Good subsequence( RMQ+二分)

Description

Give you a sequence of n numbers, and a number k you should find the max length of Good subsequence. Good subsequence is a continuous subsequence of the given sequence and its maximum value - minimum value<=k. For example n=5, k=2, the sequence ={5, 4, 2, 3, 1}. The answer is 3, the good subsequence are {4, 2, 3} or {2, 3, 1}.

Input

There are several test cases.
Each test case contains two line. the first line are two numbers
indicates n and k (1<=n<=10,000, 1<=k<=1,000,000,000). The
second line give the sequence of n numbers a[i] (1<=i<=n,
1<=a[i]<=1,000,000,000).
The input will finish with the end of file.

Output

For each the case, output one integer indicates the answer.

Sample Input

5 2
5 4 2 3 1
1 1
1

Sample Output

3
1
 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <sstream>
 #include <iomanip>
 using namespace std;
 const int INF=0x4fffffff;
 const int EXP=1e-;
 const int MS=;
 const int MS2=;
 
 int a[MS];
 int n,k;
 
 int minv[MS][];
 int maxv[MS][];
 
 void RMQ_init()
 {
       for(int i=;i<n;i++)
       {
             minv[i][]=a[i];
             maxv[i][]=a[i];
       }
 
       for(int j=;(<<j)<=n;j++)
       {
             for(int i=;i+(<<j)-<n;i++)
             {
                   minv[i][j]=min(minv[i][j-],minv[i+(<<(j-))][j-]);
                   maxv[i][j]=max(maxv[i][j-],maxv[i+(<<(j-))][j-]);
             }
       }
 }
 
 int RMQ(int l,int r)
 {
       int k=;
       while(<<(k+)<=r-l+)
             k++;
       return   max(maxv[l][k],maxv[r-(<<k)+][k])-min(minv[l][k],minv[r-(<<k)+][k]);
 }
 
 int main()
 {
 
       while(scanf("%d%d",&n,&k)!=EOF)
       {
             for(int i=;i<n;i++)
                   scanf("%d",&a[i]);
             RMQ_init();
             int ans=;
             for(int i=;i<n;i++)
             {
                   int l=i;
                   int r=n-;
                   while(l<=r)
                   {
                         int mid=(l+r)/;
                         int t=RMQ(i,mid);
                         if(t>k)
                               r=mid-;
                         else
                               l=mid+;
                   }
                   if(ans<l-i)
                         ans=l-i;
             }
             cout<<ans<<endl;
       }
       return ;
 }