結果

問題 No.2380 Sylow P-subgroup
ユーザー nikaj
提出日時 2023-07-14 22:03:51
言語 C++11(廃止可能性あり)
(gcc 13.3.0)
結果
AC  
実行時間 60 ms / 2,000 ms
コード長 2,451 bytes
コンパイル時間 1,426 ms
コンパイル使用メモリ 162,192 KB
実行使用メモリ 7,176 KB
最終ジャッジ日時 2024-09-16 07:05:22
合計ジャッジ時間 3,011 ms
ジャッジサーバーID
(参考情報)
judge5 / judge6
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 14
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#define F0(i,n) for (int i=0; i<n; i++)
#define F1(i,n) for (int i=1; i<=n; i++)
#define CL(a,x) memset(x, a, sizeof(x));
#define SZ(x) ((int)x.size())
const int inf = 1000009;
const double pi = acos(-1.0);
typedef pair<int, int> pii;
typedef long long ll;
const double EPS = 1e-9;
const int MOD = 998244353;
#define PR(x) cerr << #x << "=" << x << endl
template<class A, class B>
ostream& operator<<(ostream& os, const pair<A, B>& p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
template<class A, class B, class C>
ostream& operator<<(ostream& os, const tuple<A, B, C>& p) { os << "(" << get<0>(p) << ", " << get<1>(p) << ", " << get<2>(p) << ")"; return os; }
istream& operator>>(istream& is, pii& p) { is>>p.first>>p.second; return is; }
template<class T>
ostream& operator<<(ostream& os, const vector<T>& v) { os << "["; F0(i,SZ(v)) { if (i>0) os << ","; os << v[i]; } os << "]"; return os; }
template<class T>
ostream& operator<<(ostream& os, const set<T>& v) { os << "{"; int f=1; for(auto i:v) { if(f)f=0;else os << ","; cerr << i; } os << "}" << endl;
    return os; }
template<class T, class R>
ostream& operator<<(ostream& os, const map<T,R>& v) { os << "{"; int f=1; for(auto i:v) { if(f)f=0;else os << ", "; cerr << i.first << ":" << i.second
    ; } os << "}" << endl; return os; }
int i, j, k;
ll n, m, ans;
const int N = 200005;
const int M = 435;
int dp[N];
const int DX[]={-1,0,1,0};
const int DY[]={0,1,0,-1};
const string CS="nesw";
const string HS="URDL";
void Add(int& x, int y) {
x += y;
if (x >= MOD) x -= MOD;
}
ll Mult(int x, int y) {
return 1LL * x * y % MOD;
}
ll modpow(ll x, ll n) {
if (n == 0) return 1;
ll y = modpow(x, n / 2); y = (y * y) % MOD;
if (n & 1) y = (y * x) % MOD;
return y;
}
const int MAXN = 200001;
ll f[MAXN], rf[MAXN];
ll C(int n, int k) {
return f[n] * rf[k] % MOD * rf[n - k] % MOD;
}
void Solve() {
ll s = m;
ans = 0;
while (n >= s) {
ans += n / s;
s *= m;
}
cout << modpow(m, ans) << endl;
}
int main() {
//ignore = freopen("x.in", "r", stdin);
//freopen("x.out", "w", stdout);
//int tn; cin >> tn;
//F1(ti, tn) {
f[0] = rf[0] = 1;
for (int i = 1; i <= 200000; i++) {
f[i] = (f[i - 1] * i) % MOD;
rf[i] = modpow(f[i], MOD - 2);
}
while (cin >> n >> m) {
//PR(n);
Solve();
}
return 0;
}
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