原题链接：https://ac.nowcoder.com/acm/contest/9981/H

前置知识

1. 欧拉函数φ(x) 代表[1, x]之间的数中gcd(i, x)的个数
2. 定义：
   .

题意

由于这个题n和a都是非常巨大的，因此考虑怎么去简化

题中求的是个位数，所以模数mod只有10，在欧拉降幂中也只要进行三次就结束了，为了精确大于100时，我们设为100就行

然后考虑怎么取a，从模数的角度出发，如果模数为2，我们只要知道个位数即可。如果模数为4，我们只要知道个位数和十位数即可，因为百位肯定能被4整除…

接着就是套一个板子

Code

#include <bits/stdc++.h>
using namespace std;
//#define ACM_LOCAL
#define fi first
#define se second
#define il inline
#define re register
typedef long long ll;
typedef pair<int, int> PII;
typedef unsigned long long ull;
const int N = 2e7 + 10;
const int M = 1e6 + 10;
const ll INF = 1e18 + 5;
const double eps = 1e-5;
const int MOD = 998244353;
int prime[N], phi[N], k;
bool is_prime[N];
void init(int n) {
    memset(is_prime, true, sizeof is_prime);
    phi[1] = 1;
    for (int i = 2; i < n; i++) {
        if (is_prime[i]) prime[++k] = i, phi[i] = i - 1;
        for (int j = 1; j <= k && i * prime[j] < n; j++) {
            is_prime[i * prime[j]] = false;
            if (i % prime[j] == 0) {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        }
    }
}
ll mod(ll a, ll mm) {return a >= mm ? a % mm + mm : a;}
ll ksm(ll a, ll b, ll mm) {
    ll res = 1, base = a;
    while (b) {
        if (b & 1) res = mod(res * base, mm);
        base = mod(base * base, mm);
        b >>= 1;
    }
    return res;
}
ll calc(ll a, ll b, ll mm) {
    if (b == 0 || mm == 1) return 1;
    else return ksm(a, calc(a, b-1, phi[mm]), mm);
}
ll f(ll a, ll b, ll mm) {
    if (a == 0) return 0;
    ll ans = calc(a, b, mm) % mm;
    return ans;
}
void solve() {
    init(N);
    string a, b; cin >> a >> b;
    int base = 0;
    for (int i = max(0, (int)a.size() - 3); i < a.size(); i++) {
        base = base * 10 + (a[i] - '0');
    }
    int Pow = 100;
    if (b.size() <= 2) {
        Pow = 0;
        for (int i = 0; i < b.size(); i++) Pow = Pow * 10 + (b[i] - '0');
    }
    cout << f(base, Pow, 10) << endl;
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif
    solve();
}