結果

問題 No.2528 pop_(backfront or not)
ユーザー woodywoodywoodywoody
提出日時 2023-11-03 21:47:55
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 419 ms / 2,000 ms
コード長 5,474 bytes
コンパイル時間 3,689 ms
コンパイル使用メモリ 235,708 KB
実行使用メモリ 144,044 KB
最終ジャッジ日時 2024-09-25 19:56:53
合計ジャッジ時間 8,817 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge3
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 115 ms
81,536 KB
testcase_01 AC 114 ms
81,464 KB
testcase_02 AC 114 ms
81,300 KB
testcase_03 AC 115 ms
81,384 KB
testcase_04 AC 114 ms
81,464 KB
testcase_05 AC 115 ms
81,536 KB
testcase_06 AC 115 ms
81,712 KB
testcase_07 AC 120 ms
82,432 KB
testcase_08 AC 122 ms
82,816 KB
testcase_09 AC 138 ms
86,124 KB
testcase_10 AC 190 ms
96,896 KB
testcase_11 AC 196 ms
98,048 KB
testcase_12 AC 230 ms
105,504 KB
testcase_13 AC 251 ms
107,864 KB
testcase_14 AC 273 ms
114,048 KB
testcase_15 AC 362 ms
132,352 KB
testcase_16 AC 415 ms
143,960 KB
testcase_17 AC 419 ms
144,044 KB
testcase_18 AC 414 ms
144,016 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>
#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 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 true;} else {cout << "No" << endl; return false;}}
#define retval(x) {cout << #x << endl; return;}
#define cout2(x,y) cout << x << " " << y << endl;
#define coutp(p) cout << p.fi << " " << p.se << 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 assertmle(x) if (!(x)) {vi v(3e8);}
#define asserttle(x) if (!(x)) {while(1){}}
#define coutv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define coutv2(v) fore(vy , v) cout << vy << "\n";
#define coutvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define coutvm2(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 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 mint = modint1000000007;
//using mint = dynamic_modint<0>;
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 no {cout <<"No"<<endl;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#ifdef MY_LOCAL_DEBUG
#include "./debug/localDebug.cpp"
#define showp(p) cerr<<#p<<" = "<<p.fi<<" : "<<p.se<<endl
#define show1(a) cerr<<#a<<" = "<<a<<endl
#define show2(a,b) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<endl
#define show3(a,b,c) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<endl
#define show4(a,b,c,d) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<" : "<<#d<<" = "<<d<<endl
#define show5(a,b,c,d,e) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<" : "<<#d<<" = "<<d<<" : "<<#e<<" = "<<e<<endl
#define DEBUG_LINE cout << "DEBUG_LINE : " << __LINE__ << endl
#define showv(v) {cout<<#v<<" : "; fore(vy , v) {cout << vy << " ";} cout << endl;}
#define showv2(v) {cout<<#v<<endl; fore(vy , v) cout << vy << "\n";}
#define showvm(v) {cout<<#v<<" : "; fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define showvm2(v) {cout<<#v<<endl; fore(vy , v) cout << vy.val() << "\n";}
#define showmat(v) {cout<<#v<<endl; fore(row , v) { fore(seg , row) cout << seg << " "; cout << endl;}}
#define showmatm(v) {cout<<#v<<endl; fore(row , v) { fore(seg , row) cout << seg.val() << " "; cout << endl;}}
#else
#define showp(p)
#define show1(a)
#define show2(a,b)
#define show3(a,b,c)
#define show4(a,b,c,d)
#define show5(a,b,c,d,e)
#define DEBUG_LINE
#define showv(v)
#define showv2(v)
#define showvm(v)
#define showvm2(v)
#define showmat(v)
#define showmatm(v)
#endif
#define overload5(a,b,c,d,e,f, ...) f
#define show(...) overload5(__VA_ARGS__, show5, show4, show3, show2, show1)(__VA_ARGS__)

struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    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);

void solve(){
    int n; cin>>n;
    vvm dp(2*n+2,vm(2*n+2));
    dp[0][0] = 1;
    rep(i,n*2+2) rep(j,n*2+2){
        if ((i+j)%2) continue;
        if (i&&j){
            dp[i][j] += dp[i-1][j-1];
            dp[i][j] += com(i-1,1) * com(j-1,1) * dp[i-1][j-1];
        }
        if (i>=3) dp[i][j] += com(i-1,2)*dp[i-2][j];
        if (j>=3) dp[i][j] += com(j-1,2)*dp[i][j-2];
    }
    vm ans(n*2+1);
    rep(i,n*2+1){
        int a = i;
        int b = n*2+1-a-1;
        ans[i] = dp[a][b];
    }
    fore(y , ans) cout << y.val() << endl;

    return;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    int t = 1;
    //cin>>t;

    rep(i,t){
        solve();
    }

    return 0;
}
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