CF1868E.Min-Sum-Max

普及/提高-

通过率:0%

AC君温馨提醒

该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。

题目描述

The first line of input contains a single integer tt ( 1t501\le t\le 50 ) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer nn ( 1n3001\le n\le 300 ) — the length of the array aa .

The second line of each test case contains nn integers a1,a2,,ana_1,a_2,\ldots,a_n ( 109ai109-10^9\le a_i\le 10^9 ) — the elements of the array aa .

It is guaranteed that the sum of nn over all test cases does not exceed 10001000 .

输入格式

For each test case, output a single integer — the maximum number of subsegments among all possible ways to divide the array aa .

输出格式

In the first test case, Daniel can divide the array into [1][-1] and [5,4][5,4] , and s=[1,9]s=[-1,9] . It can be shown that for any i=ji=j , the condition in the statement is already satisfied, and for i=1,j=2i=1,j=2 , we have min(1,9)(1)+9max(1,9)\min(-1,9)\le (-1)+9\le \max(-1,9) .

In the second test case, if Daniel divides the array into [2023][2023] and [2043][2043] , then for i=1,j=2i=1,j=2 we have 2023+2043>max(2023,2043)2023+2043>\max(2023,2043) , so the maximum number of subsegments is 11 .

In the third test case, the optimal way to divide the array is [1,4,7],[1],[5,4][1,4,7],[-1],[5,-4] .

In the fourth test case, the optimal to divide the array way is [4,0,3,18],[10][-4,0,3,-18],[10] .

In the fifth test case, Daniel can only get one subsegment.

输入输出样例

  • 输入#1

    8
    3
    -1 5 4
    2
    2023 2043
    6
    1 4 7 -1 5 -4
    5
    -4 0 3 -18 10
    1
    998244853
    10
    -4 2 5 -10 4 8 2 9 -15 7
    7
    -7 3 8 -9 -2 2 4
    4
    -5 5 -2 -5

    输出#1

    2
    1
    3
    2
    1
    6
    5
    3

说明/提示

In the first test case, Daniel can divide the array into [1][-1] and [5,4][5,4] , and s=[1,9]s=[-1,9] . It can be shown that for any i=ji=j , the condition in the statement is already satisfied, and for i=1,j=2i=1,j=2 , we have min(1,9)(1)+9max(1,9)\min(-1,9)\le (-1)+9\le \max(-1,9) .

In the second test case, if Daniel divides the array into [2023][2023] and [2043][2043] , then for i=1,j=2i=1,j=2 we have 2023+2043>max(2023,2043)2023+2043>\max(2023,2043) , so the maximum number of subsegments is 11 .

In the third test case, the optimal way to divide the array is [1,4,7],[1],[5,4][1,4,7],[-1],[5,-4] .

In the fourth test case, the optimal to divide the array way is [4,0,3,18],[10][-4,0,3,-18],[10] .

In the fifth test case, Daniel can only get one subsegment.

首页