結果

問題 No.1830 Balanced Majority
ユーザー milkcoffeemilkcoffee
提出日時 2022-02-04 22:02:11
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 28 ms / 2,000 ms
コード長 8,275 bytes
コンパイル時間 4,228 ms
コンパイル使用メモリ 242,424 KB
実行使用メモリ 28,248 KB
平均クエリ数 7.54
最終ジャッジ日時 2024-06-11 11:51:33
合計ジャッジ時間 5,769 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 26 ms
27,856 KB
testcase_01 AC 25 ms
27,856 KB
testcase_02 AC 24 ms
27,600 KB
testcase_03 AC 25 ms
27,856 KB
testcase_04 AC 24 ms
27,856 KB
testcase_05 AC 24 ms
27,984 KB
testcase_06 AC 23 ms
27,984 KB
testcase_07 AC 25 ms
27,864 KB
testcase_08 AC 25 ms
27,992 KB
testcase_09 AC 25 ms
27,864 KB
testcase_10 AC 24 ms
27,352 KB
testcase_11 AC 25 ms
27,768 KB
testcase_12 AC 25 ms
28,120 KB
testcase_13 AC 28 ms
27,864 KB
testcase_14 AC 27 ms
27,608 KB
testcase_15 AC 26 ms
27,736 KB
testcase_16 AC 26 ms
27,864 KB
testcase_17 AC 26 ms
27,608 KB
testcase_18 AC 25 ms
27,608 KB
testcase_19 AC 25 ms
27,864 KB
testcase_20 AC 25 ms
27,608 KB
testcase_21 AC 26 ms
27,736 KB
testcase_22 AC 26 ms
28,248 KB
testcase_23 AC 26 ms
27,608 KB
testcase_24 AC 26 ms
27,992 KB
testcase_25 AC 27 ms
27,864 KB
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>
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#define ll long long
#define forin(in, n)           \
    for (ll i = 0; i < n; i++) \
    cin >> in[i]
#define forout(out)                         \
    for (ll i = 0; i < (ll)out.size(); i++) \
    cout << out[i] << endl
#define rep(i, n) for (ll i = 0; i < n; ++i)
#define rep_up(i, a, n) for (ll i = a; i < n; ++i)
#define rep_down(i, a, n) for (ll i = a; i >= n; --i)
#define P pair<ll, ll>
#define pb push_back
#define endl "\n"

#define all(v) v.begin(), v.end()
#define fi first
#define se second
#define vvvvll vector< vector <vector <vector<ll> > > >
#define vvvll vector< vector< vector<ll> > >
#define vvll vector< vector<ll> >
#define vll vector<ll>
#define pqll priority_queue<ll>
#define pqllg priority_queue<ll, vector<ll>, greater<ll>>

template<class T> inline void vin(vector<T>& v) { rep(i, v.size()) cin >> v.at(i); }
template <class T>
using V = vector<T>;

constexpr ll INF = (1ll << 60);
//constexpr ll mod = 1000000007;
constexpr ll mod = 998244353;

constexpr double pi = 3.14159265358979323846;
template <typename T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
void pt(T val)
{
    cout << val << "\n";
}
template <typename T>
void pt_vll(vector<T> &v)
{
    ll vs = v.size();
    rep(i, vs)
    {
        cout << v[i];

        if (i == vs - 1)
            cout << "\n";
        else
            cout << " ";
    }
}
ll mypow(ll a, ll n)
{
    ll ret = 1;
    if (n == 0)
        return 1;
    if (a == 0)
        return 0;
    rep(i, n)
    {
        if (ret > (ll)(9e18 + 10) / a)
            return -1;
        ret *= a;
    }
    return ret;
}
long long modpow(long long a, long long n, long long mod)
{
    long long res = 1;
    while (n > 0)
    {
        if (n & 1)
            res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
long long modinv(long long a, long long m)
{
    long long b = m, u = 1, v = 0;
    while (b)
    {
        long long t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}

struct mint {
    ll x;
    constexpr mint(ll x = 0) noexcept : x((x % mod + mod) % mod) {}

    constexpr mint& operator+=(const mint& a) noexcept {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    constexpr mint& operator-=(const mint& a) noexcept {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    constexpr mint& operator*=(const mint& a) noexcept { (x *= a.x) %= mod; return *this; }
    constexpr mint& operator/=(const mint& a) noexcept { return *this *= a.inv(); }

    constexpr mint operator-() const noexcept { return mint(-x); }
    constexpr mint operator+(const mint& a) const noexcept { return mint(*this) += a; }
    constexpr mint operator-(const mint& a) const noexcept { return mint(*this) -= a; }
    constexpr mint operator*(const mint& a) const noexcept { return mint(*this) *= a; }
    constexpr mint operator/(const mint& a) const noexcept { return mint(*this) /= a; }
    constexpr bool operator==(const mint& a) const noexcept { return x == a.x; }
    constexpr bool operator!=(const mint& a) const noexcept { return x != a.x; }

    constexpr mint pow(ll n) const {
        if (n == 0) return 1;
        mint res = pow(n >> 1);
        res *= res;
        if (n & 1) res *= *this;
        return res;
    }
    constexpr mint inv() const { return pow(mod - 2); }

    friend istream& operator>>(istream& is, mint& a) noexcept {
        ll v; is >> v;
        a = mint(v);
        return is;
    }
    friend ostream& operator<<(ostream& os, const mint& a) noexcept {
        return os << a.x;
    }
};

struct UnionFind
{
    vector<ll> par, size;

    UnionFind(ll N) : par(N)
    { //最初はすべてが根であるとして初期化
        size.resize(N, 1);
        for (ll i = 0; i < N; i++)
            par[i] = i;
    }

    ll root(ll x)
    { //データxの木の根を再帰で得る
        if (par[x] == x)
            return x;
        return par[x] = root(par[x]);
    }

    void unite(ll x, ll y)
    { //xとyの木を併合
        ll rx = root(x);
        ll ry = root(y);
        if (rx == ry)
            return; //同じ木にあるときはそのまま
        if (size[rx] > size[ry])
        {
            par[ry] = rx;
            size[rx] += size[ry];
        }
        else
        {
            par[rx] = ry;
            size[ry] += size[rx];
        }
        return;
    }

    bool same(ll x, ll y)
    { //2つのデータx,yが属する木が同じならtrue
        ll rx = root(x);
        ll ry = root(y);
        return rx == ry;
    }

    ll treesize(ll x) { return size[root(x)]; }
};

const int MAX = 2010000;
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit()
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; 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
long long 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;
}

vector<ll> enum_div(ll n)
{ //約数全列挙
    vector<ll> ret;
    for (ll i = 1; i * i <= n; ++i)
    {
        if (n % i == 0)
        {
            ret.push_back(i);
            if (i * i != n)
            {
                ret.push_back(n / i);
            }
        }
    }
    return ret;
}

void make_prime(vector<ll> &ret, ll n)
{ //素因数分解
    ll x = n;
    for (ll i = 2; i * i <= x; i++)
    {
        while (n % i == 0)
        {
            n /= i;
            ret.push_back(i);
        }
    }
    if (n != 1)
    {
        ret.push_back(n);
    }
    return;
}

vector<bool> prime(1000010, true);
vector<ll> pri(1000010);
vector<bool> isprime(int N)
{ //素数判定
    if (N >= 0)
        prime[0] = false;
    if (N >= 1)
        prime[1] = false;
    for (ll i = 2; i * i <= N; i++)
    {
        if (!prime[i])
        {
            continue;
        }
        for (ll j = i * i; j <= N; j += i)
        {
            if (prime[j])
                pri[j] = i;
            prime[j] = false;
        }
    }
    return prime;
}

map<ll,ll> compression(vector<ll> v){
    map<ll,ll> ret;
    ll cnt = 0;
    sort(v.begin(),v.end());
    for(ll i=0; i<v.size(); i++){
        if(!ret.count(v[i])){
            ret[v[i]] = cnt;
            cnt++;
        }
    }
    return ret;
}

//素因数分解: void make_prime(vector<ll> &ret, ll n)
//素数判定　: vector<bool> isprime(int N)
//約数全列挙: vector<ll> enum_div(ll n)

void solve(){
    ll n, m, k, cnt = 0, sum = 0, ans = 0;
    cin>>n;
    ll x,y;
    cout<<"? "<<1<<endl;
    cout.flush();
    cin>>x;
    cout<<"? "<<n-1<<endl;
    cout.flush();
    cin>>y;
    ll flag=0;
    if(y-x==(n-2)/2){
        cout<<"! 2 "<<n-1<<endl;
        cout.flush();
        return;
    }
    if(y-x<(n-2)/2) flag=1;
    ll mid,s,t;
    s=1;
    t=n-1;
    ll z;
    cnt=0;
    while(t-s>1){
        mid=(t+s)/2;
        cout<<"? "<<mid<<endl;
        cout.flush();
        cin>>z;
        if(2*z==mid){
            if(mid<n/2){
                cout<<"! "<<mid+1<<" "<<n<<endl;
                cout.flush();
            }else{
                cout<<"! 1 "<<mid<<endl;
                cout.flush();
            }
            return;
        }
        if(2*z>mid){
            if(flag==1) s=mid;
            else t=mid;
        }else{
            if(flag==0) s=mid;
            else t=mid;
        }
        cnt++;
        if(cnt==18) break;
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    //cout << fixed << setprecision(16);
    //ll T;
    //cin>>T;
    //rep(ca,T)
    solve();
}
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