結果

問題 No.2382 Amidakuji M
ユーザー cn_449
提出日時 2023-07-14 21:59:36
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 59 ms / 2,000 ms
コード長 4,047 bytes
コンパイル時間 1,320 ms
コンパイル使用メモリ 130,240 KB
最終ジャッジ日時 2025-02-15 14:02:38
ジャッジサーバーID
（参考情報） 		judge4 / judge3
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 * 3
other AC * 19
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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <iomanip>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cassert>
#include <complex>
#include <stdio.h>
#include <time.h>
#include <numeric>
#include <random>
#include <unordered_set>
#include <unordered_map>
#define all(a) (a).begin(), (a).end()
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define req(i, a, b) for (ll i = (a); i < (b); i++)
#define pb push_back
#define debug(x) cerr << __LINE__ << ' ' << #x << ':' << (x) << '\n'
#define debug2(x, y) cerr << __LINE__ << ' ' << #x << ':' << (x) << ',' << #y << ':' << (y) << '\n'
#define debug3(x, y, z) cerr << __LINE__ << ' ' << #x << ':' << (x) << ',' << #y << ':' << (y) << ',' << #z << ':' << (z) << '\n'
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef long double ld;
template<class T> using P = pair<T, T>;
template<class T> using pri_s = priority_queue<T, vector<T>, greater<T>>;
template<class T> using pri_b = priority_queue<T>;
constexpr int inf = 1000000010;
constexpr int inf2 = 2000000010;
constexpr ll INF = 1000000000000000010;
constexpr ll INF4 = 4000000000000000010;
constexpr int mod1e9 = 1000000007;
constexpr int mod998 = 998244353;
constexpr ld eps = 1e-12;
constexpr ld pi = 3.141592653589793238;
constexpr ll ten(int n) { return n ? 10 * ten(n - 1) : 1; };
int dx[] = { 1,0,-1,0,1,1,-1,-1,0 }; int dy[] = { 0,1,0,-1,1,-1,1,-1,0 };
ll mul(ll a, ll b) { return (b != 0 && a > INF / b ? INF : a * b); }
void fail() { cout << "-1\n"; exit(0); } void no() { cout << "No\n"; exit(0); }
template<class T> void er(T a) { cout << a << '\n'; exit(0); }
template<class T, class U> inline bool chmax(T &a, const U &b) { if (a < b) { a = b; return true; } return false; }
template<class T, class U> inline bool chmin(T &a, const U &b) { if (a > b) { a = b; return true; } return false; }
template<class T> istream &operator >>(istream &s, vector<T> &v) { for (auto &e : v) s >> e; return s; }
template<class T> ostream &operator <<(ostream &s, const vector<T> &v) { for (auto &e : v) s << e << ' '; return s; }
template<class T, class U> ostream &operator << (ostream &s, const pair<T, U> &p) { s << p.first << ' ' << p.second; return s; }
struct fastio {
    fastio() {
        cin.tie(0); cout.tie(0);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(20);
        cerr << fixed << setprecision(20);
    }
}fastio_;
template<class T> class segtree {
    int n;
    vector<T> data;
    T id = 0;
    T operation(T a, T b) { return a + b; }
public:
    segtree(ll _n) {
        n = 1;
        while (n < _n + 2) n <<= 1;
        data = vector<T>(2 * n, id);
    }
    segtree(vector<T> vec) {
        int _n = vec.size();
        n = 1;
        while (n < _n + 2) n <<= 1;
        data = vector<T>(2 * n, id);
        for (int i = 0; i < _n; i++) data[i + n] = vec[i];
        for (int i = n - 1; i >= 1; i--) data[i] = operation(data[i << 1], data[i << 1 | 1]);
    }
    void change(int i, T x) {
        i += n;
        data[i] = x;
        while (i > 1) {
            i >>= 1;
            data[i] = operation(data[i << 1], data[i << 1 | 1]);
        }
    }
    void add(int i, T x) { change(i, data[i + n] + x); }
    T get(int a, int b) {
        T left = id; T right = id;
        a += n; b += n;
        while (a < b) {
            if (a & 1) left = operation(left, data[a++]);
            if (b & 1) right = operation(data[--b], right);
            a >>= 1; b >>= 1;
        }
        return operation(left, right);
    }
    T get_all() { return data[1]; }
    T operator[](int i) { return data[i + n]; }
};
int main() {
    ll n, m;
    cin >> n >> m;
    vector<ll> p(n);
    cin >> p;
    ll cnt = 0;
    segtree<ll> seg(n);
    rep(i, n){
        p[i]--;
        cnt += seg.get(p[i] + 1, n);
        seg.add(p[i], 1);
    }
    if(m % 2 == 0 && cnt % 2 == 1) fail();
    else if(m % 2 == 0){
        ll r = cnt % m;
        if(r) cnt += m - r;
        cout << cnt << '\n';
    }
    else {
        ll r = cnt % m;
        if(r){
            if(r % 2 == 0) cnt += 2 * m - r;
            else cnt += m - r;
        }
        cout << cnt << '\n';
    }
}
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