結果

問題 No.1220 yukipoker
ユーザー karinohito
提出日時 2023-07-25 22:07:01
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 171 ms / 2,000 ms
コード長 2,817 bytes
コンパイル時間 2,198 ms
コンパイル使用メモリ 200,628 KB
最終ジャッジ日時 2025-02-15 19:07:05
ジャッジサーバーID
(参考情報)
judge5 / judge1
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 1
other AC * 25
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;
//#include <atcoder/all>
//using namespace atcoder;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vd = vector<double>;
using vvd = vector<vd>;
using vvvd = vector<vvd>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vvvb = vector<vvb>;
#define all(A) A.begin(),A.end()
#define rep(i, n) for (ll i = 0; i < (ll) (n); i++)
template<class T>
bool chmax(T& p, T q, bool C = 1) {
if (C == 0 && p == q) {
return 1;
}
if (p < q) {
p = q;
return 1;
}
else {
return 0;
}
}
template<class T>
bool chmin(T& p, T q, bool C = 1) {
if (C == 0 && p == q) {
return 1;
}
if (p > q) {
p = q;
return 1;
}
else {
return 0;
}
}
ll modPow(long long a, long long n, long long p) {
if (n == 0) return 1; // 0
if (n == 1) return a % p;
if (n % 2 == 1) return (a * modPow(a, n - 1, p)) % p;
long long t = modPow(a, n / 2, p);
return (t * t) % p;
}
ll gcd(ll(a), ll(b)) {
if (a == 0)return b;
if (b == 0)return a;
ll c = a;
while (a % b != 0) {
c = a % b;
a = b;
b = c;
}
return b;
}
ll sqrtz(ll N) {
ll L = 0;
ll R = sqrt(N) + 10000;
while (abs(R - L) > 1) {
ll mid = (R + L) / 2;
if (mid * mid <= N)L = mid;
else R = mid;
}
return L;
}
/*
using mint = modint998244353;
using vm = vector<mint>;
using vvm = vector<vm>;
using vvvm = vector<vvm>;
using vvvvm = vector<vvvm>;
using vvvvvm = vector<vvvvm>;
using vvvvvvm = vector<vvvvvm>;
vector<mint> fact, factinv, inv;
const ll mod = 998244353;
void prenCkModp(ll n) {
fact.resize(n + 5);
factinv.resize(n + 5);
inv.resize(n + 5);
fact[0] = fact[1] = 1;
factinv[0] = factinv[1] = 1;
inv[1] = 1;
for (ll i = 2; i < n + 5; i++) {
fact[i] = (fact[i - 1] * i);
inv[i] = (mod - ((inv[mod % i] * (mod / i))));
factinv[i] = (factinv[i - 1] * inv[i]);
}
}
mint nCk(ll n, ll k) {
if (n < k || k < 0) return 0;
return (fact[n] * ((factinv[k] * factinv[n - k])));
}
*/
vvb mul(vvb A,vvb B){
ll N=A.size();
vvb res(N,vb(N,0));
rep(i,N)rep(j,N){
rep(k,N){
if(A[k][j]&&B[i][k])res[i][j]=1;
}
}
return res;
}
int main() {
cin.tie(nullptr);
ios::sync_with_stdio(false);
ll N=1e5+10,M,K;
vd fl(N,0);
for(ll i=1;i<N;i++){
fl[i]=fl[i-1]+log(double(i));
}
ll T;
cin>>T;
rep(t,T){
cin>>N>>M>>K;
double F=log(double(M))+(fl[N]-fl[N-K]-fl[K]);
double S=double(log(double(N-K+1)))+double(K)*log(double(M));
cout<<(F>S?"Straight":"Flush")<<endl;
}
}
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