题干：
A function is called Lipschitz continuous if there is a real constant Ksuch that the inequality |f(x) - f(y)| ≤ K·|x - y| holds for all
. We'll deal with a more... discrete version of this term.
For an array , we define it's Lipschitz constant
as follows:
- if n < 2,
- if n ≥ 2,
over all 1 ≤ i < j ≤ n
In other words, is the smallest non-negative integer such that |h[i] - h[j]| ≤ L·|i - j| holds for all 1 ≤ i, j ≤ n.
You are given an array of size n and q queries of the form [l, r]. For each query, consider the subarray
; determine the sum of Lipschitz constants of all subarrays of
.
Input
The first line of the input contains two space-separated integers n and q (2 ≤ n ≤ 100 000 and 1 ≤ q ≤ 100) — the number of elements in array and the number of queries respectively.
The second line contains n space-separated integers (
).
The following q lines describe queries. The i-th of those lines contains two space-separated integers li and ri (1 ≤ li < ri ≤ n).
Output
Print the answers to all queries in the order in which they are given in the input. For the i-th query, print one line containing a single integer — the sum of Lipschitz constants of all subarrays of .
Examples
Input
10 4 1 5 2 9 1 3 4 2 1 7 2 4 3 8 7 10 1 9
Output
17 82 23 210
Input
7 6 5 7 7 4 6 6 2 1 2 2 3 2 6 1 7 4 7 3 5
Output
2 0 22 59 16 8
Note
In the first query of the first sample, the Lipschitz constants of subarrays of with length at least 2 are:
The answer to the query is their sum.
题目大意：
给定n(n<=1e5)个整数的数组h[i],现在定义一个函数
L(h)的值是区间[L,R]内，abs ( h[i]-h[j] ) / ( i-j )的最大值。
现在有q个询问，每个询问表示询问区间[L,R]内，所有连续子序列的L(h)的值的和。
解题报告：
首先分析L（h）函数，从定义上看就是斜率的绝对值的最大值。
一下一段来自题解：
观察到是斜率的最大值，所以能够很容易考虑到有决策单调性。（比如位子i所选择的最优的位子是j，那么对于位子i+1来讲，肯定选取的位子是大于等于j的）；而又考虑是两个值的差除以距离差，那么我们很容易考虑到问题选择的单调性还是相邻的（位子i所选取的最优一定是i-1）；
（但是我感觉并没有决策单调性呀、、、）
所以讲斜率弄成一个新数组，然后单调栈乱搞一下就行了。去重方法和51nod那个题一样。。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int L[MAX],R[MAX];
int n,q;
int a[MAX],b[MAX];
stack<int> sk;
ll ans;
int main()
{cin>>n>>q;for(int i = 1; i<=n; i++) cin>>a[i];for(int i = 1; i<n; i++) b[i] = abs(a[i+1] - a[i]);n--;for(int i = 1; i<=n; i++) {while(!sk.empty() && b[sk.top()] <= b[i]) sk.pop();L[i] = sk.empty() ? 0 : sk.top();sk.push(i);}while(sk.size()) sk.pop();for(int i = n; i>=1; i--) {while(!sk.empty() && b[sk.top()] < b[i]) sk.pop();R[i] = sk.empty() ? n+1 : sk.top();sk.push(i);}while(q--) {ans=0;int l,r,LL,RR;scanf("%d%d",&l,&r);for(int i = l; i<=r-1; i++) {LL=max(L[i],l-1);RR=min(R[i],r);ans += 1LL*b[i]*(RR-i)*(i-LL);//b[i] * (RR - LL - 1);}printf("%lld\n",ans);}return 0 ;
}
