Codeforces788A Functions again 2017-04-12 18:22 56人阅读 评论(0) 收藏

C. Functions again

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

3

input

4
1 5 4 7

output

6

Note

——————————————————————————————————————

题目的意思是选取一段区间，使得满足关系式的值最大

思路：观察关系式我们可以知道，它由相邻两项之差，由某项开始奇数项取正偶数项取负求和而得，我们可以预处理出相邻的差，构造两个序列，一个奇正偶负，一个相反，然后求最大字段和

#include <iostream>
#include<queue>
#include<cstdio>
#include<cmath>
#include<set>
using namespace std;
#define LL long long
LL a[100005];
LL b[100005];
LL c[100005];

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        scanf("%lld",&a[i]);
    }
    for(int i=1; i<n; i++)
    {
        LL x=fabs(a[i]-a[i-1]);
        if(i%2) b[i]=x;
        else b[i]=-x;
        c[i]=-b[i];
    }
    LL s=0,mx=0;
    for(int i=1; i<n; i++)
    {
        if(s+c[i]<0) s=0;
        else s+=c[i];
        if(s>mx) mx=s;
    }
    s=0;
    for(int i=1; i<n; i++)
    {
        if(s+b[i]<0)  s=0;
        else  s+=b[i];
        if(s>mx)  mx=s;

    }
    printf("%lld\n",mx);
    return 0;
}