Codeforces Round 1101 (Div. 2) 题解

nike0good2026-08-08 9:39

A. Convergence

Alice is inviting her friends to a party to eat cakes. However, each friend may not be at the same place, so everyone has to meet up at the same location first.

Alice has nnn friends, where the iii-th friend is at position aia_iai. To make everyone be at the same place, Alice has to make multiple group calls. Unfortunately, the signal is weak, and Alice can only call 222 other people at a time.

Being a good person, Alice doesn't want her friends to walk too far. So, for each group call containing the iii-th friend and the jjj-th friend, Alice will tell both of them to meet at some integer location between min⁡(ai,aj)\min(a_i, a_j)min(ai,aj) and max⁡(ai,aj)\max(a_i, a_j)max(ai,aj) inclusive. After that, both of them will move to that location so quickly that Alice cannot make any group call during their movement. Please note that Alice can call these friends again once they reach that location.

The party is starting soon, so Alice needs to make group calls fast. Help her find the minimum number of group calls she needs to make.

cpp 复制代码 
#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 (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#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);
#define sqr(x) (x*x)
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;}
ll gcd(ll a,ll b){return (!a)?b:a%b;}
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);
	}
}
mi f[412345]={};
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T;
	cin>>T;
	while(T--) {
		int n;
		cin>>n;
		vector<ll> a(n);
		Rep(i,n) cin>>a[i];
		sort(ALL(a));
		int ans=1e9;
		Rep(i,n) {
			int p=0,q=0;
			Rep(j,n) if(a[j]<a[i])++p;
			else if(a[j]>a[i]) ++q;
			gmin(ans,max(p,q))
		}
		cout<<ans<<endl;
	}
	
	return 0;
}

B. Cake Leveling

Alice is preparing a cake for her party. However, she is in a rush, so the frosting on the cake is uneven. To quickly solve this issue, Alice will put her knife at some integer height and then sweep the frosting from left to right to make the frosting level.

Formally, let aia_iai be the height of the frosting at the iii-th position. Suppose that Alice puts her knife at some integer height hhh. If the height of frosting at position iii is greater than hhh, the excess frosting will be pushed to position i+1i + 1i+1. Excess frosting on position nnn will be pushed off the cake completely.

Alice and her friends really love cake frosting. Since Alice might decide to serve some prefix of the cake instead of the whole cake, help her find the maximum height the frosting can be while keeping the frosting level for the first iii positions for i=1,2,...,ni = 1, 2, \ldots, ni=1,2,...,n.

cpp 复制代码 
#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 (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#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);
#define sqr(x) (x*x)
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;}
ll gcd(ll a,ll b){return (!a)?b:a%b;}
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);
	}
}
mi f[412345]={};
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T;
	cin>>T;
	while(T--) {
		int n;
		cin>>n;
		vector<ll> a(n),s(n);
		Rep(i,n) cin>>a[i];
		For(i,n-1) a[i]+=a[i-1];
		ll ans=1e9;
		Rep(i,n) {
			ans=min(ans,a[i]/(i+1));
			cout<<ans<<' ';
		}
		cout<<endl;
	}
	
	return 0;
}

C1. Seating Arrangement (Easy Version)

派对上有 x 张桌子，每张桌子上有 s 个座位。每个座位只能坐一个人。

有一个长度为n的序列包含AIE，I必须在空桌，E 必须坐在非空桌，A可以坐在任何一张桌子。

最初，每个座位都是空的。但是，由于爱丽丝正在吃蛋糕，她的朋友们已经排成了一队，爱丽丝不能改变他们的顺序。对于队伍中的每个人，爱丽丝都必须给他们分配一张桌子，否则就把他们踢出聚会。每个人坐下后，下一个人才会被分配到一张桌子。最初，每个座位都是空的。对于序列中的每个人，爱丽丝都必须给他们分配一张桌子，否则就把他们踢出。每个人坐下后，下一个人才会被分配到一张桌子。爱丽丝希望在派对上玩得尽兴，她需要在派对上安排尽可能多的人就座，求聚会上最多可容纳的朋友人数。

cpp 复制代码 
#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 (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#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);
#define sqr(x) (x*x)
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;}
ll gcd(ll a,ll b){return (!a)?b:a%b;}
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);
	}
}
mi f[412345]={};
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T;
	cin>>T;
	while(T--) {
		int n,x,s,ans=0;
		cin>>n>>x>>s;
		string a;
		cin>>a;
		int left=x,p=0,r=0;
		Rep(i,n) {
			if(a[i]=='I') {
				if(left) left--,ans++,p+=s-1;
			}else if(a[i]=='E') {
				if(p) p--,ans++;
				else if(r>0 && left>0) {
					r--,left--,p+=s-1;
					++ans;
				}
			}else {
				if(p) --p,++ans,++r;
				else if(left) --left,++ans,p+=s-1;
			}
		}
		cout<<ans<<endl;
	}
	
	return 0;
}

D. Magical Tiered Cake

Alice has finished a magical tiered cake with nnn magical layers, where each layer is smaller than all layers below it (i.e. the first layer is the smallest, the nnn-th layer is the largest). Now, she needs your help to transport it from her kitchen to the party site. Since moving the whole cake at once is impossible, she also prepares a warehouse for you to ease the transportation process.

Since the cake is magical, the iii-th layer of cake is movable if and only if there are exactly aia_iai layers of cake above it.

In each move, you can choose exactly one movable layer of cake from any location and move it on top of the tiered cake in any other location. However, to preserve a structure of the cake, the moved layer has to be on the layer that is strictly larger than it if there is a tiered cake at the destination. For example, you cannot move a layer with size 444 on top of a location where there is already a layer with size 333.

The party is starting soon, so we need to move fast. Help Alice transport the cake to the party within 2n2^n2n moves or report that it is impossible.

考虑hanoi塔的答案，如何才能保证限制条件的前提下，用递归解决搬运问题。

cpp 复制代码 
#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 (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#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);
#define sqr(x) (x*x)
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;}
ll gcd(ll a,ll b){return (!a)?b:a%b;}
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);
	}
}
mi f[412345]={};
int n,a[101010]={},S;
bool flag=1;
vector<tuple<int,int,int> > ans;
void solve(int i,int f,int p,int t) {
	if(i<=0) return ;
	solve(i-1-a[i],f,t,p);
    ans.pb({i,f,t});
    if(a[i]) {
	    solve(i-1-a[i],p,t,f);
		solve(i-1,f,p,t);
	}else {
		solve(i-1,p,f,t);
	}
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T;
	cin>>T;
	while(T--) {
		flag=1;
		cin>>n;
		For(i,n) cin>>a[i];
		For(i,n) flag&=a[i]<i;
		
		if(!flag) {
			cout<<"NO"<<endl;
			continue;
		}
		S=1<<n;
		solve(n,1,2,3);
		
		if(!flag || SI(ans)>S) {
			cout<<"NO"<<endl;
			continue;
		}
		cout<<"YES"<<endl;
		cout<<SI(ans)<<endl;
		for(auto [a,b,c]:ans) {
			cout<<a<<' '<<b<<' '<<c<<endl;
		}
		ans.resize(0);
	}
	
	return 0;
}

E Snaking Arrangement

给一个n*n的矩阵，每条蛇为一个长为1,3,5/cdots,2n−11,3,5/cdots,2n-11,3,5/cdots,2n−1的路径，它们不能重叠，只能向下或右，其中一些蛇已知位置。问合法解数。

cpp 复制代码 
#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 (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#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);
#define sqr(x) (x*x)
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;}
ll gcd(ll a,ll b){return (!a)?b:a%b;}
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=1000000007; 
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()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T;
	cin>>T;
	while(T--) {
		int n,k;
		cin>>n>>k;
		if(!k) {
			cout<<simp::gfac(n).v<<endl;
			continue;
		}
		vector<vector<int> > a(n+1,vi(n+1,0));
		vector<bool> b(n+1,0);
		Rep(i,k) {
			int s,r,c;
			cin>>s>>r>>c;
			b[n-(s+1)/2+1]=1;
			if(s>1) {
				string p;cin>>p;
				for(auto ch:p) {
					a[r][c]=ch;					
					if(ch=='R') {
						++c;
					}else ++r;
				}
			}
		}
		mi ans=1;
		For(i,n) if(!b[i]){
			int l=1,r=i;
			For(x,i-1) {
				int y=i-x;
				if(a[x][y]=='D') gmin(r,x)
				if(a[x][y]=='R') gmax(l,x+1)
			}
			int valid_st_point=r-l+1;
			if(l<=r) ans*=valid_st_point;
		}
		cout<<ans.v<<endl;
	}
	return 0;
}
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