1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 #define ll long long
 5 ll k,n,a[3000005],q_mx[3000005],q_mn[3000005];
 6 ll head_mx,head_mn,tail_mx,tail_mn,len,pre;
 7 int main(){
 8 //    freopen("1.in","r",stdin);
 9     scanf("%lld%lld",&k,&n);
10     len=0;
11     for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
12     q_mx[1]=1;q_mn[1]=1;pre=1;
13     head_mx=1;tail_mx=1;head_mn=1;tail_mn=1;
14     for(int i=2;i<=n;i++){
15         while(head_mx<=tail_mx&&a[q_mx[tail_mx]]<a[i])tail_mx--;
16         while(head_mn<=tail_mn&&a[q_mn[tail_mn]]>a[i])tail_mn--;
17         q_mx[++tail_mx]=i;q_mn[++tail_mn]=i;
18         while(a[q_mx[head_mx]]-a[q_mn[head_mn]]>k){
19             if(q_mx[head_mx]<q_mn[head_mn]){pre=q_mx[head_mx]+1;head_mx++;}
20             else {pre=q_mn[head_mn]+1;head_mn++;}
21         }
22         //pre=min(q_mx[head_mx],q_mn[head_mn]);
23         len=max(len,i-pre+1);
24     }
25     printf("%lld",len);
26 }

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<deque>
 6 using namespace std;
 7 const int MAXN=3000010;
 8 inline void read(int &n)
 9 {
10     char c=getchar();n=0;bool flag=0;
11     while(c<'0'||c>'9')    c=='-'?flag=1,c=getchar():c=getchar();
12     while(c>='0'&&c<='9')    n=n*10+c-48,c=getchar();flag==1?n=-n:n=n;
13 }
14 struct node
15 {
16     int pos,val;
17     node(){pos=val=0;}
18     node(int a,int b)    {    pos=a;val=b;    }
19 };
20 int a[MAXN];
21 int n,k;
22 int check(int mid)//按顺序的性质二分
23 {
24     deque<node>qmax;// 升
25     deque<node>qmin;// 降
26     for(int i=1;i<=n;i++)
27     {
28         while(qmax.size()>0&&qmax.front().pos<=(i-mid))    qmax.pop_front();
29         while(qmin.size()>0&&qmin.front().pos<=(i-mid))    qmin.pop_front();
30         while(qmax.size()>0&&a[i]>qmax.back().val)    qmax.pop_back();
31         qmax.push_back(node(i,a[i]));
32         while(qmin.size()>0&&a[i]<qmin.back().val)    qmin.pop_back();
33         qmin.push_back(node(i,a[i]));
34     //    printf("%d %d\n",qmax.front().val,qmin.front().val);
35         if(i>=mid&&qmax.front().val-qmin.front().val<=k)
36             return 1;
37     }
38     return 0;
39 }
40 int main()
41 {
42     read(k);read(n);
43     for(int i=1;i<=n;i++)    read(a[i]);
44     int l=1,r=n;
45     int ans=0;
46     while(l<=r)
47     {
48         int mid=l+r>>1;
49         if(check(mid))    ans=mid,l=mid+1;
50         else r=mid-1;
51     }
52     printf("%d",ans);
53     return 0;
54 }