Functions again CodeForces - 789C

最新推荐文章于 2022-05-09 01:15:19 发布

原创最新推荐文章于 2022-05-09 01:15:19 发布 · 257 阅读

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本文介绍了一种解决特定数学问题的方法——即寻找一对(l, r)，求解给定公式中F的最大值。该问题通过将数组分为奇数起始和偶数起始两种情况来解决，并采用动态规划的思想来高效地计算结果。

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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.

题意：
给定了一个公式，让你找到一对（l,r）,求解出公式给定的F值
思路：
这道题是看了题解才懂的，它分为以奇数起始和偶数起始两类；
如果以奇数起始，则偶数项为原来的相反数。处理以后求一个maxx；
反之，以偶数起始，则奇数项为原来的相反数。处理一次再求一个maxx；
求两者的较大值即可。


#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
typedef long long ll;
ll a[100010],b[100010];
int n;
int main()
{
    scanf("%d",&n);
    for(ll i=1; i<=n; i++)
        scanf("%lld",&a[i]);
    for(ll i=1; i<n; i++)
        b[i]=abs(a[i]-a[i+1]);
    ll ans=0;
    for(ll i=1; i<n; i++)
    {
        if(i%2)
            b[i]=-b[i];//区间是从偶数位置的数字开始的
    }
    ll maxx,maxn;
    maxx=maxn=b[1];
    for(ll i=2; i<n; i++)
    {
        if(maxn<=0)
            maxn=b[i];
        else
            maxn+=b[i];
        maxx=max(maxx,maxn);
    }
    ans=maxx;
    for(ll i=1; i<n; i++)
        b[i]=-b[i];//区间是从奇数位置的数字开始的
    maxx=maxn=b[1];
    for(ll i=2; i<n; i++)
    {
        if(maxn<=0)//若现有的和为负的
            maxn=b[i];//则当前最大和为元素本身
        else
            maxn+=b[i];
        maxx=max(maxx,maxn);
    }
    ans=max(ans,maxx);
    printf("%lld\n",ans);
    return 0;

}