D - Functions again CodeForces - 789C

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.

题目意思:就是数组的最大和只不过这个和是是要满足公式的和

公式的意思就是从   l->r 区间前一个数减后一个数差乘上(-1)^ i -l 就是1 和-1交替

求这个公式的最大值

解题思路:这个题刚开开始认为是 dp但二维的数组太大,一维的又不会,不过看题和之前求数列最大和很像就试着套了一下,测试的救过了,提交错了,看别人的和自己写的一样,认真看发现求得时候有了个小错误 。后来改完就过了,不过时间有点长应该有更快得算法;

代码

  1.  #include <stdio.h>
  2. #include <string.h>
  3. #include<algorithm>
  4. long long int a[100010];
  5. long long int b[100010];
  6. int main()
  7. {
  8.     int n;
  9.     scanf("%d",&n);
  10.     {
  11.         memset(a,0,sizeof(a));
  12.         memset(b,0,sizeof(b));
  13.         for(int i=1; i<=n; i++)
  14.             scanf("%lld",&a[i]);
  15.         for(int i=1; i<n; i++)
  16.             b[i] = abs(a[i]-a[i+1]);// 求相邻两个数的差值
  17.         long long sum=0,maxx=0;
  18.         int dd=0;
  19.         int k=1;
  20.         for(int i=k; i<n; i++)
  21.         {
  22.             if(dd %2 == 0 )       //第一个正值
  23.                 sum+=b[i];
  24.             else
  25.                 sum-=b[i];      //负值
  26.           dd++;
  27.             if(sum>maxx)
  28.                 maxx=sum;      //更新最大值
  29.             if(sum<0)   
  30.             {
  31.                 sum=0;        //重新计数
  32.                 dd=0;  
  33.                 i=k++;        //之前写的 k=i++;错了
  34.             }
  35.         }
  36.         printf("%lld\n",maxx);
  37.     }
  38.     return 0;
  39. }

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转载自blog.csdn.net/TANG3223/article/details/81842115

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