CodeForces 789C Functions again_789 c,c c-优快云博客

本文链接：https://blog.youkuaiyun.com/Vace___yun/article/details/81841871

本文介绍了一种解决特定数学函数最大值的方法。该函数基于一个数组的相邻元素差值，并通过计算最大子序列之和来确定最大值。文章详细解释了算法流程并提供了实现代码。

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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.

题意：给你一个数组a，再给你一个关于这个数组的函数，求函数的最大值。

分析：题意看了半天才看懂，求函数的最大值，很像求数组的最大子序列之和，需要开个数组b来存入a数组中每俩个相邻元素的差值，这可以从给的函数看出，然后就是求最大子序列之和，需要注意b数组的每一项有正有负，根据首项判断，奇数位的为正，偶数位的为负，数组要开long long 型。

代码：
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int N=100010;
long long int a[N],b[N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        for(int i=1;i<n;i++)
            b[i]=abs(a[i]-a[i+1]);
        long long int sum,maxx,k,v;
        sum=maxx=k=0;
        v=1;
        for(int i=v;i<n;i++)
        {
            if(k%2==0)
            {
                sum+=b[i];
                k++;
            }
            else
            {
                sum-=b[i];
                k++;
            }
            if(sum>maxx)
                maxx=sum;
            if(sum<0)
            {
                sum=0;
                k=0;
                i=v++;
            }
        }
        printf("%lld\n",maxx);
    }
}