题意

求 (b-1) * b ^n-1 % c，b∈[2,10^1000000 ]，n∈[1, 10^1000000 ]

分析

非常裸的欧拉降幂

根据定义可以化简n-1，特殊的b小于φ(m)的情况注意一下

b直接边读入边取膜即可

Code

#include 
   
    using namespace std;//#define ACM_LOCAL#define fi first#define se second#define il inline#define re registertypedef long long ll;typedef pair
    
      PII;typedef unsigned long long ull;const int N = 2e5 + 10;const int M = 1e6 + 10;const ll INF = 1e18 + 5;const double eps = 1e-5;const int MOD = 998244353;ll init(ll n) {       ll m = (int)sqrt(n + 0.5);    ll ans = n;    for (ll i = 2; i <= m; ++ i) {           if (n % i == 0) {               ans = ans / i * (i - 1);            while(n % i == 0) n /= i;        }    }    if (n > 1) ans = ans / n * (n - 1);    return ans;}ll ksm(ll a, ll b, ll mm) {       ll res = 1, base = a;    while (b) {           if (b & 1) res = res * base % mm;        base = base * base % mm;        b >>= 1;    }    return res;}void solve() {       string b, n; cin >> b >> n;    ll c; cin >> c;    ll mm = init(c);    ll b_1, b_ = 0;//取b-1,b的值    for (int i = 0; i < b.size(); i++) b_ = b_ * 10 + (b[i] - '0'), b_ %= c;    b_1 = (b_ - 1 + c) % c;    ll n_1 = 0, flag = 0;//取n-1的值    for (int i = 0; i < n.size(); i++) {           n_1 = n_1 * 10 + (n[i] - '0');        if (n_1 - 1 >= mm) n_1 %= mm, flag = 1;    }    n_1 = (n_1 - 1 + mm) % mm;    if (flag) n_1 += mm;    ll ans = ksm(b_, n_1, c);    ans = ans * b_1 % c;    if (ans == 0) cout << c << endl;    else cout << ans << 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();}

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