結果

問題 No.1832 NAND Reversible
ユーザー snukesnuke
提出日時 2022-02-13 02:52:22
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 12 ms / 2,000 ms
コード長 5,296 bytes
コンパイル時間 2,235 ms
コンパイル使用メモリ 208,192 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-29 05:23:27
合計ジャッジ時間 3,151 ms
ジャッジサーバーID
(参考情報)
judge2 / judge1
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 2 ms
6,812 KB
testcase_01 AC 2 ms
6,944 KB
testcase_02 AC 1 ms
6,940 KB
testcase_03 AC 12 ms
6,940 KB
testcase_04 AC 2 ms
6,940 KB
testcase_05 AC 2 ms
6,940 KB
testcase_06 AC 11 ms
6,940 KB
testcase_07 AC 11 ms
6,944 KB
testcase_08 AC 11 ms
6,940 KB
testcase_09 AC 2 ms
6,944 KB
testcase_10 AC 11 ms
6,940 KB
testcase_11 AC 1 ms
6,940 KB
testcase_12 AC 2 ms
6,940 KB
testcase_13 AC 2 ms
6,944 KB
testcase_14 AC 2 ms
6,940 KB
testcase_15 AC 5 ms
6,940 KB
testcase_16 AC 8 ms
6,944 KB
testcase_17 AC 11 ms
6,944 KB
testcase_18 AC 11 ms
6,944 KB
testcase_19 AC 10 ms
6,940 KB
testcase_20 AC 10 ms
6,940 KB
testcase_21 AC 11 ms
6,944 KB
testcase_22 AC 11 ms
6,940 KB
testcase_23 AC 2 ms
6,940 KB
testcase_24 AC 2 ms
6,940 KB
testcase_25 AC 8 ms
6,940 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>
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rep1(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < (t); ++i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define pob pop_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define newline puts("")
using namespace std;
template<typename T> using vc = vector<T>;
template<typename T> using vv = vc<vc<T>>;
template<typename T> using PQ = priority_queue<T,vc<T>,greater<T>>;
using uint = unsigned; using ull = unsigned long long;
using vi = vc<int>; using vvi = vv<int>; using vvvi = vv<vi>;
using ll = long long; using vl = vc<ll>; using vvl = vv<ll>; using vvvl = vv<vl>;
using P = pair<int,int>; using vp = vc<P>; using vvp = vv<P>;
int getInt(){int x;scanf("%d",&x);return x;}
vi pm(int n, int s=0) { vi a(n); iota(rng(a),s); return a;}
template<typename T>istream& operator>>(istream&i,vc<T>&v){rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const T&v,const string& d=""){stringstream s;rep(i,sz(v))(i?s<<d:s)<<v[i];return s.str();}
template<typename T>ostream& operator<<(ostream&o,const vc<T>&v){if(sz(v))o<<join(v," ");return o;}
template<typename T1,typename T2>istream& operator>>(istream&i,pair<T1,T2>&v){return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,const pair<T1,T2>&v){return o<<v.fi<<","<<v.se;}
template<typename T1,typename T2>bool mins(T1& x,const T2&y){if(x>y){x=y;return true;}else return false;}
template<typename T1,typename T2>bool maxs(T1& x,const T2&y){if(x<y){x=y;return true;}else return false;}
template<typename Tx, typename Ty>Tx dup(Tx x, Ty y){return (x+y-1)/y;}
template<typename T>ll suma(const vc<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;}
template<typename T>ll suma(const vv<T>&a){ll res(0);for(auto&&x:a)res+=suma(x);return res;}
template<typename T>void uni(T& a){sort(rng(a));a.erase(unique(rng(a)),a.end());}
template<typename T>void prepend(vc<T>&a,const T&x){a.insert(a.begin(),x);}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return;}
#define yes { puts("Yes"); return;}
#define no { puts("No"); return;}
#define ret(x) { cout<<(x)<<endl; return;}
#define yn {puts("Yes");}else{puts("No");}
const int MX = 200005;

// Mod int
const uint mod = 998244353;
struct mint {
  uint x;
  mint(): x(0) {}
  mint(ll x):x((x%mod+mod)%mod) {}
  mint operator-() const { return mint(0) - *this;}
  mint operator~() const { return mint(1) / *this;}
  mint& operator+=(const mint& a) { if((x+=a.x)>=mod) x-=mod; return *this;}
  mint& operator-=(const mint& a) { if((x+=mod-a.x)>=mod) x-=mod; return *this;}
  mint& operator*=(const mint& a) { x=(ull)x*a.x%mod; return *this;}
  mint& operator/=(const mint& a) { x=(ull)x*a.pow(mod-2).x%mod; return *this;}
  mint operator+(const mint& a) const { return mint(*this) += a;}
  mint operator-(const mint& a) const { return mint(*this) -= a;}
  mint operator*(const mint& a) const { return mint(*this) *= a;}
  mint operator/(const mint& a) const { return mint(*this) /= a;}
  mint pow(ll t) const {
    mint res = 1; for (mint p=x;t;p*=p,t>>=1) if (t&1) res *= p; return res;
  }
  mint ppow(ll t) const { int p=mod-1; return pow((t%p+p)%p);}
  bool operator<(const mint& a) const { return x < a.x;}
  bool operator==(const mint& a) const { return x == a.x;}
  bool operator!=(const mint& a) const { return x != a.x;}
};
mint ex(mint x, ll t) { return x.pow(t);}
istream& operator>>(istream&i,mint&a) {i>>a.x;return i;}
//*
ostream& operator<<(ostream&o,const mint&a) {o<<a.x;return o;}
/*/
ostream& operator<<(ostream&o, const mint&x) {
  int a = x.x, b = 1;
  rep(s,2)rep1(i,1000) {
    int y = ((s?-x:x)*i).x; if (abs(a)+b > y+i) a = s?-y:y, b = i;
  }
  o<<a; if (b != 1) o<<'/'<<b; return o;
}//*/
using vm = vector<mint>;
using vvm = vector<vm>;
struct modinv {
  int n; vm d;
  modinv(): n(2), d({0,1}) {}
  mint operator()(int i) { while (n <= i) d.pb(-d[mod%n]*(mod/n)), ++n; return d[i];}
  mint operator[](int i) const { return d[i];}
} invs;
struct modfact {
  int n; vm d;
  modfact(): n(2), d({1,1}) {}
  mint operator()(int i) { while (n <= i) d.pb(d.back()*n), ++n; return d[i];}
  mint operator[](int i) const { return d[i];}
} facs;
struct modfactinv {
  int n; vm d;
  modfactinv(): n(2), d({1,1}) {}
  mint operator()(int i) { while (n <= i) d.pb(d.back()*invs(n)), ++n; return d[i];}
  mint operator[](int i) const { return d[i];}
} ifacs;
mint comb(int a, int b) {
  if (a < b || b < 0) return 0;
  return facs(a)*ifacs(b)*ifacs(a-b);
}
//

struct Solver {
  void solve() {
    int n,k;
    scanf("%d%d",&n,&k);
    mint ans;
    if (k == 1) {
      if (n&1) ans += n-2;
      else ans += 2;
    }

    if (k >= 2) {
      rep(i,n-1) {
        int r = n-i-2;
        if (r%2 == 0) {
          ans += comb(i,k-2)*(r+1);
        }
      }
    }
    if (k == 0) ans += 1;
    cout<<ans<<endl;
  }
};

int main() {
  int ts = 1;
  // scanf("%d",&ts);
  rep1(ti,ts) {
    Solver solver;
    solver.solve();
  }
  return 0;
}
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