Codeforces LATOKEN Round 1 (Div. 1 + Div. 2)

Posted 2022-12-03 nike0good

tags:

篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了Codeforces LATOKEN Round 1 (Div. 1 + Div. 2)相关的知识，希望对你有一定的参考价值。

A Colour the Flag

#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>0;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;
 
char s[100][100];
int main()

//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	int t=read();
	while(t--) 
		int n=read(),m=read();
		For(i,n) cin>>(s[i]+1);
		char h[3]='R','W';
		bool fl1=0,fl2=0;
		For(i,n) 
			For(j,m) 
				if(s[i][j]==h[(i+j)&1]) fl2=1;
				if(s[i][j]==h[((i+j)&1)^1]) fl1=1;
			
		
		if(!fl1) 
			puts("YES");
			For(i,n) 
				For(j,m) 
					cout<<h[(i+j)&1];
				cout<<endl;
			
				
		else if(!fl2) 
			puts("YES");
			For(i,n) 
				For(j,m) 
					cout<<h[((i+j)&1)^1];
				cout<<endl;
			
		else 
			puts("NO");
		
	
	return 0;

B Histogram Ugliness

#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>0;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 t,n,a[412345];
int main()

//	freopen("B.in","r",stdin);
//	freopen(".out","w",stdout);
	
	int t=read();
	while(t--) 
		n=read();
		For(i,n)
			a[i]=read();
        a[n+1]=0;
        ll ans=0;
        For(i,n) 
            ans+=max(0,a[i]-a[i+1])+max(0,a[i]-a[i-1]);
            if (a[i]>a[i+1]&&a[i]>a[i-1]) 
				ans-=a[i]-max(a[i-1],a[i+1]);
        
        cout<<ans<<endl;
	
	return 0;

C Little Alawn’s Puzzle

#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>0;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 a[3][412345];
class bingchaji

public:
	int father[412345],n,cnt;
	void mem(int _n)
	
		n=cnt=_n;
		For(i,n) father[i]=i;
	
	int getfather(int x) 
	
		if (father[x]==x) return x;
		
		return father[x]=getfather(father[x]);
	
	void unite(int x,int y)
	
		x=getfather(x);
		y=getfather(y);
		if (x^y) 
			--cnt;
			father[x]=y;
		
	
	bool same(int x,int y)
	
		return getfather(x)==getfather(y);
	
S;
ll work() 
	int n=2,m=read();
	For(i,n)For(j,m) a[i][j]=read();
	For(j,m) if(a[1][j]==a[2][j]) return 0;
	S.mem(m);
	For(j,m) 
		S.unite(a[1][j],a[2][j]);
	
	ll p=1;
	For(j,m) if(S.getfather(j)==j) p=p*2%F;return p;

int main()

//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	int t=read();
	while(t--) 
		cout<<work()<<endl;	
	
	return 0;

E Lost Tree

交互题，有n个数，每次可以询问一个大小恰为k的子集的数的xor和。最小步数求n个数的xor和。无解输-1.

dp， $f [i]$ 表示知道i个数的xor和的最小步数。

#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>0;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 f[600],pr[600];
bool b[600]=;
int need[600]=以上是关于Codeforces LATOKEN Round 1 (Div. 1 + Div. 2)的主要内容，如果未能解决你的问题，请参考以下文章 
 Codeforces LATOKEN Round 1 (Div. 1 + Div. 2)
 Codeforces LATOKEN Round 1 (Div. 1 + Div. 2) E. Lost Array(构造)
 Codeforces LATOKEN Round 1 (Div. 1 + Div. 2) B. Histogram Ugliness（贪心）
 Codeforces LATOKEN Round 1 (Div. 1 + Div. 2) D. Lost Tree(交互,二分/图,思维)
 D. Lost Tree树的二分图性质——Codeforces LATOKEN Round 1 (Div. 1 + Div. 2)
 D. Lost Tree树的二分图性质——Codeforces LATOKEN Round 1 (Div. 1 + Div. 2)