The nod 1628 non-bourbon tree

Original title Link: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1628

It took a morning + half an afternoon to finally cut off the problem ...

God problem (film title)

We consider the general formula of the Fibonacci sequence:

$$ F (i) =[(\frac{1+\sqrt{5}}{2}) ^{i}-(\frac{1-\sqrt{5}}{2}) ^{i}]*\frac{1}{\sqrt{5}} $$

The last $ \frac{1}{\sqrt{5}} $ we can take it out and throw it back in the answer.

Then consider separating the previous two items separately.

Set $g (i) = (\frac{1+\sqrt{5}}{2}) ^{i}$ then it can be found that $G (a+b) =g (a) *g (b) $ with this nature, we can count things and multiply them.

Specifically, pre-$2^{i}$ from a step up, the point on the chain from the bottom to go up can get the contribution, from the top down the same, and the chain within the point 22 of the contribution of the answer (in fact, the previous statistics from the top down only for this part of the service).

For each group of inquiries, the first to find the two-point LCA, and then the two points to the LCA chain of the answer, and then merge together, the merger process has many details, see the code.

Lucky #

#include <cstdio>#include<algorithm>#defineMN 100001using namespacestd;intRead_p,read_ca;inlineintRead () {read_p=0; read_ca=GetChar ();  while(read_ca<'0'|| Read_ca>'9') read_ca=GetChar ();  while(read_ca>='0'&&read_ca<='9') read_p=read_p*Ten+read_ca- -, read_ca=GetChar (); returnread_p;}Const ints1=691504013, s2=308495997, mod=1e9+9, s3=276601605;structna{intY,ne;} b[mn<<1];intn,m,x,y,f[mn][ +],g[mn][ +],g[mn][ +],a[mn][ +],k[mn][ +],l[mn],num=0, s[2][mn],x[mn],y[mn],z[mn],s[mn],de[mn],mmh[mn],lx[mn],ly[mn],ap[mn];inlinevoid inch(intXintY) {b[++num].y=y;b[num].ne=l[x];l[x]=num;} InlineintMintx) { while(X>=mod) X-=mod; while(x<0) X+=mod;returnx;}voidDFS (intx) {S[x]=1; Registerinti;  for(i=1; i<= -; i++)if(f[f[x][i-1]][i-1]) f[x][i]=f[f[x][i-1]][i-1];Else  Break;  for(i=1; i<= -; i++)if(f[k[x][i-1]][i]) k[x][i]=f[k[x][i-1]][i];Else  Break;  for(; i;i--) k[x][i]=k[x][i-1]; k[x][0]=x;  for(i=l[x];i;i=b[i].ne)if(b[i].y!=f[x][0]) {DE[B[I].Y]=de[k[b[i].y][0]=f[b[i].y][0]=x]+1;        DFS (B[I].Y); AP[X]=m (ap[x]+1ll*s[x]*s[b[i].y]%MOD); S[X]+=S[B[I].Y]; }}inlineintLcaintXintYint&a,int&b) {     for(RegisterintI= -; i>=0; i--)    if(De[f[x][i]]>=de[y]) x=F[x][i]; if(x==y)returnx;  for(RegisterintI= -; i>=0; i--)    if(F[x][i]!=f[y][i]) x=f[x][i],y=F[y][i]; A=x;b=y; returnf[x][0];}voidDfsintXinto) {a[x][0]=1ll*ap[x]*s[o][1]%MOD; g[x][0]=g[x][0]=1ll*s[x]*s[o][1]%MOD;  for(RegisterintI=1; i<= -; i++)    if(K[x][i]) g[x][i]=m ((1ll* (g[x][i-1]-s[k[x][i-1]]) *s[o][1<< (I-1)]+g[f[x][i-1]][i-1])%MOD), G[x][i]=m ((1ll* (g[f[x][i-1]][i-1]-g[k[x][i-1]][0]) *s[o][1<< (I-1)]+g[x][i-1])%MOD), A[x][i]=m ((1ll*g[x][i-1]* (g[f[x][i-1]][i-1]-g[k[x][i-1]][0]) +a[x][i-1]+a[f[x][i-1]][i-1]-1ll*s[k[x][i-1]]*m (g[f[x][i-1]][i-1]-g[k[x][i-1]][0]))%MOD);  for(RegisterintI=l[x];i;i=b[i].ne)if(b[i].y!=f[x][0]) DFS (b[i].y,o);} InlinevoidWorkintXintZint&_a,int&_g,into) {    intp=0; _a=0; _g=0;  for(RegisterintI= -; i>=0; i--)    if(de[k[x][i]]>=De[z]) {_a=m ((1ll*_g* (g[x][i]-g[p][0]) +_a+a[x][i]-1ll*s[p]*m (g[x][i]-g[p][0]))%MOD); _g=m ((1ll* (_g-s[p]) *s[o][1<<i]+g[x][i])%MOD); if(k[x][i]==z)return; P=K[x][i]; X=F[x][i]; }}inlinevoidCalcintx) {    intAx,ay,gx,gy; Registerinti; DFS (1, x);  for(i=1; i<=m;i++)    if(y[i]==Z[i])        {work (x[i],z[i],ax,gx,x); Mmh[i]=m (mmh[i]+ (X?-1LL:1LL) * (1ll*gx* (N-s[z[i]]) +ax)%MOD); }Else{work (x[i],lx[i],ax,gx,x), work (Y[I],LY[I],AY,GY,X); Mmh[i]=m (mmh[i]+ (X?-1LL:1LL) * (1ll* (gx+gy) * (N-s[lx[i]]-s[ly[i]])%mod*s[x][1]+1ll*gx*gy%mod*s[x][1]+ax+ay+a[z[i]][0]-(1ll*s[lx[i]]* (S[z[i]]-s[lx[i]) +1ll*s[ly[i]]* (S[z[i]]-s[lx[i]]-s[ly[i]))%mod*s[x][1]+ (1ll* (S[z[i]]-s[lx[i]]-s[ly[i]) * (N-s[z[i]])%mod*s[x][1])%MOD); }}intMain () {registerinti; N=read ();  for(i=1; i<n;i++) X=read (), Y=read (),inch(x, Y),inch(Y,X); s[0][0]=s[1][0]=1;  for(i=1; i<=n;i++) s[0][i]=1ll*s[0][i-1]*s1%mod,s[1][i]=1ll*s[1][i-1]*s2%MOD; de[1]=1;D FS (1); M=read ();  for(i=1; i<=m;i++) {X[i]=read (); y[i]=read (); if(de[x[i]]<De[y[i]) swap (x[i],y[i]); Z[i]=LCA (X[i],y[i],lx[i],ly[i]); } Calc (0); Calc (1);  for(i=1; i<=m;i++) printf ("%d\n", 1ll*mmh[i]*s3%MOD);}

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The nod 1628 non-bourbon tree