codeforces 407 div1 A题(Functions again)

最新推荐文章于 2025-09-09 08:37:13 发布

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最新推荐文章于 2025-09-09 08:37:13 发布

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原文链接：http://www.cnblogs.com/zsyacm666666/p/6792152.html

本文介绍了CodeForces407Div1A题的两种解题思路，一种是通过转换数列并求最大子序列和，另一种是通过动态规划计算最大序列和与最小序列和。

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?<?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

题解：存在两种解决方案。

方案1：

如果把转化后的数列看成一个新的数列，观察可得：

相邻位的数字总是异号的
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<n; i++){
                  a[i]=cin.nextInt();
              }
              int tmp;
              for(int i=0;i<n-1;i++){
                  tmp=Math.abs(a[i]-a[i+1]);
                  if(i%2==0){
                      b[i]=tmp;
                      c[i]=-tmp;
                  }else{
                      b[i]=-tmp;
                      c[i]=tmp;
                  }
              }
              long ans1=-inf,ans2=-inf;
              long sum=0;
              for(int i=0;i<n-1;i++){
                  sum+=b[i];
                  ans1=Math.max(ans1, sum);
                  if(sum<0){
                      sum=0;
                  }
              }
              sum=0;
              for(int i=0;i<n-1;i++){
                  sum+=c[i];
                  ans2=Math.max(ans2, sum);
                  if(sum<0){
                      sum=0;
                  }
              }
              System.out.println(Math.max(ans1, ans2));
          }
          cin.close();
      }
}

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<n; i++){
                  a[i]=cin.nextInt();
              }
              b[n-1]=c[n-1]=0;
              long ans=0;
              for(int i=n-2;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();
      }
}

转载于:https://www.cnblogs.com/zsyacm666666/p/6792152.html