結果

問題 No.2068 Restricted Permutation
ユーザー woodywoodywoodywoody
提出日時 2022-09-02 22:54:27
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 4,573 bytes
コンパイル時間 4,589 ms
コンパイル使用メモリ 233,840 KB
実行使用メモリ 81,572 KB
最終ジャッジ日時 2024-11-16 05:25:02
合計ジャッジ時間 8,662 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
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 114 ms
81,448 KB
testcase_01 AC 115 ms
81,280 KB
testcase_02 AC 115 ms
81,488 KB
testcase_03 AC 114 ms
81,500 KB
testcase_04 AC 114 ms
81,408 KB
testcase_05 AC 114 ms
81,408 KB
testcase_06 AC 115 ms
81,280 KB
testcase_07 AC 116 ms
81,280 KB
testcase_08 WA -
testcase_09 AC 114 ms
81,408 KB
testcase_10 WA -
testcase_11 AC 114 ms
81,408 KB
testcase_12 AC 113 ms
81,408 KB
testcase_13 WA -
testcase_14 AC 115 ms
81,408 KB
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 AC 147 ms
81,408 KB
testcase_20 AC 148 ms
81,280 KB
testcase_21 AC 150 ms
81,280 KB
testcase_22 AC 148 ms
81,280 KB
testcase_23 WA -
testcase_24 WA -
testcase_25 WA -
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 #

// #define _GLIBCXX_DEBUG
#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,b) for(int i=0;i<b;i++)
#define rrep(i,b) for(int i=b-1;i>=0;i--)
#define rep1(i,b) for(int i=1;i<b;i++)
#define repx(i,x,b) for(int i=x;i<b;i++)
#define rrepx(i,x,b) for(int i=b-1;i>=x;i--)
#define fore(i,a) for(auto i:a)
#define fore1(i,a) for(auto &i:a)
#define rng(x) (x).begin(), (x).end()
#define rrng(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define fi first
#define se second
#define pcnt __builtin_popcountll

using namespace std;
using namespace atcoder;

using ll = long long;
using ld = long double;
template<typename T> using mpq = priority_queue<T, vector<T>, greater<T>>;
template<typename T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<typename T> bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T> ll sumv(const vector<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;}
bool yn(bool a) { if(a) {cout << "Yes" << endl; return 1;} else {cout << "No" << endl; return 0;}}
#define dame { cout << "No" << endl; return 0;}
#define dame1 { cout << -1 << endl; return 0;}
#define test(x) cout << "test" << x << endl;
#define deb(x,y) cout << x << " " << y << endl;
#define debp(p) cout << p.fi << " " << p.se << endl;
#define deb3(x,y,z) cout << x << " " << y << " " << z << endl;
#define deb4(x,y,z,x2) cout << x << " " << y << " " << z << " " << x2 << endl;
#define out cout << ans << endl;
#define outd cout << fixed << setprecision(20) << ans << endl;
#define outm cout << ans.val() << endl;
#define outv fore(yans , ans) cout << yans << "\n";
#define outdv fore(yans , ans) cout << yans.val() << "\n";
#define show(x) cerr<<#x<<" = "<<x<<endl;
#define showdeb(x,y) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<endl;
#define showv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define showv2(v) fore(vy , v) cout << vy << "\n";
#define showvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define showvm2(v) fore(vy , v) cout << vy.val() << "\n";
using pll = pair<ll,ll>;using pil = pair<int,ll>;using pli = pair<ll,int>;using pii = pair<int,int>;using pdd = pair<ld,ld>;
using tp = tuple<int ,int ,int>;
using vi = vector<int>;using vd = vector<ld>;using vl = vector<ll>;using vs = vector<string>;using vb = vector<bool>;
using vpii = vector<pii>;using vpli = vector<pli>;using vpll = vector<pll>;using vpil = vector<pil>;
using vvi = vector<vector<int>>;using vvl = vector<vector<ll>>;using vvs = vector<vector<string>>;using vvb = vector<vector<bool>>;
using vvpii = vector<vector<pii>>;using vvpli = vector<vector<pli>>;using vvpll = vector<vpll>;using vvpil = vector<vpil>;
using mint = modint998244353;
using vm = vector<mint>;
using vvm = vector<vector<mint>>;
vector<int> dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1};
ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
#define yes {cout <<"Yes"<<endl;}
#define yesr { cout <<"Yes"<<endl; return 0;}
#define no {cout <<"No"<<endl;}
#define nor { cout <<"No"<<endl; return 0;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;

const int mod = 998244353;

struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
} com(10000005);

int main(){
    int n,k,x; cin>>n>>k>>x;

    auto f = [&](mint y)->mint{
        // if (y == 0) return 0; 
        mint t = y;
        mint ret = t * (t-1) / 2;

        return ret;
    };

    mint ans = 0;
    k--;

    rep(i,n){
        mint tmp = 0;
        if (i <= k){
            tmp = f(com(n-1,i)*com.fact[i]) * com.fact[n-i-1] * com.fact[n-i-1];
            if (i != k) tmp += mint(n-x) * com(n-2 , i) * com.fact[i] * com.fact[n-i-2] * com.fact[n-i-1];
            else tmp += com(n-1,i) * com.fact[n-i-1] * (com.fact[n-i-1]-1) / 2;
        }else{
            tmp = f(com(n-1,k)*com.fact[k]) * com.fact[n-k-1] * com.fact[n-k-1];
            tmp += mint(x-1) * com(n-2 , k) * com.fact[k] * com.fact[n-k-2] * com.fact[n-k-1];
        }
        ans += tmp;
        // cout << tmp.val() << endl;
        // show(i);
    }

    outm;

// 消せ!!!!!! #define _GLIBCXX_DEBUG
// priority_queue でしぬ

    return 0;
}
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