Codeforces Round #407 (Div. 2) C Functions again(最大子序列和)

Posted Wally的博客

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Codeforces Round #407 (Div. 2) C Functions again(最大子序列和)相关的知识,希望对你有一定的参考价值。

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:

技术分享

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.

题意:给你一个长度为n的序列,问你在 l 与 r 取不同值时技术分享最大值是多少?

思路:能够看出这就是在abs(a[i]-a[i+1])取正负正负···和取负正负正···然后求子序列的最大值

AC代码:

技术分享
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=100005;
 4 long long a[maxn],b[maxn],c[maxn];
 5 int main()
 6 {
 7 
 8     int n;
 9     while(~scanf("%d",&n))
10     {
11         for(int i=0; i<n; i++)
12         {
13             scanf("%I64d",&c[i]);
14         }
15         for(int i=0; i<n-1; i++)
16         {
17             if(i%2==0)
18             {
19                 a[i]=abs(c[i]-c[i+1]);
20                 b[i]=-a[i];
21             }
22             else
23             {
24                 b[i]=abs(c[i]-c[i+1]);
25                 a[i]=-b[i];
26             }
27         }
28         long long ans=0,num=0;
29         for(int i=0; i<n-1; i++)
30         {
31             num+=a[i];
32             if(num>ans)
33             {
34                 ans=num;
35             }
36             else if(num<0)
37             {
38                 num=0;
39             }
40         }
41         num=0;
42         for(int i=0; i<n-1; i++)
43         {
44             num+=b[i];
45             if(num>ans)
46             {
47                 ans=num;
48             }
49             else if(num<0)
50             {
51                 num=0;
52             }
53         }
54         printf("%I64d\n",ans);
55     }
56     return 0;
57 }
View Code


以上是关于Codeforces Round #407 (Div. 2) C Functions again(最大子序列和)的主要内容,如果未能解决你的问题,请参考以下文章

Codeforces Round #407 (Div. 2)解题报告

Codeforces Round #407 (Div. 2) C Functions again(最大子序列和)

Codeforces Round #407 (Div. 2)(争取明天24点前补掉)

Codeforces Round #436 E. Fire(背包dp+输出路径)

[ACM]Codeforces Round #534 (Div. 2)

codeforces 407C Curious Array