AtCoder Beginner Contest 430 题解

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F - Almost Sorted 2

问题陈述

给你一个长度为 $N$ 的整数序列 $A=(A1,A2,…,AN)A=(A_1,A_2,\ldots,A_N)$ 和一个正整数 $D$ 。

求通过重新排列 $A$ 得到的满足以下条件的整数序列 $B=(B1,B2,…,BN)B=(B_1, B_2, \ldots, B_N)$ 的个数（模为 $998244353$ ）：

$Bi+1≥Bi−DB_{i+1}\geq B_i-D$ 对所有 $(1≤i≤N−1)i\ (1\leq i\leq N-1)$ 都成立。

考虑插入法，每次按 $B_i$ 从小到大插入下一个数，一个数可以被插在某个位置等价于后面那个数 $≥Bi−D\geq B_i-D$

#include<bits/stdc++.h> 
using namespace std;

#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define vii(f,n,m) vector<int> f(n,vi(m)) 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define is_ok(a) puts(a?"Yes":"No")
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#endif

#define DEBUG
int recur_depth = 0;
#ifdef DEBUG
#define dbg(x) {++recur_depth; auto x_=x; --recur_depth; cerr<<string(recur_depth, '\t')<<"\e[91m"<<__func__<<":"<<__LINE__<<"\t"<<#x<<" = "<<x_<<"\e[39m"<<endl;}
#else
#define dbg(x)
#endif
template<typename Ostream, typename Cont>
typename enable_if<is_same<Ostream,ostream>::value, Ostream&>::type operator<<(Ostream& os,  const Cont& v){
	os<<"[";
	for(auto& x:v){os<<x<<", ";}
	return os<<"]";
}
template<typename Ostream, typename ...Ts>
Ostream& operator<<(Ostream& os,  const pair<Ts...>& p){
	return os<<"{"<<p.first<<", "<<p.second<<"}";
}

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 

template<int MOD, int RT> struct mint {
	static const int mod = MOD;
	static constexpr mint rt() { return RT; } // primitive root for FFT
	int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
	mint():v(0) {}
	mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
		if (v < 0) v += MOD; }
	bool operator==(const mint& o) const {
		return v == o.v; }
	friend bool operator!=(const mint& a, const mint& b) { 
		return !(a == b); }
	friend bool operator<(const mint& a, const mint& b) { 
		return a.v < b.v; }
   
	mint& operator+=(const mint& o) { 
		if ((v += o.v) >= MOD) v -= MOD; 
		return *this; }
	mint& operator-=(const mint& o) { 
		if ((v -= o.v) < 0) v += MOD; 
		return *this; }
	mint& operator*=(const mint& o) { 
		v = int((ll)v*o.v%MOD); return *this; }
	mint& operator/=(const mint& o) { return (*this) *= inv(o); }
	friend mint pow(mint a, ll p) {
		mint ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans; }
	friend mint inv(const mint& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mint operator-() const { return mint(-v); }
	mint& operator++() { return *this += 1; }
	mint& operator--() { return *this -= 1; }
	friend mint operator+(mint a, const mint& b) { return a += b; }
	friend mint operator-(mint a, const mint& b) { return a -= b; }
	friend mint operator*(mint a, const mint& b) { return a *= b; }
	friend mint operator/(mint a, const mint& b) { return a /= b; }
};
const int MOD=998244353; 
using mi = mint<MOD,5>; // 5 is primitive root for both common mods

namespace simp {
	vector<mi> fac,ifac,invn;
	void check(int x) {
		if (fac.empty()) {
			fac={mi(1),mi(1)};
			ifac={mi(1),mi(1)};
			invn={mi(0),mi(1)};
		}
		while (SI(fac)<=x) {
			int n=SI(fac),m=SI(fac)*2;
			fac.resize(m);
			ifac.resize(m);
			invn.resize(m);
			for (int i=n;i<m;i++) {
				fac[i]=fac[i-1]*mi(i);
				invn[i]=mi(MOD-MOD/i)*invn[MOD%i];
				ifac[i]=ifac[i-1]*invn[i];
			}
		}
	}
	mi gfac(int x) {
		assert(x>=0);
		check(x); return fac[x];
	}
	mi ginv(int x) {
		assert(x>0);
		check(x); return invn[x];
	}
	mi gifac(int x) {
		assert(x>=0);
		check(x); return ifac[x];
	}
	mi binom(int n,int m) {
		if (m < 0 || m > n) return mi(0);
		return gfac(n)*gifac(m)*gifac(n - m);
	}
}
int main()
{
	int n=read(),d=read();
	vi a(1e6+1,0),s(1e6+1,0);
	For(i,n) {
		a[read()]++;
	}
	For(i,1e6) s[i]+=a[i]+s[i-1];
	mi ans=1;
	For(i,1e6) if(a[i]){
		ll k=s[i-1]-((i-d-1>=0)?s[i-d-1]:0)+1;
		ll s=a[i];
		ans*=simp::binom(k-1+s,s);
	}
	cout<<ans.v<<endl;
	return 0;
}

G - Range Set Modifying Query

问题陈述

有 $N$ 个集合 $S1,…,SNS_1, \ldots,S_N$ 。最初，所有集合都是空的。

你会得到以下格式的 $Q$ 个查询。按顺序处理它们。

类型 $1$ ：给出的格式为 1 L R x对于满足 $\leq i \leq R$ 的每个 $S_i$ ，添加 $x$ 。
类型 $2$ ：指定为 2 L R x 为满足 $\leq i \leq R$ 的每个 $S_i$ ，删除 $x$ 。
类型 $3$ ：给定为 3 L R 求满足 $\leq i \leq R$ 的 $S_i$ 中元素的最大个数，以及达到这个最大个数的集合的个数。

主席树板子题

#include<bits/stdc++.h> 
 using namespace std;

#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define vii(f,n,m) vector<int> f(n,vi(m)) 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define is_ok(a) puts(a?"Yes":"No")
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#endif

#define DEBUG
int recur_depth = 0;
#ifdef DEBUG
#define dbg(x) {++recur_depth; auto x_=x; --recur_depth; cerr<<string(recur_depth, '\t')<<"\e[91m"<<__func__<<":"<<__LINE__<<"\t"<<#x<<" = "<<x_<<"\e[39m"<<endl;}
#else
#define dbg(x)
#endif
template<typename Ostream, typename Cont>
typename enable_if<is_same<Ostream,ostream>::value, Ostream&>::type operator<<(Ostream& os,  const Cont& v){
	os<<"[";
	for(auto& x:v){os<<x<<", ";}
	return os<<"]";
}
template<typename Ostream, typename ...Ts>
Ostream& operator<<(Ostream& os,  const pair<Ts...>& p){
	return os<<"{"<<p.first<<", "<<p.second<<"}";
}

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 

template<int MOD, int RT> struct mint {
	static const int mod = MOD;
	static constexpr mint rt() { return RT; } // primitive root for FFT
	int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
	mint():v(0) {}
	mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
		if (v < 0) v += MOD; }
	bool operator==(const mint& o) const {
		return v == o.v; }
	friend bool operator!=(const mint& a, const mint& b) { 
		return !(a == b); }
	friend bool operator<(const mint& a, const mint& b) { 
		return a.v < b.v; }
   
	mint& operator+=(const mint& o) { 
		if ((v += o.v) >= MOD) v -= MOD; 
		return *this; }
	mint& operator-=(const mint& o) { 
		if ((v -= o.v) < 0) v += MOD; 
		return *this; }
	mint& operator*=(const mint& o) { 
		v = int((ll)v*o.v%MOD); return *this; }
	mint& operator/=(const mint& o) { return (*this) *= inv(o); }
	friend mint pow(mint a, ll p) {
		mint ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans; }
	friend mint inv(const mint& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mint operator-() const { return mint(-v); }
	mint& operator++() { return *this += 1; }
	mint& operator--() { return *this -= 1; }
	friend mint operator+(mint a, const mint& b) { return a += b; }
	friend mint operator-(mint a, const mint& b) { return a -= b; }
	friend mint operator*(mint a, const mint& b) { return a *= b; }
	friend mint operator/(mint a, const mint& b) { return a /= b; }
};
const int MOD=F; 
using mi = mint<MOD,5>; // 5 is primitive root for both common mods

namespace simp {
	vector<mi> fac,ifac,invn;
	void check(int x) {
		if (fac.empty()) {
			fac={mi(1),mi(1)};
			ifac={mi(1),mi(1)};
			invn={mi(0),mi(1)};
		}
		while (SI(fac)<=x) {
			int n=SI(fac),m=SI(fac)*2;
			fac.resize(m);
			ifac.resize(m);
			invn.resize(m);
			for (int i=n;i<m;i++) {
				fac[i]=fac[i-1]*mi(i);
				invn[i]=mi(MOD-MOD/i)*invn[MOD%i];
				ifac[i]=ifac[i-1]*invn[i];
			}
		}
	}
	mi gfac(int x) {
		assert(x>=0);
		check(x); return fac[x];
	}
	mi ginv(int x) {
		assert(x>0);
		check(x); return invn[x];
	}
	mi gifac(int x) {
		assert(x>=0);
		check(x); return ifac[x];
	}
	mi binom(int n,int m) {
		if (m < 0 || m > n) return mi(0);
		return gfac(n)*gifac(m)*gifac(n - m);
	}
}
int n,q;
struct BIT{
	#define MAXN (212345+10)
	ll f[MAXN]={}; 
	int n;
	void add(int x,ll v) {
		for(int i=x;i<=n;i+=i&(-i)) 
		 		f[i]+=v;
	}
	ll qur(int x) {
		ll v=0;
		for(int i=x;i;i-=i&(-i)) 
		 		v+=f[i];
		return v;
	}
}T;

namespace Seg{
const int N = MAXN, M = MAXN, LOG2N = 32-__builtin_clz(N-1);
int roots[N+1], allo;
struct Segment { int ch[2], cnt; } seg[N * 4 + M * LOG2N];

void build(int &t, int l, int r) {
  t = ++allo;
  if (l < r-1) {
    int m = l+r >> 1;
    build(seg[t].ch[0], l, m);
    build(seg[t].ch[1], m, r);
  }
}

void add(int *t, int u, int l, int r, int v) {
  while (l < r-1) {
    *t = ++allo;
    seg[*t].cnt = seg[u].cnt+1;
    int m = l+r >> 1, d = v >= m;
    if (d) l = m;
    else r = m;
    seg[*t].ch[d^1] = seg[u].ch[d^1];
    t = &seg[*t].ch[d];
    u = seg[u].ch[d];
  }
  *t = ++allo;
  seg[*t].cnt = seg[u].cnt+1;
}

int kth(int t, int u, int l, int r, int k) {
  while (l < r-1) {
    int m = l+r >> 1, lcnt = seg[seg[t].ch[0]].cnt-seg[seg[u].ch[0]].cnt, d = k >= lcnt;
    if (d) l = m, k -= lcnt;
    else r = m;
    t = seg[t].ch[d];
    u = seg[u].ch[d];
  }
  return l;
}

int count_less_then_k(int t, int u, int l, int r, int v) {
  int ret=0;
  while (l < r-1) {
    int m = l+r >> 1, lcnt = seg[seg[t].ch[0]].cnt-seg[seg[u].ch[0]].cnt, d = v >= m;
    if (d) l = m, ret += lcnt;
    else r = m;
    t = seg[t].ch[d];
    u = seg[u].ch[d];
  }
  return ret;
}
};


ll a[MAXN];
void work() {
	n=read(),q=read();
	vector<ll> a(n),u(n),d(n);
	Rep(i,n) cin>>a[i];
	T.n=n;
	Seg::build(Seg::roots[n], 0, n+2);
	RepD(i,n-1) {
		u[i]=T.qur(n)-T.qur(a[i]);
		d[i]=T.qur(a[i]-1);
		T.add(a[i],1);
		Seg::add(&Seg::roots[i], Seg::roots[i+1], 0, n+2, a[i]);
	}
	map<int,vi> h;
	int id1=-1,id2=-1;
	Rep(i,n) {
		if(id1==-1 && h.count(a[i])) {
			id2=i,id1=h[a[i]][0];
		}
		if(!h.count(a[i])) {
			h[a[i]]={i};
		}
		else h[a[i]].pb(i);
	}
	
	For(i,n-1) {
		d[i]+=d[i-1];
		u[i]+=u[i-1];
	}
	while(q--) {
		ll k;cin>>k;
		if(k<=d[n-1]){
			int i;
			auto it=lower_bound(ALL(d),k);
			if((*it)==k) {
				i=lower_bound(ALL(d),k)-d.begin();
			}else i=upper_bound(ALL(d),k)-d.begin();
			ll newk=k-(i-1>=0?d[i-1]:0);
			int j=Seg::kth(Seg::roots[i+1], Seg::roots[n], 0, n+2,newk-1 );
			int cnt=Seg::count_less_then_k(Seg::roots[i+1], Seg::roots[n], 0, n+2,j );
			auto iter=h[j].end()-(newk-cnt);
			cout<<i+1<<' '<<(*iter)+1<<endl;
			continue;
		}
		else if((ll)n*(n-1)/2-k+1<=u[n-1]){
			k=(ll)n*(n-1)/2-k+1;
			int i;
			auto it=lower_bound(ALL(u),k);
			if((*it)==k) {
				i=lower_bound(ALL(u),k)-u.begin();
			}else i=upper_bound(ALL(u),k)-u.begin();
			ll newk=k-(i-1>=0?u[i-1]:0);
			int j=Seg::kth(Seg::roots[i+1], Seg::roots[n], 0, n+2,(n-i)-newk-1 );
			int cnt=n-i-1-Seg::count_less_then_k(Seg::roots[i+1], Seg::roots[n], 0, n+2,j+1 );
			auto iter=h[j].end()-(newk-cnt);
			cout<<i+1<<' '<<(*iter)+1<<endl;
			continue;
		}
		else cout<<id1+1<<' '<<id2+1<<endl;
	}
}
int main()
{
//	freopen("g.in","r",stdin);
//	freopen(".out","w",stdout);
	work();
	
	return 0;
}