結果

問題 No.275 中央値を求めよ
ユーザー soraie_soraie_
提出日時 2020-11-02 22:56:05
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 1,000 ms
コード長 4,810 bytes
コンパイル時間 1,969 ms
コンパイル使用メモリ 180,844 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-07-22 08:15:13
合計ジャッジ時間 2,748 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
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 * 3
other AC * 38
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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
/*----------------------------------ここからマクロ----------------------------------*/
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define vecin(a) rep(i,a.size())cin >> a[i]
#define overload4(_1,_2,_3,_4,name,...) name
/*#define rep1(n) for(int i=0;i<(int)n;++i)
#define rep2(i,n) for(int i=0;i<(int)n;++i)
#define rep3(i,a,b) for(int i=(int)a;i<(int)b;++i)
#define rep4(i,a,b,c) for(int i=(int)a;i<(int)b;i+=(int)c)*/
#define rep1(n) for(ll i=0;i<(ll)n;++i)
#define rep2(i,n) for(ll i=0;i<(ll)n;++i)
#define rep3(i,a,b) for(ll i=(ll)a;i<(ll)b;++i)
#define rep4(i,a,b,c) for(ll i=(ll)a;i<(ll)b;i+=(ll)c)
#define rep(...) overload4(__VA_ARGS__,rep4,rep3,rep2,rep1)(__VA_ARGS__)
#ifdef _DEBUG
#define debug1(a) cerr << #a << ": " << a << "\n"
#define debug2(a,b) cerr << #a << ": " << a << ", " << #b << ": " << b << "\n"
#define debug3(a,b,c) cerr << #a << ": " << a << ", " << #b << ": " << b << ", " << #c << ": " << c << "\n"
#define debug4(a,b,c,d) cerr << #a << ": " << a << ", " << #b << ": " << b << ", " << #c << ": " << c << ", " << #d << ": " << d << "\n"
#define debug(...) overload4(__VA_ARGS__,debug4,debug3,debug2,debug1)(__VA_ARGS__)
#define vecout(a) cerr << #a << ": [";rep(i,a.size()){cerr << a[i];cerr << (i == a.size() - 1 ? "":",");}cerr << "]\n"
#else
#define debug(...)
#define vecout(a)
#endif
#define mp make_pair
#define fi first
#define se second
//struct doset{doset(int n){cout << fixed << setprecision(n);cerr << fixed << setprecision(n);}};
//struct myset{myset(){ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}};
void myset(){ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}
void doset(int n){cout << fixed << setprecision(n);cerr << fixed << setprecision(n);}
void koko(){cerr << "koko\n";}
using ll = long long;
using ld = long double;
using dou = double;
const int inf = 1 << 30;
const ll INF = 1LL << 61;
const ld pi = 3.14159265358;
const ll mod1 = 1000000007LL;
const ll mod2 = 998244353LL;
typedef pair<ll,ll> P;
template<class T, class U> inline bool chmin(T& a, const U& b){ if(a > b){ a = b; return 1; } return 0; }
template<class T, class U> inline bool chmax(T& a, const U& b){ if(a < b){ a = b; return 1; } return 0; }
//nのm乗をMODで割ったあまりO(logm)
ll modpow(ll n,ll m,ll MOD){
    if(m == 0)return 1;
    if(m < 0)return 0;
    ll res = 1;
    n %= MOD;
    while(m){
        if(m & 1)res = (res * n) % MOD;
        m >>= 1;
        n *= n;
        n %= MOD;
    }
    return res;
}

ll mypow(ll n,ll m){
    if(m == 0)return 1;
    if(m < 0)return -1;
    ll res = 1;
    while(m){
        if(m & 1)res = (res * n);
        m >>= 1;
        n *= n;
    }
    return res;
}

//素数判定O(sqrt(N))
template<class T>
inline bool isp(T n){
    bool res = true;
    if(n == 1 || n == 0)return false;
    else{
        for(ll i = 2;i * i <= n;i++){
            if(n % i == 0){
                res = false;
                break;
            }
        }
        return res;
    }
}


inline bool Yes(bool b = 1){cout << (b ? "Yes\n":"No\n");return b;}
inline bool YES(bool b = 1){cout << (b ? "YES\n":"NO\n");return b;}


map<ll,ll> primefactor(ll n){
    map<ll,ll> ma;
    if(n <= 1)return ma;
    ll m = n;
    for(ll i = 2;i * i <= n;i++){
        while(m % i == 0){
            ma[i]++;
            m /= i;
        }
    }
    if(m != 1)ma[m]++;
    return ma;
}

ll __lcm(ll a,ll b){
    return a / __gcd(a,b) * b;
}


inline void in(){return;}
template<class First,class...Other>inline void in(First& F,Other&...O){cin >> F;in(O...);}



template<ll MOD = mod1>
struct fac_solver{
private:
    ll cmax;
    vector<ll> fac,finv,inv;
public:
    fac_solver(ll n = 100000):cmax(n),fac(n),finv(n),inv(n){
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        inv[1] = 1;
        for (ll i = 2; i < cmax; i++){
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }

    //二項係数計算nCk
    ll COM(ll n, ll k){
        if (n < k) return 0;
        if (n < 0 || k < 0) return 0;
        return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
    }

    ll PER(ll n,ll k){
        if(n < k || n < 0 || k < 0)return 0;
        return fac[n] * finv[n - k] % MOD;
    }

    ll stair_pow(ll n){return fac[n];}
};

/*----------------------------------マクロここまで----------------------------------*/


int main(){
    myset();
    int N;
    cin >> N;
    vector<int> vec(N);
    rep(i,N)cin >> vec[i];
    sort(all(vec));
    if(N & 1)cout << vec[N / 2] << "\n";
    else cout << ld(vec[N / 2 - 1] + vec[N / 2]) / 2.0 << "\n";
}
0