題意：給你n,m,k以及一陣列，n個元素，求一個序列使得序列裡最大值減去最小值>=m且<k，求這個序列的最大長度，序列必須連續。

思路：用單調棧維護一個單調遞增序列和一個單調遞減序列，計算兩個單調棧的棧底元素的差值分別跟m，k 作比較得出序列的最大值。詳細看程式碼，註釋很詳細

#include<stdio.h>
#include<algorithm>
using namespace std;
const int N=1e5+5;
int a[N],maxx[N],minn[N],n,m,k;//maxx陣列維護單調遞減序列，存的是下標，棧底元素最大，minn陣列反之
int main()
{
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        //front為頭指標，tail尾指標
        int front1=0,tail1=-1,front2=0,tail2=-1,flag=-1,ans=0;
        for(int i=0;i<n;i++)
        {
            //頭指標小於尾指標且當前元素大於棧頂元素時讓尾指標減一，即讓使棧頂元素出棧，因為要維護一個單調遞減序列
            while(front1<=tail1&&a[i]>a[maxx[tail1]]) tail1--;
            maxx[++tail1]=i;
            while(front2<=tail2&&a[i]<a[minn[tail2]]) tail2--;
            minn[++tail2]=i;
            //兩個棧的棧頂元素，一個是存的是陣列中最大的元素下標一個是陣列中最小的元素下標
            //用最大的減去最小的分別跟m，k比較
            while(a[maxx[front1]]-a[minn[front2]]>k)
            {
                flag=min(maxx[front1],minn[front2]);//選擇更小的下標讓其++
                if(flag==maxx[front1]) front1++;
                if(flag==minn[front2]) front2++;
            }
            if(a[maxx[front1]]-a[minn[front2]]>=m)//記錄答案
                ans=max(ans,i-flag);
        }
        printf("%d\n",ans);
    }
}
 

There is a sequence of integers. Your task is to find the longest subsequence that satisfies the following condition: the difference between the maximum element and the minimum element of the subsequence is no smaller than m and no larger than k.

Input

There are multiple test cases.
For each test case, the first line has three integers, n, m and k. n is the length of the sequence and is in the range [1, 100000]. m and k are in the range [0, 1000000]. The second line has n integers, which are all in the range [0, 1000000].
Proceed to the end of file.

Output

For each test case, print the length of the subsequence on a single line.

Sample Input

5 0 0
1 1 1 1 1
5 0 3
1 2 3 4 5

Sample Output

5
4