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CF1039E Summer Oenothera Exhibition

阿新 • • 發佈：2018-12-11

Summer Oenothera Exhibition

While some people enjoy spending their time solving programming contests, Dina prefers taking beautiful pictures. As soon as Byteland Botanical Garden announced Summer Oenothera Exhibition she decided to test her new camera there.

The exhibition consists of l=10100l=10^{100}

0100 Oenothera species arranged in a row and consecutively numbered with integers from 00 to l−1l−1. Camera lens allows to take a photo of ww species on it, i.e. Dina can take a photo containing flowers with indices from xx to x+w−1x+w−1 for some integer xx between 00 and l−wl−w. We will denote such photo with [
x,x+w−1][x,x+w−1].

She has taken nn photos, the ii-th of which (in chronological order) is [xi,xi+w−1][x_i,x_i+w−1] in our notation. She decided to build a time-lapse video from these photos once she discovered that Oenothera blossoms open in the evening.

Dina takes each photo and truncates it, leaving its segment containing exactly k

k flowers, then she composes a video of these photos keeping their original order and voilà, a beautiful artwork has been created!

A scene is a contiguous sequence of photos such that the set of flowers on them is the same. The change between two scenes is called a cut. For example, consider the first photo contains flowers [1,5][1,5], the second photo contains flowers [3,7][3,7] and the third photo contains flowers [8,12][8,12]. If k=3k=3, then Dina can truncate the first and the second photo into [3,5][3,5], and the third photo into [9,11][9,11]. First two photos form a scene, third photo also forms a scene and the transition between these two scenes which happens between the second and the third photos is a cut. If k=4k=4, then each of the transitions between photos has to be a cut.

Dina wants the number of cuts to be as small as possible. Please help her! Calculate the minimum possible number of cuts for different values of kk.

Input

The first line contains three positive integer n,w,q(1≤n,q≤100000,1≤w≤109)n, w, q (1≤n,q≤100000, 1≤w≤10^9) — the number of taken photos, the number of flowers on a single photo and the number of queries.

Next line contains nn non-negative integers xi(0≤xi≤109)x_i (0≤x_i≤10^9) — the indices of the leftmost flowers on each of the photos.

Next line contains qq positive integers ki(1≤ki≤w)k_i (1≤k_i≤w) — the values of kk for which you have to solve the problem.

It’s guaranteed that all kik_i are distinct.

Output

Print qq integers — for each width of the truncated photo kik_i, the minimum number of cuts that is possible.

Examples
input

3 6 5 2 4 0 1 2 3 4 5

output

0 0 1 1 2

input

6 4 3 1 2 3 4 3 2 1 2 3

output

0 1 2

題目大意

直接看英文是真的不知道在說個什麼**玩意兒，但是修修告訴我題意是這樣的：

給定長度為nn的序列，常數ww，詢問次數qq。

接下來qq組詢問，每次給出一個常數kik_i，求最少把序列分為多少段使得每段序列中數的極差不超過w−kiw-k_i，輸出最小的段數−1-1。

題解

讓我們先考慮一下單次詢問怎麼做：直接貪心，每次儘量往後跳，直到極差不滿足條件為止，這樣我們就O(n)O(n)解決了一次詢問。

暴力跳太low\mathcal{low}了，不如我們倍增吧，先處理出最大/最小值的ST\mathcal{ST}表，我們就可以愉快的倍增往後跳了。

線上一看就很不可做的樣子，那就把所有詢問單增離線吧，離線之後，跳的距離也會單增，考慮維護每個點向後能跳到的最遠點nxt[v]nxt[v]。

暴力一點的做法：對於一個新的極差rr，我們把所有點都嘗試往後挪來更新nxt[v]nxt[v]，然後從頭開始往後沿著nxt[v]nxt[v]跳。

感覺這個操作修改和查詢都很爆炸，自然想到分塊一下，塊內路徑壓縮，更新路徑也僅限於塊內，塊之間跳用倍增來找。

複雜度？大概是O(n32log⁡2n)O(n^{\frac{3}{2}}\log_2n)的吧。

據說修修有個O(O(n53+n43log⁡n))O(O(n^{\frac{5}{3}} + n^{\frac{4}{3}}\log n))的優秀做法，然而我DZYO的O(n32log⁡2n)O(n^{\frac{3}{2}}\log_2n)全CFCF最快，一手LCT\mathcal{LCT}彈飛綿羊代替路徑壓縮跑的飛快，吊錘所有演算法。

程式碼

ST\mathcal{ST}表表示2i2^i的維度一定要開在前面，這樣就能訪問連續記憶體，這大概就是AC\mathcal{AC}與TLE\mathcal{TLE}的距離吧。

#include<bits/stdc++.h>
using namespace std;
const int M=1e5+5;
struct sd{int id,r;}ask[M<<6];
bool operator <(sd a,sd b){return a.r==b.r?a.id>b.id:a.r<b.r;}
int st[20][M][2],que[M],nxt[M],blo[M][2],jmp[M][2],ans[M],n,w,q,siz,tot;
void in()
{
	scanf("%d%d%d",&n,&w,&q);
	for(int i=1;i<=n;++i)scanf("%d",&que[i]),st[0][i][0]=st[0][i][1]=que[i];
	for(int i=1,k;i<=q;++i)scanf("%d",&k),ask[++tot]=(sd){i-q,w-k};
}
void up(int v){(nxt[v]==n+1||(nxt[v]-1)%siz==0)?(jmp[v][0]=1,jmp[v][1]=v):(jmp[v][0]=jmp[nxt[v]][0]+1,jmp[v][1]=jmp[nxt[v]][1]);}
void ac()
{
	for(;siz*siz*siz<6*n;++siz);
	for(int j=1;(1<<j)<=n;++j)for(int i=1;i+(1<<j)<=n+1;++i)
	st[j][i][0]=min(st[j-1][i][0],st[j-1][i+(1<<j-1)][0]),
	st[j][i][1]=max(st[j-1][i][1],st[j-1][i+(1<<j-1)][1]);
	for(int i=n;i>=1;--i)
	{
		nxt[i]=i+1,blo[i][0]=blo[i][1]=que[i],up(i);
		for(int j=i+1,mn=que[i],mx=que[i];j<=n&&(j-1)%siz>0;++j)mn=min(mn,que[j]),mx=max(mx,que[j]),ask[++tot]=(sd){i,mx-mn};
	}
	sort(ask+1,ask+1+tot);
	for(int i=1,pos,cst,mn,mx,k,to,maxn,minn;i<=tot;++i)
	if(ask[i].id<=0)for(pos=1,cst=-1;pos<=n;ans[ask[i].id+q]=cst)
	{
		cst+=jmp[pos][0],pos=jmp[pos][1],mn=blo[pos][0],mx=blo[pos][1],to=pos,k=0;
		for(;pos+(1<<k)<=n+1&&(maxn=max(mx,st[k][pos][1]))-(minn=min(mn,st[k][pos][0]))<=ask[i].r;++k)mx=maxn,mn=minn;
		for(;k--;)if(to+(1<<k)<=n+1&&(maxn=max(mx,st[k][to
            
  

           

              
              

            

            
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