結果

問題 No.2326 Factorial to the Power of Factorial to the...
ユーザー deuteridayodeuteridayo
提出日時 2023-05-28 14:22:09
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 10 ms / 2,000 ms
コード長 2,740 bytes
コンパイル時間 7,084 ms
コンパイル使用メモリ 265,812 KB
実行使用メモリ 9,600 KB
最終ジャッジ日時 2024-12-27 00:46:02
合計ジャッジ時間 7,315 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge5
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 * 2
other AC * 20
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 # 

#include"bits/stdc++.h"
#include"atcoder/all"
using namespace std;
using namespace atcoder;
using lint = long long;
using ulint = unsigned long long;
#define endl '\n'
int const INF = 1<<30;
lint const INF64 = 1LL<<61;
lint const mod107 = 1e9+7;
using mint107 = modint1000000007;
long const mod = 998244353;
using mint = modint998244353;
lint ceilDiv(lint x, lint y){if(x >= 0){return (x+y-1)/y;}else{return x/y;}}
lint floorDiv(lint x, lint y){if(x >= 0){return x/y;}else{return (x-y+1)/y;}}
lint Sqrt(lint x){lint upper = 1e9;lint lower = 0;while(upper - lower > 0){lint mid = (1+upper + lower)/2;if(mid * mid > x){upper = mid-1;}else{lower = mid;}}return upper;}
lint gcd(lint a,lint b){if(a<b)swap(a,b);if(a%b==0)return b;else return gcd(b,a%b);}
lint lcm(lint a,lint b){return (a / gcd(a,b)) * b;}
lint chmin(vector<lint>&v){lint ans = INF64;for(lint i:v){ans = min(ans, i);}return ans;}
lint chmax(vector<lint>&v){lint ans = -INF64;for(lint i:v){ans = max(ans, i);}return ans;}
double dist(double x1, double y1, double x2, double y2){return sqrt(pow(x1-x2, 2) + pow(y1-y2,2));}
string toString(lint n){string ans = "";if(n == 0){ans += "0";}else{while(n > 0){int a = n%10;char b = '0' + a;string c = "";c += b;n /= 10;ans = c + ans;}}return ans;}
string toString(lint n, lint k){string ans = toString(n);string tmp = "";while(ans.length() + tmp.length() < k){tmp += "0";}return tmp + ans;}
vector<lint>prime;void makePrime(lint n){prime.push_back(2);for(lint i=3;i<=n;i+=2){bool chk = true;for(lint j=0;j<prime.size() && prime[j]*prime[j] <= i;j++){if(i % prime[j]==0){chk=false;break;}}if(chk)prime.push_back(i);}}
lint Kai[20000001]; bool firstCallnCr = true; 
lint ncrmodp(lint n,lint r,lint p){ if(firstCallnCr){ Kai[0] = 1; for(int i=1;i<=20000000;i++){ Kai[i] = Kai[i-1] * i; Kai[i] %= mod;} firstCallnCr = false;} if(n<0)return 0;
if(n < r)return 0;if(n==0)return 1;lint ans = Kai[n];lint tmp = (Kai[r] * Kai[n-r]) % p;for(lint i=1;i<=p-2;i*=2){if(i & p-2){ans *= tmp;ans %= p;}tmp *= tmp;tmp %= p;}return ans;}
#define rep(i, n) for(int i = 0; i < n; i++)
#define repp(i, x, y) for(int i = x; i < y; i++)
#define vec vector
#define pb push_back
#define se second
#define fi first
#define al(x) x.begin(),x.end()

struct edge{
    int to;
};
using graph = vector<vec<int>>;


	graph g(200000);
	vec<lint>L(200000,-1);
	vec<bool>went(200000,false);
	
lint f(int l,lint x){
	if(went[l])return 0;
	went[l]=true;
	L[l]=max(L[l],x);
	for(int v:g[l])f(v,L[l]);
	return 0;
}
	
int main(){
	lint n,p;
	cin>>n>>p;
	mint107 ans=0;
	for(lint pp=p;pp<=n;pp *=p){
		ans+=n/pp;
	}
	lint q=1;
	for(int i=2;i<=n;i++){
		q=(q*i)%(mod107-1);
	}
	lint r=1;
	repp(i,1,n+1)r=(r*i)%mod107;
	ans *= pow_mod(r,q,mod107);
	cout<<ans.val()<<endl;
}
0