結果

問題 No.1785 Inequality Signs
コンテスト
ユーザー eQe
提出日時 2021-12-18 13:59:19
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 296 ms / 2,000 ms
コード長 4,738 bytes
コンパイル時間 4,803 ms
コンパイル使用メモリ 276,272 KB
最終ジャッジ日時 2025-01-27 03:17:59
ジャッジサーバーID
（参考情報） 		judge1 / 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 * 2
other AC * 52
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 <atcoder/all>
using namespace atcoder;
#include <bits/stdc++.h>
#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
using namespace std;
using ll=long long;
using ld=long double;
using pll=pair<ll,ll>;
template<class T>bool chmax(T &a, const T &b) {if(a<b) {a=b; return 1;} return 0;}
template<class T>bool chmin(T &a, const T &b) {if(b<a) {a=b; return 1;} return 0;}
istream& operator >> (istream& is, modint1000000007& x) { unsigned int t; is >> t; x=t; return is; }
istream& operator >> (istream& is, modint998244353& x) { unsigned int t; is >> t; x=t; return is; }
istream& operator >> (istream& is, modint& x) { unsigned int t; is >> t; x=t; return is; }
ostream& operator << (ostream& os, const modint1000000007& x) { os << x.val(); return os; }
ostream& operator << (ostream& os, const modint998244353& x) { os << x.val(); return os; }
ostream& operator << (ostream& os, const modint& x) { os << x.val(); return os; }
template<class T1, class T2>istream& operator >> (istream& is, pair<T1,T2>& p) { is >> p.first >> p.second; return is; }
template<class T1, class T2>ostream& operator << (ostream& os, const pair<T1,T2>& p) { os << p.first << " " << p.second; return os;}
template<class T>istream& operator >> (istream& is, vector<T>& v) { for(T& x:v) is >> x; return is; }
template<class T>ostream& operator << (ostream& os, const vector<T>& v) {for(int i=0;i<(int)v.size();i++) {os << v[i] << (i+1 == v.size() ? "":" ");} return os;}
template<class... A> void pt()      { std::cout << "\n"; }
template<class... A> void pt_rest() { std::cout << "\n"; }
template<class T, class... A> void pt_rest(const T& first, const A&... rest) { std::cout << " " << first; pt_rest(rest...); }
template<class T, class... A> void pt(const T& first, const A&... rest)      { std::cout << first; pt_rest(rest...); }
template<typename V, typename H> void resize(vector<V>& vec, const H head){ vec.resize(head); }
template<typename V, typename H, typename ... T> void resize(vector<V>& vec, const H& head, const T ... tail){ vec.resize(head); for(auto& v: vec) resize(v, tail...); }
template<typename V, typename T> void fill(V& x, const T& val){ x = val; }
template<typename V, typename T> void fill(vector<V>& vec, const T& val){ for(auto& v: vec) fill(v, val); }
template<typename H> void vin(istream& is, const int idx, vector<H>& head){ is >> head[idx]; }
template<typename H, typename ... T> void vin(istream& is, const int idx, vector<H>& head, T& ... tail){ vin(is >> head[idx], idx, tail...); }
template<typename H, typename ... T>
void vin(istream& is, vector<H>& head, T& ... tail){ for(int i=0; i<(int)head.size(); i++) vin(is, i, head, tail...); }
template<typename H, typename ... T> void vin(vector<H>& head, T& ... tail){ vin(cin, head, tail...); }
map<ll,ll>divisor(ll n){map<ll,ll>res;for(ll i=1;i*i<=n;i++)if(n%i==0){res[i]++;res[n/i]++;}return res;}
map<ll,ll>factrization(ll n){map<ll,ll>res;for(ll i=2;i*i<=n;i++)while(n%i==0){res[i]++;n/=i;}if(n>1)res[n]++;return res;}
static const ll inf=1e18+7;
static const ld pi=acos(-1);
static const ld eps=1e-10;
using v1=vector<ll>;
using v2=vector<v1>;
using v3=vector<v2>;
using v4=vector<v3>;
using S=ll;
S op(S a,S b){return a+b;}
S e(){return 0;}

template<class T>struct Combination{
  vector<T>fa,rfa,bs;
  vector<vector<pair<bool,T>>>mp;
  ll LIM=5000;
  Combination(ll n){
    fa.resize(n+1);
    rfa.resize(n+1);
    fa[0]=rfa[0]=1;
    for(ll i=1;i<=n;i++)fa[i]=fa[i-1]*i;
    for(ll i=1;i<=n;i++)rfa[i]=rfa[i-1]/i;
    
    bs.resize(n+1);
    bs[0]=1;
    for(ll i=1;i<=n;i++)bs[i]=bs[i-1]+T(-1).pow(i&1)*rfa[i];
    
    if(n<=LIM)mp.resize(n+1,vector<pair<bool,T>>(n+1));
  }
  T C(ll n,ll k){return n<k||n<0||k<0?0:fa[n]*rfa[k]*rfa[n-k];}
  T P(ll n,ll k){return C(n,k)*fa[k];}
  T H(ll n,ll k){return C(n-1+k,k);}
  T S(ll n,ll k){
    T res=0;
    for(ll i=0;i<=k;i++)res+=T(-1).pow((k-i)&1)*C(k,i)*T(i).pow(n);
    res*=rfa[k];
    return res;
  }
  T B(ll n,ll k){
    T res=0;
    for(ll i=0;i<=k;i++)res+=T(i).pow(n)*rfa[i]*bs[k-i];
    return res;
  }
  T p(ll n,ll k){
    if(n==0 and k==0)return 1;
    if(k==0)return 0;
    if(mp[n][k].first)return mp[n][k].second;
    if(n-k>=0)return (mp[n][k]={true,p(n,k-1)+p(n-k,k)}).second;
    return (mp[n][k]={true,p(n,k-1)}).second;
  }
};


int main(void) {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  
  ll N,K;cin>>N>>K;
  using mint=modint1000000007;
  Combination<mint>com(N);
  
  map<ll,mint>mem;
  function<mint(ll)>f=[&](ll x)->mint{
    if(x==0 or x==K-N)return mem[x]=1;
    return mem[x]=x*f(x-1);
  };
  f(K+N-1);
  
  mint ans=0;
  for(ll i=0;i<N;i++){
    if(K+i-N<0)continue;
    ans+=com.C(N-1,i)*mem[K+i]/mem[K+i-N]*com.rfa[N];
  }
  pt(ans);
  
  

}
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