Codeforces Round #407(Div. 2)C. Functions again【思维+最大连续子序列的和】

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

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.

题目大意：

给你N个数，其中有：



让你找到一个最大的f（l，r）；

思路：

首先做出相邻两个数的差，设定为b【i】；

那么很显然，b【i】的正负取决于选取起点的奇偶性来决定。

那么如果我们选1.3.5.7............的点作为起点的话，那么2.4.6........的位子上的b【i】都是负数了。

那么我们此时对于b【i】可以做一次最大连续子序列的和的dp，时间复杂度O（n）；

同理，选择2.4.6...........的点作为起点的话，那么1.3.5...............的位子上的b【i】就是负数了。

那么再做一次b【i】的dp，过程维护最大值即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define ll __int64
ll a[100800];
ll b[100800];
ll dp[100800];
ll abss(ll num)
{
if(num<0)return -num;
else return num;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
{
scanf("%I64d",&a[i]);
}
for(int i=0;i<n-1;i++)
{
b[i]=abss(a[i]-a[i+1]);
if(i%2==1)b[i]=-b[i];
}
dp[0]=b[0];
ll output=0;
output=max(output,dp[0]);
for(int i=1;i<n-1;i++)
{
dp[i]=max(b[i],dp[i-1]+b[i]);
output=max(dp[i],output);
}
for(int i=1;i<n-1;i++)
{
b[i]=abss(a[i]-a[i+1]);
if(i%2==0)b[i]=-b[i];
}
dp[1]=b[1];
output=max(output,dp[1]);
for(int i=2;i<n-1;i++)
{
dp[i]=max(b[i],dp[i-1]+b[i]);
output=max(dp[i],output);
}
printf("%I64d\n",output);
}
}