codeforces 601B Lipshitz Sequence

题目链接

B. Lipshitz Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A function is called Lipschitz continuous if there is a real constant K such 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 l_i and r_i (1 ≤ l_i < r_i ≤ 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 .

Sample test(s)

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.

题解：把i看成横坐标，h[i]看成纵坐标，那么L(h)其实就是一些点中两点之间斜率绝对值的最大值。那么，可以证明L(h)一定是坐标相邻的点的斜率的绝对值。那么L(h)=max(abs(a[i+1]-a[i])),1<=i<n。

询问只有100次，那么对于每次询问我们可以用O(n)的方法求解。

令D[i]=abs(a[i]-a[i-1])

对于每次询问，从L到R进行一次枚举，每次枚举到i，求解以i问结尾的所有区间的L值之和。在枚举的过程中维护一个单调栈，栈内元素下标i递增，D[i]递减，即可统计我们需要的值。详情见代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
typedef __int64 LL;
typedef unsigned __int64 LLU;
const int nn=110000;
const int inf=0x3fffffff;
const LL inf64=(LL)inf*inf;
const double eps = 1e-12;
const int mod = 1000000007;
using namespace std;
int n,q;
int a[nn];
int l,r;
LL tem;
pair<int,int>que[nn];
int tou,wei;
void add(int id,int val)
{
    while(wei-2>=tou)
    {
        if(val>=que[wei-1].second)
        {
            tem-=1LL*que[wei-1].second*(que[wei-1].first-que[wei-2].first);
            wei--;
        }
        else
            break;
    }
    que[wei++]=make_pair(id,val);
    tem+=1LL*val*(id-que[wei-2].first);
}
int main()
{
    int i;
    while(scanf("%d%d",&n,&q)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        while(q--)
        {
            scanf("%d%d",&l,&r);
            LL ans=0;
            tem=0;
            tou=wei=0;
            que[wei++]=make_pair(l,inf);
            for(i=l+1;i<=r;i++)
            {
                add(i,abs(a[i]-a[i-1]));
                ans+=tem;
            }
            printf("%I64d\n",ans);
        }
    }
    return 0;
}