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2021-07-07 集训题解

自动机

题目传送门

Description

Solution

可以想到一个 dp，设 \(f_{u,s,i}\) 表示起点在 \(u\)，现在在 \(s\) ，考虑了前面 \(i\) 个字符时合法的方案数。可以列出 dp 转移式：

\[f_{u,d_{r,1},i}\to \sum_{j=1}^{i-1} f_{u,l,j}\times e_{l,0} \times f_{d_{l,0},r,i-j-1}\times e[r][1] \]

\[f_{u,d_{r,2},i}\to f_{u,r,i-1} \]

然后这个用分治 fft 优化即可。复杂度 \(\Theta(n\log^2n)\)。

Code

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

#define SZ(A) ((int)A.size())
#define Int register int
#define mod 998244353
#define MAXN 400005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> void chkmax (T &a,T b){a = max (a,b);}
template <typename T> void chkmin (T &a,T b){a = min (a,b);}

int n,q,V,S[MAXN],T[MAXN],N[MAXN],d[2][3],e[2][3];
int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int k){
	int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod;
	return res;
}
int inv (int x){return qkpow (x,mod - 2);}
void Add (int &a,int b){a = add (a,b);}

typedef vector <int> poly;

int w[MAXN],rev[MAXN];

void init_ntt (){
	int lim = 1 << 18;
	for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 17);
	int Wn = qkpow (3,(mod - 1) / lim);w[lim >> 1] = 1;
	for (Int i = lim / 2 + 1;i < lim;++ i) w[i] = mul (w[i - 1],Wn);
	for (Int i = lim / 2 - 1;i;-- i) w[i] = w[i << 1];
}
void ntt (poly &a,int lim,int type){
#define G 3
#define Gi 332748118
	static int d[MAXN];
	for (Int i = 0,z = 18 - __builtin_ctz(lim);i < lim;++ i) d[rev[i] >> z] = a[i];
	for (Int i = 1;i < lim;i <<= 1)
		for (Int j = 0;j < lim;j += i << 1)
			for (Int k = 0;k < i;++ k){
				int x = mul (w[i + k],d[i + j + k]);
				d[i + j + k] = dec (d[j + k],x),d[j + k] = add (d[j + k],x);
			}
	for (Int i = 0;i < lim;++ i) a[i] = d[i] % mod;
	if (type == -1){
		reverse (a.begin() + 1,a.begin() + lim);
		for (Int i = 0,Inv = inv (lim);i < lim;++ i) a[i] = mul (a[i],Inv);
	}
#undef G
#undef Gi 
}
poly operator + (poly a,poly b){
	a.resize (max (SZ (a),SZ (b)));
	for (Int i = 0;i < SZ (b);++ i) a[i] = add (a[i],b[i]);
	return a;
}
poly operator - (poly a,poly b){
	a.resize (max (SZ (a),SZ (b)));
	for (Int i = 0;i < SZ (b);++ i) a[i] = dec (a[i],b[i]);
	return a;
}
poly operator * (poly a,int b){
	for (Int i = 0;i < SZ (a);++ i) a[i] = mul (a[i],b);
	return a;
}
poly operator + (poly a,int b){
	Add (a[0],b);
	return a;
}
poly operator * (poly a,poly b){
	int d = SZ (a) + SZ (b) - 1,lim = 1;while (lim < d) lim <<= 1;
	a.resize (lim),b.resize (lim);
	ntt (a,lim,1),ntt (b,lim,1);
	for (Int i = 0;i < lim;++ i) a[i] = mul (a[i],b[i]);
	ntt (a,lim,-1),a.resize (d);
	return a;
}
poly inv (poly a,int n){
	poly b(1,inv (a[0])),c;
	for (Int l = 4;(l >> 2) < n;l <<= 1){
		c.resize (l >> 1);
		for (Int i = 0;i < (l >> 1);++ i) c[i] = i < n ? a[i] : 0;
		c.resize (l),b.resize (l);
		ntt (c,l,1),ntt (b,l,1);
		for (Int i = 0;i < l;++ i) b[i] = mul (b[i],dec (2,mul (b[i],c[i])));
		ntt (b,l,-1),b.resize (l >> 1);
	}
	b.resize (n);
	return b;
}
poly inv (poly a){return inv (a,SZ (a));}

poly F[2][2];

void cdq (int l,int r){
	if (l > r) return ;
	if (l == r){
		if (l == 0){
			for (Int u = 0;u < V;++ u) F[u][u][l] = 1;
		}
		else{
			for (Int u = 0;u < V;++ u)
				for (Int now = 0;now < V;++ now)
					Add (F[u][d[now][2]][l],mul (e[now][2],F[u][now][l - 1]));
		}
		return ;
	}
	if (l + 1 == r){
		int mid = l + r >> 1;
		cdq (l,mid),cdq (mid + 1,r);
		return ;
	}
	int mid = l + r >> 1,len = r - l + 1,len1 = mid - l + 1,len2 = r - l;
	cdq (l,mid);
	if (l){
		for (Int u = 0;u < V;++ u){
			for (Int L = 0;L < V;++ L)
				for (Int R = 0;R < V;++ R){
					poly F1,G1;
					F1.resize (len1),G1.resize (len);
					for (Int i = 0;i < len1;++ i) F1[i] = F[u][L][l + i];
					for (Int i = 0;i < len;++ i) G1[i] = F[d[L][0]][R][i];
					F1 = F1 * G1;
					for (Int i = mid + 1;i <= r;++ i) if (i - l - 2 >= 0) Add (F[u][d[R][1]][i],mul (mul (e[L][0],e[R][1]),F1[i - l - 2]));
					F1.resize (len1),G1.resize (len);
					for (Int i = 0;i < len1;++ i) F1[i] = F[d[L][0]][R][l + i];
					for (Int i = 0;i < len;++ i) G1[i] = F[u][L][i];
					F1 = F1 * G1;
					for (Int i = mid + 1;i <= r;++ i) if (i - l - 2 >= 0) Add (F[u][d[R][1]][i],mul (mul (e[L][0],e[R][1]),F1[i - l - 2]));
				}
		} 
	}
	else{
		for (Int u = 0;u < V;++ u){
			for (Int L = 0;L < V;++ L)
				for (Int R = 0;R < V;++ R){
					poly F1,G1;
					F1.resize (len1),G1.resize (len1);
					for (Int i = 0;i < len1;++ i) F1[i] = F[u][L][i],G1[i] = F[d[L][0]][R][i];
					F1 = F1 * G1;
					for (Int i = mid + 1;i <= r;++ i) if (i - l - 2 >= 0) Add (F[u][d[R][1]][i],mul (mul (e[L][0],e[R][1]),F1[i - l - 2]));
				}
		}
	}
	cdq (mid + 1,r);
}

signed main(){
	freopen( "dfa.in", "r", stdin );
	freopen( "dfa.out", "w", stdout );
	read (V);
	for (Int u = 0;u < V;++ u) for (Int i = 0;i < 3;++ i) read (d[u][i],e[u][i]);
	read (q);
	for (Int i = 1;i <= q;++ i) read (S[i],T[i],N[i]),chkmax (n,N[i]);
	for (Int u = 0;u < V;++ u) for (Int now = 0;now < V;++ now) F[u][now].resize (n + 1);
	init_ntt(),cdq (0,n);
	for (Int i = 1;i <= q;++ i) write (F[S[i]][T[i]][N[i]]),putchar ('\n');
 	return 0;
}

神

题目传送门

Description

Solution

考虑将两个序列放在一起排序，在原序列中属于第一个的染成白色，另一种染成黑色。可以发现，因为逆序对个数为 \(\Theta(n)\) 级别，那么段数就会是 \(\Theta(\sqrt n)\) 级别的。

那么就可以一段一段地暴力考虑了，复杂度 \(\Theta(n\sqrt n\log n)\)。

Code

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

#define Int register int
#define mod 1000000007
#define MAXN 100005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> void chkmax (T &a,T b){a = max (a,b);}
template <typename T> void chkmin (T &a,T b){a = min (a,b);}

int n,q,a[MAXN],rt[MAXN],fac[MAXN],ifac[MAXN];
int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int k){
	int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod;
	return res;
}
int inv (int x){return qkpow (x,mod - 2);}
void Add (int &a,int b){a = add (a,b);}

struct Segment{
#define LOGN 31
#define ls(x) son[x][0]
#define rs(x) son[x][1]
	int cnt,sum[MAXN * LOGN],son[MAXN * LOGN][2];
	void clear (){cnt = 0;}
	void pushup (int x){sum[x] = sum[ls(x)] + sum[rs(x)];}
	void modify (int &x,int y,int l,int r,int pos){
		x = ++ cnt,sum[x] = sum[y] + 1,ls(x) = ls(y),rs(x) = rs(y);
		//cout << l << " -> " << r << " " << x << ": " << sum[x] << endl;
		if (l == r) return ;
		int mid = l + r >> 1;
		if (pos <= mid) modify (ls(x),ls(y),l,mid,pos);
		else modify (rs(x),rs(y),mid + 1,r,pos);
	}
	int query (int x,int y,int l,int r,int v){
		//cout << l << " -> " << r << ": " << x << " " << y << " " << sum[x] - sum[y] << endl;
		if (l > v || sum[x] == sum[y]) return 0;
		if (l == r) return l;
		int mid = l + r >> 1;
		if (v <= mid) return query (ls(x),ls(y),l,mid,v);
		else{
			int tmp = query (rs(x),rs(y),mid + 1,r,v);
			if (tmp) return tmp;
			else return query (ls(x),ls(y),l,mid,v);	
		}
	}
	int getit (int x,int y,int l,int r,int ql,int qr){
		if (l >= ql && r <= qr) return sum[y] - sum[x];
		int mid = l + r >> 1,res = 0;
		if (ql <= mid) res += getit (ls(x),ls(y),l,mid,ql,qr);
		if (qr > mid) res += getit (rs(x),rs(y),mid + 1,r,ql,qr);
		return res;
	}
}Tree;

void Work (){
	read (n,q),Tree.clear();
	fac[0] = 1;for (Int i = 1;i <= n;++ i) fac[i] = mul (fac[i - 1],i);
	ifac[n] = qkpow (fac[n],mod - 2);for (Int i = n;i;-- i) ifac[i - 1] = mul (ifac[i],i);
	for (Int i = 1;i <= n;++ i) read (a[i]),rt[i] = 0,Tree.modify (rt[i],rt[i - 1],1,n,a[i]);
	//cout << Tree.query (rt[0],rt[1],1,n,n) << endl;
	//return ;
	while (q --> 0){
		int l1,r1,l2,r2;
		read (l1,r1,l2,r2);
		int x = Tree.query (rt[l1 - 1],rt[r1],1,n,n),y = Tree.query (rt[l2 - 1],rt[r2],1,n,n),ans = 1,tot1 = 0,tot2 = 0;
		while (x || y){
			//cout << x << " , " << y << endl;
			if (x < y){
				tot2 += Tree.getit (rt[l2 - 1],rt[r2],1,n,x + 1,y),
				y = Tree.query (rt[l2 - 1],rt[r2],1,n,x - 1);
			}
			else{
				int sum = Tree.getit (rt[l1 - 1],rt[r1],1,n,y + 1,x);
				if (tot2 >= tot1 + sum - 1) ans = mul (ans,mul (fac[tot2 - tot1],ifac[tot2 - tot1 - sum]));
				else ans = 0;
				tot1 += sum,x = Tree.query (rt[l1 - 1],rt[r1],1,n,y - 1);
			}
		}
		write (ans),putchar ('\n');
	}
}

signed main(){
	freopen( "god.in", "r", stdin );
	freopen( "god.out", "w", stdout );
	int T;read (T);
	while (T --> 0) Work ();
 	return 0;
}

我会彻查

题目传送门

Description

Solution

可以想到，假设设 \(f(k)\) 表示每个人选 \(k\) 个的时候的最大贡献和，那么这个东西是一个没有台阶的单峰函数，也就是说可以三分。

考虑如何计算 \(f(k)\)，可以想到一定是左边一段，右边一段，中间一部分，所以直接二分在哪里分界即可。

Code

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

#define Int register int
#define int long long
#define MAXN 200005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> void chkmax (T &a,T b){a = max (a,b);}
template <typename T> void chkmin (T &a,T b){a = min (a,b);}

int n,m,a[MAXN];

int getsum (int i,int a,int b){//计算对于第i个人，前面a个，后面b个时候的贡献
	int A = a * i + n * (a - 1) * a / 2,B = b * (m - n + i) - n * (b - 1) * b / 2;
	return A + B;
}

map <int,int> mp;
int f (int k){
	if (mp.find (k) != mp.end()) return mp[k];
	int &res = mp[k] = 0;int flg = 1;
	for (Int i = 1;i <= n;++ i)
		if (getsum(i,k,0) > a[i]){flg = -1;break;}
		else if (getsum (i,0,k) > a[i]) flg = 0;
	if (flg == 1) return res = n * m * k - (k * n) * (k * n - 1) / 2;
	else if (flg == -1) return res = -k;
	int l = 1,r = k;
	while (l < r){
		int mid = l + r + 1 >> 1;flg = 1;
		for (Int i = n;i >= 1;-- i) if (getsum (i,mid - 1,k - mid + 1) > a[i]){flg = 0;break;}
		if (flg) r = mid - 1;
		else l = mid;
	}
	int j = l;
	for (Int i = n;i >= 1;-- i) if (getsum (i,j - 1,k - j + 1) > a[i]){
		for (Int t = 1;t <= i;++ t) res += getsum (t,j,k - j);
		for (Int t = i + 1;t <= n;++ t) res += getsum (t,j - 1,k - j + 1);
		int p = n * (j - 1) + i,q = m - n * (k - j) - (n - i);
		for (Int i1 = i;i1 >= 1;-- i1){
			chkmin (q,a[i1] - getsum (i1,j - 1,k - j));
			res += q - p,-- q,-- p;
		}
		-- j;
		if (j){
			for (Int i1 = n;i1 > i;-- i1){
				chkmin (q,a[i1] - getsum (i1,j - 1,k - j));
				res += q - p,-- q,-- p;
			}
		}
		return res;
	}
	return -1;
}

signed main(){
	freopen("investigate.in","r",stdin);
	freopen("investigate.out","w",stdout);
	read (n,m);
	for (Int i = 1;i <= n;++ i) read (a[i]);
	int l = 0,r = m / n;
	while (l < r){
		int mid = l + r >> 1,det = f (mid + 1) - f (mid);
		if (det == 0) l = r = mid;
		else if (det < 0) r = mid;
		else l = mid + 1;
	}
	write (f(l)),putchar ('\n');
 	return 0;
}
posted @ 2021-07-07 21:55  Dark_Romance  阅读(54)  评论(0)  编辑  收藏  举报