結果

問題 No.75 回数の期待値の問題
ユーザー pionepione
提出日時 2020-12-20 09:41:13
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 5,000 ms
コード長 2,631 bytes
コンパイル時間 1,742 ms
コンパイル使用メモリ 167,832 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-09-21 11:34:41
合計ジャッジ時間 2,436 ms
ジャッジサーバーID
(参考情報)
judge1 / judge4
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 * 4
other AC * 16
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;
// #define int long long
#define rep(i, n) for (long long i = (long long)(0); i < (long long)(n); ++i)
#define reps(i, n) for (long long i = (long long)(1); i <= (long long)(n); ++i)
#define rrep(i, n) for (long long i = ((long long)(n)-1); i >= 0; i--)
#define rreps(i, n) for (long long i = ((long long)(n)); i > 0; i--)
#define irep(i, m, n) for (long long i = (long long)(m); i < (long long)(n); ++i)
#define ireps(i, m, n) for (long long i = (long long)(m); i <= (long long)(n); ++i)
#define irreps(i, m, n) for (long long i = ((long long)(n)-1); i > (long long)(m); ++i)
#define SORT(v, n) sort(v, v + n);
#define REVERSE(v, n) reverse(v, v+n);
#define vsort(v) sort(v.begin(), v.end());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) (long long)(x.size())
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vs = vector<string>;
using vpll = vector<pair<ll, ll>>;
using vtp = vector<tuple<ll,ll,ll>>;
using vb = vector<bool>;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;
const ll INF = 1e9+10;
const ll MOD = 1e9+7;
const ll LINF = 1e18;
//
// dp[i]+=(dp[i+j]+1.0)/6
// k
// +1
// dp[i] i
// k
//
// dp[k]
// dp[k]
ll K;
double kari;
double dp[205];
double rec(ll k){
if(k==0) return 0;
if(dp[k]) return dp[k];
dp[k]+=1;
reps(i,6){
if(k-i<0) dp[k]+=kari/6;
else dp[k]+=rec(k-i)/6;
}
return dp[k];
}
signed main()
{
cin.tie( 0 ); ios::sync_with_stdio( false );
cin>>K;
double ok=10000, ng=0;
rep(_,100){
kari=(ok+ng)/2;
rep(i,205) dp[i]=0;
if(rec(K)<kari) ok=kari;
else ng=kari;
}
printf("%.15lf\n", dp[K]);
}
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