Neowise Labs Contest 1 (Codeforces Round 1018, Div. 1 + Div. 2) 题解

原创 于 2025-04-23 11:15:27 发布 · 1.1k 阅读 · 19 ·
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#平衡3进制

A Wonderful Sticks

#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 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);
#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;
} 
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read();
	while(T--) {
		int n=read();string s;cin>>s;
		int l=1,r=n;
		vi v;
		RepD(i,n-2) {
			if(s[i]=='<') v.pb(l++);else v.pb(r--);
		}
		v.pb(l);
		reverse(ALL(v));
		For(i,n) cout<<v[i-1]<<' ';cout<<endl;
	}
	return 0;
}

B

#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 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);
#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;
} 
ll l[212345],r[212345];
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read();
	while(T--) {
		int n=read(),k=read();
		ll s2=0;
		For(i,n) l[i]=read(),s2+=l[i];
		For(i,n) r[i]=read(),s2+=r[i];
		
		For(i,n) gmin(l[i],r[i])
		sort(l+1,l+1+n);
		ll s=0;
		For(i,n-(k-1)) s+=l[i];
		cout<<s2-s+1<<endl;
	}
	return 0;
}

C Wonderful City

给1个矩阵，第i行加1需要ci代价，第i列需要di代价，每行，每列只能加1次。问最小代价使矩阵不存在相邻元素相同。

每行每列取不取只被上行影响。行列可分离考虑。

#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 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);
#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;
} 
#define MAXN (1010)
int h[MAXN][MAXN];
bool b[MAXN][10];
ll f[MAXN][2][2],d[MAXN],c[MAXN];
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read();
	while(T--) {
		int n=read();
		For(i,n) For(j,n) h[i][j]=read();
		For(i,n) c[i]=read();
		For(i,n) d[i]=read();
		
		For(i,n) Rep(k,6) b[i][k]=0;
		For(i,n) For(j,n) {
			if(i<n && h[i][j]==h[i+1][j]) b[i][0]=1;
			if(j<n && h[i][j]==h[i][j+1]) b[j][1]=1;
			
			if(i<n && h[i][j]==1+h[i+1][j]) b[i][2]=1;
			if(j<n && h[i][j]==1+h[i][j+1]) b[j][3]=1;
			
			if(i<n && h[i][j]+1==h[i+1][j]) b[i][4]=1;
			if(j<n && h[i][j]+1==h[i][j+1]) b[j][5]=1;
		}
//		For(i,n) {
//			Rep(k,6) cout<<b[i][k]<<' ';cout<<endl;
//		}
//		
		f[1][0][0]=0;f[1][1][0]=c[1],f[1][0][1]=d[1]; f[1][1][1]=c[1]+d[1];
		Fork(i,2,n) {
			Rep(k,2) Rep(l,2){
				f[i][k][l]=-1;
				ll ci=0;if(k==1) ci+=c[i];if(l==1) ci+=d[i];
				
				Rep(j,2) Rep(t,2) if(f[i-1][j][t]!=-1){
					if(b[i-1][0]) {
						if(j==k ) continue;
					}
					if(b[i-1][1]) {
						if(t==l) continue;
					}
					if(b[i-1][2]) {
						if(k==1 && j!=1) continue;
					}
					if(b[i-1][3]) {
						if(l==1 && t!=1) continue;
					}
					if(b[i-1][4]) {
						if(j==1 && k!=1) continue;
					}
					if(b[i-1][5]) {
						if(t==1 && l!=1) continue;
					}
					if(f[i][k][l]==-1) f[i][k][l]=f[i-1][j][t]+ci;else f[i][k][l]=min(f[i][k][l],f[i-1][j][t]+ci);
				}
			}
		}
		For(i,n) {
			Rep(k,2) Rep(l,2){
				dbg(f[i][k][l])
			}
		}
		ll t=-1;
		Rep(k,2)Rep(l,2) if(f[n][k][l]!=-1) {
			if(t==-1) t=f[n][k][l];else t=min(t,f[n][k][l]);
		}
		cout<<t<<endl;
		
	}
	return 0;
}

D Wonderful Lightbulbs

You are the proud owner of an infinitely large grid of lightbulbs, represented by a Cartesian coordinate system. Initially, all of the lightbulbs are turned off, except for one lightbulb, where you buried your proudest treasure.

In order to hide your treasure’s position, you perform the following operation an arbitrary number of times (possibly zero):

  • Choose two integer numbers x x x and y y y, and switch the state of the 4 4 4 lightbulbs at ( x , y ) (x, y) (x,y), ( x , y + 1 ) (x, y + 1) (x,y+1), ( x + 1 , y − 1 ) (x + 1, y - 1) (x+1,y−1), and ( x + 1 , y ) (x + 1, y) (x+1,y). In other words, for each lightbulb, turn it on if it was off, and turn it off if it was on. Note that there are no constraints on x x x and y y y.

In the end, there are n n n lightbulbs turned on at coordinates ( x 1 , y 1 ) , ( x 2 , y 2 ) , … , ( x n , y n ) (x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n) (x1​,y1​),(x2​,y2​),…,(xn​,yn​). Unfortunately, you have already forgotten where you buried your treasure, so now you have to figure out one possible position of the treasure. Good luck!

性质1：x坐标为奇数的一定是答案。
性质2：可以把2个x坐标相同的数向左上方平移。移到同一x后，只剩下1个数
，

#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 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);
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;
} 
ll x[212345],y[212345];
int main()
{
//	freopen("D.in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read();
	while(T--) {
		int n=read();
		vector<pair<ll,ll> > v;
		ll p=0;
		For(i,n) {
			x[i]=read(),y[i]=read();
			p^=x[i];
		}
		For(i,n) {
			y[i]+=x[i]-p;
			x[i]=p;
		}
		ll q=0;
		For(i,n) q^=y[i];
		cout<<p<<' '<<q<<endl;
	} 
	return 0;
}

E Wonderful Teddy Bears

#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 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);
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;
} 
int main()
{
//	freopen("E.in","r",stdin);
//	freopen(".out","w",stdout);
	
	int T=read();
	while(T--) {
		int n=read();
		string s;cin>>s;
		ll t=0,b=0,c1=0,c2=0;
		RepD(i,n-1) {
			if(s[i]=='B') ++b;
			else {
				t+=b;
				if(i%2==1) ++c1;else ++c2;
			}
		}
		ll c=abs(c1-c2);
		cout<<c/2+(t-c/2+1)/2<<endl;
		
	} 
	
	return 0;
}

F Wonderful Impostors

G Wonderful Guessing Game

平衡三进制
看其他人的代码才发现有这么个东西

#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 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);
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;
} 
#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<<"}";
}
vi getkth(int p) {
	int f=1;
	if(p<0) p=-p,f=-1;
	vi v;
	while(p) {
		if(p%3<=1) v.pb(p%3*f),p/=3;
		else {
			v.pb(-1*f);p/=3,++p;
		}
	}
//	dbg(v);
	return v;
}
int main()
{
//	freopen("E.in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read();
	cin.tie(0);
	while(T--) {
		int n=read();
		int p3=3,k=1;while(p3<n) p3*=3,++k;
		vector<vi > a;
		if(n%2) {
			a.pb(getkth(0));
			a[SI(a)-1].resize(k);
			a[SI(a)-1].pb(0);
		}
		For(i,n/2) {
			a.pb(getkth(i));
			a[SI(a)-1].resize(k);
			int t;
			t=0;for(auto p:a.back()) t+=p;
			t=(t%3+3)%3;if(t==2) t-=3;
			a[SI(a)-1].pb(t);
			a.pb(getkth(-i));
			a[SI(a)-1].resize(k);
			t=0;for(auto p:a.back()) t+=p;
			t=(t%3+3)%3;if(t==2) t-=3;
			a[SI(a)-1].pb(t);
		}
		cout<<k+1<<endl;
		Rep(i,k+1) {
			int tot=0;
			Rep(j,n) tot+=a[j][i]!=0;
			cout<<tot;
			Rep(j,n) if(a[j][i]==-1) {
				cout<<' '<<j+1;
			}
			Rep(j,n) if(a[j][i]==1) {
				cout<<' '<<j+1;
			}
			cout<<endl;
		}
		fflush(stdout);
		string s;
		cin>>s;
		Rep(j,n) {
			bool fl=1;
			Rep(i,k+1)
				if(s[i]!='?') {
					if(s[i]=='L' && a[j][i]!=-1 || s[i]=='R' && a[j][i]!=1 || s[i]=='N' && a[j][i]!=0 ) {
						fl=0;break;
					}
				}
			if(fl) {
				cout<<j+1<<endl;
				break;
			}
		}
	}
	return 0;
}
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