codeforces 407 div1 A题(Functions again)

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?
    

题解：存在两种解决方案。

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方案1：

如果把转化后的数列看成一个新的数列，观察可得：

1. 相邻位的数字总是异号的
2. 每个数字都有正负两种可能的存在
   因此可以将这个数列转化为两种形式
   +-+-+-+-.....
   -+-+-+-+.....
   我们只要对这两个数列求最大子序列和即可

方案2：

这是一个更加优美的解法，设b[i]为l为i的最大序列和,c[i]为l为i的最小序列和

b[i]=max(-c[i+1],0)+abs(a[i]-a[i+1])

c[i]=min(-b[i+1],0)+abs(a[i]-a[i+1])

方案1：

import java.io.*;import  java.util.*;public class cf407c {      static final int N=(int)1e5+10;      static final long inf=(long)1e10;      static int a[]=new int[N];      static int b[]=new int[N];      static int c[]=new int[N];      public static void main(String[] args){          Scanner cin=new Scanner(new InputStreamReader(System.in));          while(cin.hasNext()){              int n=cin.nextInt();              for(int i =0; i

方案2：

import java.io.*;import  java.util.*;public class cf407c {      static final int N=(int)1e5+10;      static final long inf=(long)1e10;      static int a[]=new int[N];      static long b[]=new long[N];      static long c[]=new long[N];      public static void main(String[] args){          Scanner cin=new Scanner(new InputStreamReader(System.in));          while(cin.hasNext()){              int n=cin.nextInt();              for(int i =0; i
    
     =0;i--){                  b[i]=Math.max(-c[i+1],0)+Math.abs(a[i]-a[i+1]);                  c[i]=Math.min(-b[i+1],0)+Math.abs(a[i]-a[i+1]);                  ans=Math.max(ans,b[i]);              }              System.out.println(ans);          }          cin.close();      }}