集训 4 & 模拟 5

集训 4 & 模拟 5

有点唐简单，所以一起写了（其实是因为之前懒得写）

集训 4：

T1 模拟，赛时不删调试保龄了。

T2 显然贪心

T3 发现显然要两两互质，有因为父比子小，所以方案数就是将 $\phi$ 乘起来（甚至都不需要线性筛）

T4 meet in middle 板子。

模拟 5

1. T1 特殊字符串

   显然有 $n^2$ 的 $dp$，发现字符集为 $26$ 可以对结尾字符分别存储 $max$ 优化到 $n\ast 26$

   CODE

   #include<bits/stdc++.h>
   #include<sys/mman.h>
   #include<fcntl.h>
   using namespace std;
   using llt=long long;
   using llf=long double;
   using ull=unsigned long long;
   #define Ct const
   #define Il __always_inline
   #define For(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
   #define For_(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
   #define For_it(i,a,b) for(auto i=(a);i!=(b);++i)
    
   namespace IO{
   #ifdef ONLINE_JUDGE
   	int Fin=open("shiki.in",0); FILE* Fout=fopen("shiki.out","w");
   #elif defined(UN_FAST)
   	FILE *Fin=freopen("in_out/in.in","r",stdin),*Fout=freopen("in_out/out.out","w",stdout);
   #else
   	int Fin=open("in_out/in.in",0); FILE* Fout=fopen("in_out/out.out","w");
   #endif // file
   #ifdef UN_FAST
   	char cc;
   	#define G (cc=getchar())
   	#define C cc
   #else
   	const char *I=(char*)mmap(0,1<<28,1,2,Fin,0)-1; 
   	#define G (*++I)
   	#define C (*I)
   #endif // fast (mmap)
   #define P(x) putc_unlocked(x,Fout)
   	template<class T> Il void read(T &x){x=0;bool f=0; while(f|=G==45,C<48); while(x=x*10+(C&15),G>47); f?x=-x:0;}
   	template<class T> void write(T x){if(x<0)P('-'),x=-x; if(x/10)write(x/10); P('0'+x%10);}
   	Il void read(char &c){while(G<33); c=C;} void read(char *s){int p=-1; while(G<33); while(s[++p]=C,G>32); s[++p]='\0';}
   	Il void read(string &s){s.clear(); while(G<33); while(s.push_back(C),G>32);}
   	template<class T=int> Il T read(){T a; read(a); return a;}
   	template<class T,class... Argc> Il void read(T &x,Argc&... argc){read(x),read(argc...);}
   	Il void write(const char &c){P(c);} Il void write(const char *s){for(const char *c=s;*c!='\0';++c) P(*c);}
   	Il void write(const string &s){for(const char& c:s) P(c);}
   	template<class T,class... Argc> Il void write(const T &x,const Argc&... argc){write(x),write(argc...);}
   	template<class T> Il void Write(const T &x){write(x),P(' ');} Il void Write(const char &c){P(c); if(c>32)P(' ');}
   	template<class T,class... Argc> Il void Write(const T &x,const Argc&... argc){Write(x),Write(argc...);}
   #undef G
   #undef C
   #undef P
   #define endl '\n'
   }using IO::read; using IO::write; using IO::Write;
    
   constexpr int N=1e5+3;
   llt ma[30],n,m,wt[30][30],ans;
   string s;
    
   int main(){
   	read(n,s,m);
   	For(i,1,m,1){
   		char a,b; int k; read(a,b,k);
   		wt[a-'a'][b-'a']+=k;
   	}
   	memset(ma,-0x3f,sizeof(ma));
   	ma[s[0]-'a']=0;
   	For(i,2,n,1){
   		int nw=s[i-1]-'a';
   		For(j,0,25,1) ans=max(ans,ma[j]+wt[j][nw]);
   		ma[nw]=max(ma[nw],ans);
   	}
   	write(ans);
   }
   // For(i,1,n,1) For(j,1,i-1,1) dp[i]=max(dp[i],dp[j]+wt[s[j-1]-'a'][s[i-1]-'a']);

2. T2 宝可梦

   发现从一个点到自己的路径是一个环，并且每条询问路径都是环上一段，直接预处理环即可（不太好写，赛时想假了好几次）。

   CODE

   #include<bits/stdc++.h>
   #include<sys/mman.h>
   #include<fcntl.h>
   using namespace std;
   using llt=long long;
   using llf=long double;
   using ull=unsigned long long;
   #define Ct const
   #define Il __always_inline
   #define For(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
   #define For_(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
   #define For_it(i,a,b) for(auto i=(a);i!=(b);++i)
    
   namespace IO{
   #ifdef ONLINE_JUDGE
   	int Fin=open("pokemon.in",0); FILE* Fout=fopen("pokemon.out","w");
   #elif defined(UN_FAST)
   	FILE *Fin=freopen("in_out/in.in","r",stdin),*Fout=freopen("in_out/out.out","w",stdout);
   #else
   	int Fin=open("in_out/in.in",0); FILE* Fout=fopen("in_out/out.out","w");
   #endif // file
   #ifdef UN_FAST
   	char cc;
   	#define G (cc=getchar())
   	#define C cc
   #else
   	const char *I=(char*)mmap(0,1<<28,1,2,Fin,0)-1; 
   	#define G (*++I)
   	#define C (*I)
   #endif // fast (mmap)
   #define P(x) putc_unlocked(x,Fout)
   	template<class T> Il void read(T &x){x=0;bool f=0; while(f|=G==45,C<48); while(x=x*10+(C&15),G>47); f?x=-x:0;}
   	template<class T> void write(T x){if(x<0)P('-'),x=-x; if(x/10)write(x/10); P('0'+x%10);}
   	Il void read(char &c){while(G<33); c=C;} void read(char *s){int p=-1; while(G<33); while(s[++p]=C,G>32); s[++p]='\0';}
   	Il void read(string &s){s.clear(); while(G<33); while(s.push_back(C),G>32);}
   	template<class T=int> Il T read(){T a; read(a); return a;}
   	template<class T,class... Argc> Il void read(T &x,Argc&... argc){read(x),read(argc...);}
   	Il void write(const char &c){P(c);} Il void write(const char *s){for(const char *c=s;*c!='\0';++c) P(*c);}
   	Il void write(const string &s){for(const char& c:s) P(c);}
   	template<class T,class... Argc> Il void write(const T &x,const Argc&... argc){write(x),write(argc...);}
   	template<class T> Il void Write(const T &x){write(x),P(' ');} Il void Write(const char &c){P(c); if(c>32)P(' ');}
   	template<class T,class... Argc> Il void Write(const T &x,const Argc&... argc){Write(x),Write(argc...);}
   #undef G
   #undef C
   #undef P
   #define endl '\n'
   }using IO::read; using IO::write; using IO::Write;
    
   constexpr int N=1e5+3,NM=5e5+3,mx[]={0,0,-1,0,1},my[]={0,1,0,-1,0},tlt[]={0,2,3,4,1},trt[]={0,4,1,2,3};
   vector<bool> c[N];
   vector<int> pos[5][N];
   int n,m,q,cnt;
    
   void Dfs(Ct int &x,Ct int &y,Ct int &fw){
   	if(pos[fw][x][y]) return ;
   	pos[fw][x][y]=++cnt;
   	int nf=trt[fw];
   	For(i,1,4,1){
   		int nx=x+mx[nf],ny=y+my[nf];
   		if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&c[nx][ny]) Dfs(nx,ny,nf);
   		nf=tlt[nf];
   	}
   }
   Il void Init(){
   	int x,y;
   	read(n,m);
   	For(i,1,n,1){
   		c[i].resize(m+3);
   		For(k,1,4,1) pos[k][i].resize(m+3);
   		For(j,1,m,1) if(read<char>()=='.') c[i][j]=1,x=i,y=j;
   	}
   	Dfs(x,y,2);
   }
    
   Il int Get(Ct int &xa,Ct int &ya,Ct int &fa,Ct int &xb,Ct int &yb,Ct int &fb){
   	if(!pos[fa][xa][ya]||!pos[fb][xb][yb]) return 0x3f3f3f3f;
   	if(pos[fa][xa][ya]<=pos[fb][xb][yb]) return pos[fb][xb][yb]-pos[fa][xa][ya];
   	return cnt+pos[fb][xb][yb]-pos[fa][xa][ya]-1;
   }
    
   int main(){
   	Init();
   	read(q);
   	For(i,1,q,1){
   		int xa,ya,xb,yb,fa; char c; read(xa,ya,xb,yb,c);
   		if(c=='R') fa=1; else if(c=='U') fa=2; else if(c=='L') fa=3; else fa=4;
   		if(xa==xb&&ya==yb){write(0,'\n'); continue;}
   		xa+=mx[fa],ya+=my[fa];
   		int ans=0x3f3f3f3f; For(i,1,4,1) ans=min(ans,Get(xa,ya,fa,xb,yb,i)+1);
   		write(ans,'\n');
   	}
   }

3. T3 矩阵

   如果有相邻且相同的，就是 $-1$

   其他情况对公比分治，在公比 $>200$ 时最多长度为 $2$，可以直接整。

   公比 $\le 200$ 时 $O(n^2\ast \text{公比})$ 暴力 $dp$ 即可。

   也可以直接最长路。

   ps:

   谁 tm 造的数据，每个 sub 里塞一个 ans=1 ，给我卡没了。

   CODE

   #include<bits/stdc++.h>
   #include<sys/mman.h>
   #include<fcntl.h>
   using namespace std;
   using llt=long long;
   using llf=long double;
   using ull=unsigned long long;
   #define Ct const
   #define Il __always_inline
   #define For(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
   #define For_(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
   #define For_it(i,a,b) for(auto i=(a);i!=(b);++i)
    
   namespace IO{
   #ifdef ONLINE_JUDGE
   	int Fin=open("matrix.in",0); FILE* Fout=fopen("matrix.out","w");
   #elif defined(UN_FAST)
   	FILE *Fin=freopen("in_out/in.in","r",stdin),*Fout=freopen("in_out/out.out","w",stdout);
   #else
   	int Fin=open("in_out/in.in",0); FILE* Fout=fopen("in_out/out.out","w");
   #endif // file
   #ifdef UN_FAST
   	char cc;
   	#define G (cc=getchar())
   	#define C cc
   #else
   	const char *I=(char*)mmap(0,1<<28,1,2,Fin,0)-1; 
   	#define G (*++I)
   	#define C (*I)
   #endif // fast (mmap)
   #define P(x) putc_unlocked(x,Fout)
   	template<class T> Il void read(T &x){x=0;bool f=0; while(f|=G==45,C<48); while(x=x*10+(C&15),G>47); f?x=-x:0;}
   	template<class T> void write(T x){if(x<0)P('-'),x=-x; if(x/10)write(x/10); P('0'+x%10);}
   	Il void read(char &c){while(G<33); c=C;} void read(char *s){int p=-1; while(G<33); while(s[++p]=C,G>32); s[++p]='\0';}
   	Il void read(string &s){s.clear(); while(G<33); while(s.push_back(C),G>32);}
   	template<class T=int> Il T read(){T a; read(a); return a;}
   	template<class T,class... Argc> Il void read(T &x,Argc&... argc){read(x),read(argc...);}
   	Il void write(const char &c){P(c);} Il void write(const char *s){for(const char *c=s;*c!='\0';++c) P(*c);}
   	Il void write(const string &s){for(const char& c:s) P(c);}
   	template<class T,class... Argc> Il void write(const T &x,const Argc&... argc){write(x),write(argc...);}
   	template<class T> Il void Write(const T &x){write(x),P(' ');} Il void Write(const char &c){P(c); if(c>32)P(' ');}
   	template<class T,class... Argc> Il void Write(const T &x,const Argc&... argc){Write(x),Write(argc...);}
   #undef G
   #undef C
   #undef P
   #define endl '\n'
   }using IO::read; using IO::write; using IO::Write;
    
   constexpr int N=4e4+3,mx[]={0,1,-1,0,0},my[]={0,0,0,1,-1};
   vector<int> c[N],dp[N];
   int n,m,ans=1;
   int dfs(int x,int y,int k){
   	if(dp[x][y]) return dp[x][y];
   	else {
   		For(i,1,4,1){
   			int nx=x+mx[i],ny=y+my[i];
   			if(nx>n||nx<1||ny>m||ny<1||c[nx][ny]!=c[x][y]*k) continue;
   			dp[x][y]=max(dp[x][y],dfs(nx,ny,k));
   		}
   		dp[x][y]+=1;
   		return dp[x][y];
   	}
   }
    
   int main(){
   	For(i,1,N-3,1) c[i].push_back(0);
   	read(n,m);
   	For(i,1,m,1){
   		c[1].push_back(read());
   		if(c[1][i]==c[1][i-1]){write(-1); return 0;}
   		else if(i!=1&&(c[1][i]%c[1][i-1]==0||c[1][i-1]%c[1][i]==0)) ans=2;
   	}
   	For(i,2,n,1) For(j,1,m,1){
   		c[i].push_back(read());
   		if(c[i][j]==c[i][j-1]||c[i][j]==c[i-1][j]){write(-1); return 0;}
   		else if(j!=1&&(c[i][j]%c[i][j-1]==0||c[i][j-1]%c[i][j]==0||c[i][j]%c[i-1][j]==0||c[i-1][j]%c[i][j]==0)) ans=2;
   	}
   	For(i,2,200,1){
   		For(j,1,n,1) dp[j].clear(),dp[j].resize(m+1);
   		For(a,1,n,1) For(b,1,m,1) ans=max(ans,dfs(a,b,i));
   	}
   	write(ans);
   }

4. T4 乘法

   这是真唐，整个拉插板子，我也不会，题解也不会 😅😅😅