模板题。一个快速幂。

\[ a^b = (a^{b/2})^2 \]

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#define For(i,j,k) for(int i=j;i<=(k);++i)
using namespace std;
typedef long long ll;
inline int read(){
    int x=0,w=0;char c=getchar();
    for(;c<'0'||c>'9';w^=c=='-',c=getchar());
    for(;c>='0'&&c<='9';x=(x<<3)+(x<<1)+(c^48),c=getchar());
    return w?-x:x;
}
int qpow(int n,int m,int k){ //n^m%k
    if(m==0) return 1;
    long long ans=qpow(n,m/2,k)%k;
    if(m&1) return (int)(ans*ans%k*n%k);
    return (int)(ans*ans%k);
}
int main(){
    //freopen(".in","r",stdin);
    //freopen(".out","w",stdout);
    //ios::sync_with_stdio(false);
    //cin.tie(NULL);
    //cout.tie(NULL);
    int a,b;cin>>a>>b;
    cout<<qpow(a,b,2580000);
    return 0;
}