Bzoj 4408: [Fjoi 2016] Mystery number can persist line tree, divine question

4408: [Fjoi 2016] Mystery number time limit:10 Sec Memory limit:128 MB
submit:177 solved:128
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The mysterious number of a repeatable number set S is defined as the smallest positive integer that cannot be represented by a subset of S. For example, s={1,1,1,4,13},

1 = 1

2 =

3 = 1+1+1

4 = 4

5 = 4+1

6 = 4+1+1

7 = 4+1+1+1

8 cannot be represented as a subset of the set S and so the mystery number of the set S is 8.

Now given n positive integers a[1]. A[n],m a query, each time asking for a given interval [l,r] (l<=r), the mysterious number of a repeatable number set consisting of a[l],a[l+1],..., A[r].

Input

The first line is an integer n, which indicates the number of digits.
The second row of n integers, numbered from 1.
The third line, an integer m, indicates the number of queries.
The following m lines, each with a pair of integer l,r, represent a query.

Output

For each query, the output line corresponds to the answer.

Sample Input5
1 2 4) 9 10
5
1 1
1 2
1 3
1 4
1 5Sample Output2
4
8
8
8HINT

For 100% data points, n,m <= 100000,∑a[i] <= 10^9

Source

Acknowledgement Yyh Upload

Exercises

FJ God problem!!!

This problem is really psychedelic ...

I can't believe it!!! qaq!!! I've been thinking about mathematical methods ...

See the puzzle, unexpectedly is a durable line tree!!!!! (Orz the person)

All right, no more nonsense ...

Sort the sequence from small to large first.

Assuming that the current mystery number is ans, [1,ans-1] must be represented by the number of s in the set. Then, if a number A is currently added, it can be divided into two types of discussion.

1, if A<=ans, the interval can now be extended to: [1,ans+a-1], and then the mysterious number into ans+a.

2, if A>ans, the interval is empty, ans does not change.

Then ans starts from 1, each time seeking a number less than ans and get,ans into get+1.

This can be maintained with a durable line tree.

1#include <bits/stdc++.h>2 using namespacestd;3 #defineMAXN 1000104 intsum[ +*maxn],a[maxn],root[maxn],size;5 structnode6 {7     intLeft,right;8}tree[ +*MAXN];9 intRead ()Ten { One     ints=0, fh=1;CharCh=GetChar (); A      while(ch<'0'|| Ch>'9'){if(ch=='-') fh=-1; ch=GetChar ();} -      while(ch>='0'&&ch<='9') {s=s*Ten+ (ch-'0'); ch=GetChar ();} -     returns*fh; the } - voidADD (intXint&y,intLintRintadd) - { -y=++SIZE; +sum[y]=sum[x]+add; -     if(L==R)return; +tree[y].left=tree[x].left;tree[y].right=Tree[x].right; A     intMid= (L+R)/2; at     if(add<=mid) ADD (tree[x].left,tree[y].left,l,mid,add); -     ElseADD (tree[x].right,tree[y].right,mid+1, R,add); - } - intQueryintXintYintLintRintk) - { -     if(L==R)returnsum[y]-Sum[x]; in     intMid= (L+R)/2; -     if(K<=mid)returnquery (tree[x].left,tree[y].left,l,mid,k); to     Else returnQuery (tree[x].right,tree[y].right,mid+1, r,k) +sum[tree[y].left]-Sum[tree[x].left]; + } - intMain () the { *     intN,tot,i,m,l,r,ans,Get; $n=read ();Panax Notoginsengtot=0; -      for(i=1; i<=n;i++) A[i]=read (), tot+=A[i]; theSize=0; +      for(i=1; i<=n;i++) Add (root[i-1],root[i],1, Tot,a[i]); Am=read (); the      for(i=1; i<=m;i++) +     { -L=read (); r=read (); $ans=1; $          while(1) -         { -             Get=query (root[l-1],root[r],1, Tot,ans); the             if(Get<ans) Break; -ans=Get+1;Wuyi         } theprintf"%d\n", ans); -     } Wu     return 0; -}

Bzoj 4408: [Fjoi 2016] Mystery number can persist line tree, divine question