Hdu_5919_sequence II (Chairman tree)

Title Link: hdu_5919_sequence II

Test instructions

Give you the number of N, M inquiry, each time you ask each of the number of intervals in the interval of the first occurrence of the position of the median, forced online.

Exercises

A look is the chairman of the tree, but here to ask the first occurrence of the position, there is a technique is to pour the number in the back, if there are the same number, the position of the number of contributions canceled, so the query only need to query root[l], because Root[l] includes the l to n information, The first position of the same number has been reserved previously, so it is OK to query Root[l] 's L to R directly when querying a total number of numbers.

1#include <bits/stdc++.h>2 #defineF (I,A,B) for (int i=a;i<=b;i++)3 using namespacestd;4 5 Const intn=2e5+7;6 structnode{intL,r,sum;} t[ +*N];7 8 intT,n,m,a[n],pre[n],root[n],tot,ic=1;9 Ten voidUpdateint&x,intYintPosintValintL=1,intR=N) One { AT[++tot]=x? t[x]:t[y],t[tot].sum+=val,x=tot; -     if(L==R)return; -     intM=l+r>>1; the     if(pos<=m) update (T[X].L,T[Y].L,POS,VAL,L,M); -     ElseUpdate (t[x].r,t[y].r,pos,val,m+1, R); - } -  + intQueryintXintKintL=1,intR=N) - { +     if(L==R)returnl; A     intM=l+r>>1; at     if(k<=t[t[x].l].sum)returnquery (t[x].l,k,l,m); -     returnQuery (t[x].r,k-t[t[x].l].sum,m+1, R); - } -  - intGetsum (intXintLintRintL=1,intR=N) - { in     if(L&LT;=L&AMP;&AMP;R&LT;=R)returnt[x].sum; -     intM=l+r>>1, ans=0; to     if(l<=m) ans+=getsum (t[x].l,l,r,l,m); +     if(r>m) Ans+=getsum (t[x].r,l,r,m+1, R); -     returnans; the } *  $ intMain ()Panax Notoginseng { -scanf"%d",&t); the      while(t--) +     { Aprintf"Case #%d:", ic++); thescanf"%d%d",&n,&m); +tot=0; -F (I,1, N) scanf ("%d", A +i); $          for(intI=n;i>0; i--) $         { -             if(Pre[a[i]]) update (root[i],root[i+1],pre[a[i]],-1); -Update (root[i],root[i+1],i,1); thepre[a[i]]=i; -         }Wuyi         intans=0, L,r,sum; theF (I,1, M) -         { Wuscanf"%d%d",&l,&R); -L= (L+ans)%n+1, r= (R+ans)%n+1; About             if(L>r) l^=r,r^=l,l^=R; $sum=getsum (root[l],l,r); -printf"%d%c", Ans=query (root[l],sum+1>>1),"\ n"[i==m]); -         } -          while(TOT) t[tot].l=t[tot].r=t[tot].sum=0, tot--; AF (I,1, N) root[i]=0, pre[a[i]]=0; +     } the     return 0; -}

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Hdu_5919_sequence II (Chairman tree)