Sherlock and Cost

最新推荐文章于 2020-10-03 00:43:04 发布

zsyr

最新推荐文章于 2020-10-03 00:43:04 发布

阅读量394

点赞数

CC 4.0 BY-SA版权

分类专栏： acm

本文链接：https://blog.youkuaiyun.com/oyasherkin/article/details/54564113

acm 专栏收录该内容

35 篇文章

订阅专栏

本文介绍了一个关于寻找数组最大成本的问题，给出了具体的输入输出格式及样例，并提供了完整的C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Array A contains the elements, A1,A2…AN. And array B contains the elements, B1,B2…BN. There is a relationship between Ai and Bi, ∀ 1 ≤ i ≤ N, i.e.,
any element Ai lies between 1 and Bi.

Let the cost S of an array A be defined as:

You have to print the largest possible value of S.

Input Format

The first line contains, T, the number of test cases. Each test case contains an integer, N, in first line. The second line of each test case contains N integers that denote the array B.

Constraints

1 ≤ T ≤ 20
1 ≤ N ≤ 105
1 ≤ Bi ≤ 100

Output Format

For each test case, print the required answer in one line.

Sample Input

1
5
10 1 10 1 10
Sample Output

36
Explanation

The maximum value occurs when A1=A3=A5=10 and A2=A4=1.

#include<bits/stdc++.h>
using namespace std;
int ar[100005]={},dp[100005][2]={};
int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int n,i,j;
        cin >> n;
        for(i=0; i<n; i++)
            cin >> ar[i];
        for(i=0; i<n-1; i++)
        {
            dp[i+1][0]=max(dp[i][0],dp[i][1]+abs(ar[i]-1));
            dp[i+1][1]=max(dp[i][0]+abs(ar[i+1]-1),dp[i][1]+abs(ar[i]-ar[i+1]));
        }
        cout << max(dp[n-1][0],dp[n-1][1]) << endl;
    }
    return 0;
}