Codeforces Round #407 (Div. 2) C. Functions again_codeforces round #407 functions again-优快云博客

探讨了如何通过计算特定数组的Uzhlyandian函数最大值来解决数学难题，运用了最大连续子序列和算法。

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Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:

$\text{[math]}$

In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.

Input

The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.

The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.

Output

Print the only integer — the maximum value of f.

Examples

input
5
1 4 2 3 1

output
3

input
4
1 5 4 7

output
6

Note

In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].

In the second case maximal value of f is reachable only on the whole array.

题意：给你n个数，然后求f（l，r）的最大值，f（l，r）的公式在题目里。。。

思路：其实基本上就是裸的最大连续子序列和，首先先把数组相邻的差值求出来，然后很明显就只有两种情况，正负正负正负和负正负正负正，把两种的最大值都算一遍取最大就好了。下面给代码：

#include<iostream>  
#include<cmath>  
#include<queue>  
#include<cstdio>  
#include<queue>  
#include<algorithm>  
#include<cstring>  
#include<string>  
#include<utility>
#include<map>
#include<vector>
#define maxn 100005
#define inf 0x3f3f3f3f  
using namespace std;
typedef long long LL;
const double eps = 1e-8;
LL a[maxn], b[maxn], c[maxn];
int main(){
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++){
		scanf("%lld", &a[i]);
	}
	for (int i = 0; i < n - 1; i++){
		b[i] = abs(a[i] - a[i + 1]);
		c[i] = b[i];
		if (i & 1)
			b[i] = -b[i];
		else
			c[i] = -c[i];
	}
	LL ans = max(b[0], c[0]);
	for (int i = 1; i < n - 1; i++){
		if (b[i] + b[i - 1] > b[i])
			b[i] += b[i - 1];
		if (c[i] + c[i - 1]>c[i])
			c[i] += c[i - 1];
		ans = max(ans, b[i]);
		ans = max(ans, c[i]);
	}
	printf("%lld\n", ans);
}