10.2 Zheng Rui National Day training test 1

Directory

• 2018. Test
  • Summarize
  • A Chen Taiyang and taking the mold
  • B Chen Taiyang with path (tree DP)
  • C Chen Taiyang and switch
  • Exam Code
    • B
    • C

2018. Test

Time: 3.5h
Expected Score: 100+50+20
Actual score: 40+45+20

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Summarize

I feel like I'm a zz.

A Chen Taiyang and taking the mold

Topic links

\[\BEGIN{ALIGNED}C\%X&\EQUIV (c\%a) \%x\\c\%x&\equiv (C-\lfloor\frac ca\rfloor*a) \%x\\c\%x&\equiv c\%x- \lfloor\frac ca\rfloor*a\%x\\ (\lfloor\frac ca\rfloor*a) \%x&\equiv 0\end{aligned}\]

That is, \ (x\) should be an approximate (\lfloor\frac ca\rfloor*a\) .
\ (c\in[l,r]\), if \ (\frac la=\frac ra\) then \ (x\) can be \ (\lfloor\frac ca\rfloor\times a\) of the approximate;
Otherwise there is at least \ (\lfloor\frac la\rfloor\times a\) and \ ((\lfloor\frac la\rfloor+1) \times a\),\ (x\) It can only be an approximate of \ (a\) .
Then ask for an approximate number of numbers to be good.

//#include <cmath>#include <cstdio>#include <cctype>#include <algorithm>#define gc() getchar()typedef long long LL;inline LL read(){    LL now=0;register char c=gc();    for(;!isdigit(c);c=gc());    for(;isdigit(c);now=now*10+c-'0',c=gc());    return now;}void Div(LL x){    LL ans=1;//  for(int i=2,lim=sqrt(x); i<=lim; ++i)    for(int i=2; 1ll*i*i<=x; ++i)        if(!(x%i))        {            int cnt=1;            while(!(x%i)) x/=i, ++cnt;            ans*=cnt;        }    if(x!=1) ans<<=1ll;    printf("%lld\n",ans);}void Work(){    LL L=read(),R=read(),A=read();    if(A>R) return (void)puts("-1");    L/A==R/A ? Div(L/A*A) : Div(A);}int main(){//  freopen("ex_modulo3.in","r",stdin);    for(int T=read(); T--; Work());    return 0;}

B Chen Taiyang with path (tree DP)

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Because the height of the random tree is guaranteed, the expected maximum depth of the random tree generated by the prufer sequence is \ (o (sqrt n) \)and the average depth is \ (o (\log n) \).
So we can use two times the complexity of \ (O (N*DEP) \) the tree DP to find the answer to each point. So the spatial complexity is also \ (O (N*DEP) \) .
The useful size of a DP array is only related to depth, and can be dynamically allocated with vectors or pointers. In addition, the midpoint of the diameter is the root depth can be/2.
So the final complexity is probably \ (O (n\log n) \).

You can also use a long chain to do, for each long chain to maintain a DP array, the second time DFS only need to merge the non-long chain to the long chain.
So that any tree can do, the complexity of the (O (n) \).
--by DLs

#include <cstdio> #include <cctype> #include <algorithm>//#define GC () GetChar () #define Maxin 200000#define GC () (ss==tt&& (tt= (ss=in) +fread (In,1,maxin,stdin), SS==TT)? eof:*ss++) typedef long LONG Ll;const int n=5e5+5;int Enum,h[n],nxt[n<<1],to[n<<1],fa[n],pre[n],dis[n], Dep[n];int Pool[30000000],*now=pool,*f[n],*g[n];    LL Ans[n];char in[maxin],*ss=in,*tt=in,out[6000000],*o=out;inline int read () {int now=0;register char c=gc (); for (;!    IsDigit (c); C=GC ());    for (; IsDigit (c); now=now*10+c-' 0 ', C=GC ()); return now;}    void print (LL x) {if (x>9) print (X/10); *o++ = x%10+ ' 0 ';}    inline void AE (int u,int v) {to[++enum]=v, nxt[enum]=h[u], h[u]=enum; To[++enum]=u, Nxt[enum]=h[v], h[v]=enum;}    int BFS (int s) {static int q[n];    int h=0,t=1;    Q[0]=s, dis[s]=pre[s]=0;        while (h<t) {int x=q[h++];    for (int i=h[x],v; i; i=nxt[i]) if ((V=to[i])!=pre[x]) dis[v]=dis[x]+1, pre[v]=x, q[t++]=v; } return Q[t];}    void DFS0 (int x) {int tmp=0;    for (int i=h[x],v; i; i=nxt[i]) if ((V=to[i))!=fa[x]) fa[v]=x, DFS0 (v), Tmp=std::max (tmp,dep[v]+1);    dep[x]=tmp; F[x]=now, Now+=tmp+1, G[x]=now, now+=tmp+1;} void DFS1 (int x) {int *f=f[x];    LL res=0;            for (int i=h[x],v; i; i=nxt[i]) if ((V=to[i])!=fa[x]) {DFS1 (v); int *fv=f[v];        for (int j=0; j<=dep[v]; ++j) res+=1ll*f[j]*fv[j], f[j]+=fv[j];    } for (int i=dep[x]; i; i.) f[i]=f[i-1]; F[0]=1, Ans[x]=res;}    void DFS2 (int x) {int *g=g[x],*gf=g[fa[x]],*f=f[x],*ff=f[fa[x]];            if (x!=1) {for (int i=1; i<=dep[x]; ++i) {g[i]=gf[i-1]+ff[i-1];            if (i>=2) g[i]-=f[i-2];        Ans[x]+=1ll*g[i]*f[i]; }} for (int i=h[x],v; i; i=nxt[i]) if ((V=to[i])!=fa[x]) DFS2 (v);}    int main () {//Freopen ("Ex_diameter3.in", "R", stdin);//Freopen ("My.out", "w", stdout);    int N=read (); for (int i=1; i<n; ++i) AE (read (), ReaD ());    int S=bfs (1), RT=BFS (s), d=dis[rt]>>1;    while (DIS[RT]&GT;D) RT=PRE[RT];    DFS0 (RT), DFS1 (RT), DFS2 (RT); for (int i=1; i<n; ++i) print (ans[i]+1), *o++= ';    Print (ans[n]+1);    Fwrite (out,o-out,1,stdout); return 0;}

C Chen Taiyang and switch

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Exam Code B

Why are you running slower than ZZX? (though a penny bar)

#include <cstdio> #include <cctype> #include <algorithm>//#define GC () GetChar () #define Maxin 200000# Define GC () (ss==tt&& (tt= (ss=in) +fread (In,1,maxin,stdin), SS==TT)? eof:*ss++) typedef long LONG Ll;const int n=5e5+5;int Enum,h[n],nxt[n<<1],to[n<<1],dgr[n],f[n],numx[n], Num[n],maxd;    LL Ans[n];char in[maxin],*ss=in,*tt=in,out[6000000],*o=out;inline int read () {int now=0;register char c=gc (); for (;!    IsDigit (c); C=GC ());    for (; IsDigit (c); now=now*10+c-' 0 ', C=GC ()); return now;}    void print (LL x) {if (x>9) print (X/10); *o++ = x%10+ ' 0 ';}    inline void AE (int u,int v) {++dgr[v], to[++enum]=v, Nxt[enum]=h[u], h[u]=enum; ++dgr[u], To[++enum]=u, Nxt[enum]=h[v], h[v]=enum;    void DFS0 (int x,int fa) {int tmp=0;    for (int i=h[x],v; i; i=nxt[i]) if ((v=to[i))!=FA) DFS0 (v,x), Tmp=std::max (Tmp,f[v]); f[x]=tmp+1;} void DFS1 (int x,int fa)//{//if (dgr[x]==1) {fi[x]=0, se[x]=n; return;} int f=n,s=n;//for (int i=h[x],v; i; i=nxt[i]//if ((V=to[i])!=FA)//{//DFS1 (V,X);/if (fi[v]<f) s=f, F=fi[v], pos[x]=v;//el Se s=std::min (s,fi[v]);//}//:: F[x]=fi[x]=f+1, se[x]=s+1;//}//void DFS2 (int x,int fa)//{//for (int i=h[x],v; i; i= Nxt[i]//if ((V=to[i])!=FA)//{//if (POS[X]==V) f[v]=std::min (Se[x]+1,f[v]);/Else F[V]=STD::    Min (fi[x]+1,f[v]);//DFS2 (V,X);//}//}void DFS3 (int x,int fa,int d,int dep) {Maxd=std::max (maxd,d);    ++num[d], ++d;    if (!--dep) return; for (int i=h[x]; i; i=nxt[i]) if (TO[I]!=FA) DFS3 (TO[I],X,D,DEP);}    int main () {freopen ("ex_diameter2.in", "R", stdin);    Freopen ("My.out", "w", stdout);    int N=read ();    for (int i=1; i<n; ++i) AE (read (), read ()); DFS0 (+//DFS1), DFS2 (n/a),//for (int i=1; i<=n; ++i) printf ("f[%d]=%d fi[%d]=%d se[%d]=%d\n", I,f[i],i,fi[i],    I,se[i]);        for (int x=1; x<=n; ++x) if (dgr[x]==1) ans[x]=1; else {LL res=1; int tmp=f[x]-1; maxd=1;                for (int i=h[x]; i; i=nxt[i]) {DFS3 (to[i],x,1,tmp);            for (int j=1; j<=maxd&&num[j]; ++j) res+=1ll*numx[j]*num[j], numx[j]+=num[j], num[j]=0;            } for (int i=1; i<=tmp; ++i) numx[i]=0;        Ans[x]=res; } for (int i=1; i<n; ++i) print (Ans[i]), *o++= ';    Print (Ans[n]);    Fwrite (out,o-out,1,stdout); return 0;}

C

#include <cstdio> #include <cctype> #include <cstring> #include <algorithm> #define GC () GetChar () #define MOD 1000000007typedef long ll;const int n=1e5+5;int N,enum,h[n],nxt[n<<1],to[n<<1];bool mp[    21][21],rev0[21],vis0[21];inline int read () {int now=0;register char c=gc (); for (;!    IsDigit (c); C=GC ());    for (; IsDigit (c); now=now*10+c-' 0 ', C=GC ()); return now;}    inline void AE (int u,int v) {to[++enum]=v, nxt[enum]=h[u], h[u]=enum; To[++enum]=u, Nxt[enum]=h[v], h[v]=enum;}    int DFS0 (int x) {int res=1; vis0[x]=1;    for (int i=1; i<=n; ++i) if (!vis0[i] && (Rev0[x]^rev0[i]^mp[x][i])) res+=dfs0 (i); return res;}    BOOL Check () {memset (vis0,0,sizeof vis0); return DFS0 (1) ==n;}    void Subtask1 () {memset (mp,0,sizeof MP);    for (int x=1, x<=n; ++x) for (int i=h[x]; i; i=nxt[i]) mp[x][to[i]]=1;    int ans=1,all= (1<<n)-1; for (int s=1, s<=all; ++s) {for (int i=0; i<n; ++i) if (s>>i&1) rev0[I+1]=1;        Ans+=check ();    for (int i=0; i<n; ++i) if (s>>i&1) rev0[i+1]=0; } printf ("%d\n", ans);}    void work () {enum=0, memset (h,0,sizeof H);    N=read ();    for (int i=1; i<n; ++i) AE (read (), read ());        if (n<=20) {Subtask1 (); return;}    }int Main () {//Freopen ("Ex_trigger1.in", "R", stdin);//Freopen (". Out", "w", stdout); for (int t=read (); t--;    Work ()); return 0;}

10.2 Zheng Rui National Day training test 1