AcWing 875. 快速幂 · znzryb

## 思路讲解

参考以下题解

[https://www.acwing.com/solution/content/15293/](https://www.acwing.com/solution/content/15293/)

这里有一个非递归写法

[https://www.acwing.com/solution/content/16482/](https://www.acwing.com/solution/content/16482/)

## AC代码

[https://www.acwing.com/problem/content/submission/code_detail/40911595/](https://www.acwing.com/problem/content/submission/code_detail/40911595/)

```cpp
// Problem: 快速幂
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/877/
// Memory Limit: 64 MB
// Time Limit: 5000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(5e18)+3;

ll N,M,T;

ll binpow(ll a,ll k,const ll &p){
	ll res=1;
	while(k>0){
		if((k&1)==1) res=a*res%p;
		a=a*a%p;
		k>>=1;
	}
	return res;
}

inline void solve(){
	ll a,k,p;
	cin>>a>>k>>p;
	cout<<binpow(a,k,p)<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		solve();
	}
	// solve();
	return 0;
}
/*
AC
https://www.acwing.com/problem/content/submission/code_detail/40911595/
*/
```

## 心路历程（WA，TLE，MLE……）