Functions again

最新推荐文章于 2020-03-31 18:21:37 发布

&深藏功与名

最新推荐文章于 2020-03-31 18:21:37 发布

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本文链接：https://blog.youkuaiyun.com/x2535723242/article/details/81841288

本文探讨了一个关于Uzhlyandian函数的最大值问题。该函数定义为两个数组元素之间的差值绝对值之和，从数组的某个位置开始，奇数位置取正值，偶数位置取负值。通过动态规划的方法，文章提供了一种高效的算法来找到给定数组中该函数的最大可能值。

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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:

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

Input

4
1 5 4 7

Output

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.

思路：有关系式可知，它有xian相邻的两项的差，由l开始计数的，奇数项取正，偶数项取负，求和得到的。所以求两段最大和即可。

代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;

int n,a[100005];
long long l,r,ans,dp[100005][2];

int main()
{
	while(~scanf("%d",&n))
	{
		for(int i=1; i<=n; i++)
			scanf("%d", &a[i]);
		dp[1][0] = abs(a[1]-a[2]);
		dp[1][1] = (-1)*abs(a[1]-a[2]);
		l = dp[1][1];
		r = dp[1][0];
		ans = dp[1][0];
		for(int i=2; i<n; i++)
		{
			dp[i][0] = abs(a[i]-a[i+1]);
			dp[i][1] = r-abs(a[i]-a[i+1]);
			dp[i][0] = max(dp[i][0], dp[i][0]+l);
			l = dp[i][1];
			r = dp[i][0];
			ans = max(dp[i][0], ans);
			ans = max(dp[i][1], ans);
		}
		printf("%lld\n", ans);
	}
	return 0;
}