Functions again

最新推荐文章于 2020-12-29 17:35:23 发布

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最新推荐文章于 2020-12-29 17:35:23 发布

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分类专栏：动态规划

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本文介绍了一个关于序列处理的问题，即通过计算序列中相邻元素的差的绝对值来构造新的序列，并从中选出一段使特定组合的和达到最大值的方法。文章提供了完整的代码实现，采用动态规划策略解决该问题。

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题目：

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.

题意：

题意就是给一个序列，然后另一个序列等于这个序列相邻两个数两两之差的绝对值。你可以在第二个序列中随意选取一段，规定这段第一个数是正数，第二个数是负数，求出这一段的和，你要选取一段和最大的，输出最大值。

思路：

和求序列的最大子段和相似，

先把序列相邻两个数两两之差的绝对值算出来存下来，然后利用dp计算；

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,a[100010];
long long sum[100010];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<n;i++)
        sum[i]=abs(a[i]-a[i+1]);
    long long maxx=0,num=0,k=1,t=0;
    for(int i=1;i<n;i++)
    {
        if(t%2==0)
        {
            t++;
            num=num+sum[i];
        }
        else
        {
            t++;
            num=num-sum[i];
        }
        if(num>maxx)
            maxx=num;
        if(num<0)
        {
            num=0;
            t=0;
            i=k++;
        }
    }
    printf("%lld\n",maxx);
    return 0;
}