cho 2 số nguyên `a`, `n` Tính `a^n` giới hạn 1`<=``a,`n``<=``10^9` c++

$\color{#1fdb91}{\texttt{C}}\color{#0aaec9}{\texttt{+}}\color{#e9418e}{\texttt{+}}$

#include <bits/stdc++.h>

#define ll long long

using namespace std;

ll powMod(ll a, ll n, ll MOD) {
    if (n) return powMod(a*a % MOD, n >> 1, MOD) * (n&1? a % MOD: 1) % MOD;
    return 1;
}

int main() {
    
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    
    ll a, n; cin >> a >> n;
    cout << powMod(a, n, (ll)1e9+7) % (ll)(1e9+7);
    
}

\begin{array}{c|c|c}\color{#ffd710}{\texttt{\{}} \color{#8655d6}{\texttt{\{}}\ \ \color{#8cdcda}{\text{Daoanhviet96}}\ \ \color{#8655d6}{\texttt{\}}} \color{#ffd710}{\texttt{\}}}\end{array}

#include <bits/stdc++.h>

using namespace std;
long long mod = 1e9+7;
long long powermod(long long a, long long b, long long c){
    long long res = 1;
    while (b){
        if (b%2){
            res *= a;
            res %= c;
        }
        a *= a;
        a %= c;
        b /= 2;
    }
    return res;
}
int main(){
    long long a,b;
    cin >> a >> b;
    cout << powermod(a,b,mod);
    return 0;
}