
#include <vector>
#include <string.h>
#include <queue>
#include <stdio.h>
#include <iomanip>
#include <cstdio>
#include <algorithm>
#define int long long
#define double long double
using namespace std;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int a, b;
int power(int a, int b)
{
	int ans = 1; 
	while(b)
	{
		if(b & 1)
		{
			ans = ans * a;
		}
		a = a * a;
		b >>= 1;
	}
	return ans;
}
signed main()
{
	cin >> a >> b;
	cout << power(a, b) << endl;
	return 0;
}

当我写完题才发现要用递归...

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    long long x,n,result;
    cin>>x>>n;
    result=pow(x,n);
    cout<<result;
    return 0;
}

#include<iostream>
using namespace std;
int main()
{
	long long a,b;
	cin>>a>>b;
	int q=a;
	for(int i=1;i<b;i++)
	{
		a*=q;
	}
	cout<<a;
}