BZOJ2086: [Poi2010]blocks

Exercises

Thought to find out just ask for the longest average >k.

The problem of the mean is to subtract k from each number and determine whether a continuous >0 is possible.

Then we find that if I<j and S[i]<s[j], then J is no better than I for l>j.

Then we can maintain a monotonous s[i], and then for each L go two to distinguish its answer.

But that would be t.

Consider monotonicity again.

If l>j, and S[l]>s[i], then our answer is the smallest and l-i, so J must take the smaller than I can update the answer. Then you can sweep the two hands over it.

All right, I admit that I'm a hu ...

Code:

1#include <cstdio>2#include <cstdlib>3#include <cmath>4#include <cstring>5#include <algorithm>6#include <iostream>7#include <vector>8#include <map>9#include <Set>Ten#include <queue> One#include <string> A #defineINF 1000000000 - #defineMAXN 1000000+5 - #defineMAXM 1000000+5 the #defineEPS 1e-10 - #definell Long Long - #definePA pair<int,int> - #defineFor0 (i,n) for (int i=0;i<= (n); i++) + #defineFor1 (i,n) for (int i=1;i<= (n); i++) - #defineFor2 (i,x,y) for (int i= (x); i<= (y); i++) + #defineFor3 (i,x,y) for (int i= (x); i>= (y); i--) A #defineFor4 (i,x) for (int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go) at #defineMoD 1000000007 - using namespacestd; - inline ll read () - { -ll x=0, f=1;CharCh=GetChar (); -      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} in      while(ch>='0'&&ch<='9') {x=Ten*x+ch-'0'; ch=GetChar ();} -     returnx*F; to } + intN,M,Q[MAXN]; - ll A[MAXN],B[MAXN]; the intMain () * { $Freopen ("Input.txt","R", stdin);Panax NotoginsengFreopen ("output.txt","W", stdout); -N=read (); m=read (); theFor1 (i,n) a[i]=read (); +      while(m--) A     { thell X=read ();intR=0; +Q[r=1]=b[0]=0; - For1 (i,n) $         { $b[i]=b[i-1]+a[i]-x; -           if(B[i]<b[q[r]]) q[++r]=i; -         } the         intans=0; -          for(intI=n;i>0&&r;i--)Wuyi         { the              while(R&&b[i]>=b[q[r]]) r--; -r++; Wu             if(B[i]>=b[q[r]]) Ans=max (ans,i-q[r]); -         } Aboutprintf"%d", ans);if(m) printf (" ");Elseprintf"\ n"); $     } -     return 0; -}

View Code 2086: [poi2010]blocks time limit:20 Sec Memory limit:259 MB
submit:199 solved:91
[Submit] [Status] Description

Given N positive integers A[1..N], and then given a positive integer k, you can now do the following: Each time you select a positive integer greater than K a[i], a[i] minus 1, select a[i-1] or a[i+1] one plus 1. After a certain number of operations, ask the maximum can be selected a continuous sub-sequence, so that each number of this sub-sequence is not less than K.
Give a total of M inquiry, each asked the K different, you need to answer separately.

Input

The first line is two positive integers n (n <= 1,000,000) and M (M <= 50).
The second row n positive integers, and the first positive integers represent a[i] (A[i] <= 10^9).
The third row of M positive integers, and the first positive integers denote the K (k <= 10^9) of the I inquiry.

Output

A total of one row, output m positive integer, number I indicates the answer to the first question.

Sample Input5 6
1 2 1) 1 5
1 2 3 4 5 6




Sample Output5 5 2 1 1 0

BZOJ2086: [Poi2010]blocks