CF1392C-Omkar and Waterslide

题目:

题目描述:

Omkar is building a waterslide in his water park, and he needs your help to ensure that he does it as efficiently as possible.

Omkar currently has $n$ supports arranged in a line, the $i$ -th of which has height $a_i$ . Omkar wants to build his waterslide from the right to the left, so his supports must be nondecreasing in height in order to support the waterslide. In $1$ operation, Omkar can do the following: take any contiguous subsegment of supports which is nondecreasing by heights and add $1$ to each of their heights.

Help Omkar find the minimum number of operations he needs to perform to make his supports able to support his waterslide!

An array $b$ is a subsegment of an array $c$ if $b$ can be obtained from $c$ by deletion of several (possibly zero or all) elements from the beginning and several (possibly zero or all) elements from the end.

An array $b_1, b_2, \dots, b_n$ is called nondecreasing if $b_i\le b_{i+1}$ for every $i$ from $1$ to $n-1$ .

输入格式:

Each test contains multiple test cases. The first line contains the number of test cases $t$ ( $1 \leq t \leq 100$ ). Description of the test cases follows.

The first line of each test case contains an integer $n$ ( $1 \leq n \leq 2 \cdot 10^5$ ) — the number of supports Omkar has.

The second line of each test case contains $n$ integers $a_{1},a_{2},…,a_{n}$ $(0 \leq a_{i} \leq 10^9)$ — the heights of the supports.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$ .

输出格式:

For each test case, output a single integer — the minimum number of operations Omkar needs to perform to make his supports able to support his waterslide.

样例:

样例输入1:

1
2
3
4
5
6
7

3
4
5 3 2 5
5
1 2 3 5 3
3
1 1 1

样例输出1:

1
2
3

3
2
0

思路:

实现:

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#include "ybwhead/ios.h"
int n, TTT;
int main()
{
    yin >> TTT;
    while (TTT--)
    {
        yin >> n;
        int la = 0;
        long long ans = 0;
        for (int i = 1; i <= n; i++)
        {
            int x;
            yin >> x;
            if (i >= 2)
                ans += max(0, la - x);
            la = x;
        }
        yout << ans << endl;
    }
}