Record Breaker (Google Kickstart2020 Round D Problem A)

1：题目

Isyana Got its local theme park in a row N the number of tourists per day.

其中第 i The number of visitors for the day is Vi.

If one day can meet the following two conditions at the same time,The day is considered a record-breaking day：

The number of tourists on this day is strictly greater than the number of tourists on each previous day.
这是最后一天,Or the number of tourists on one day is strictly greater than the number of tourists on the next day.
请注意,The first day can also be a record-breaking day.

请帮助 Isyana Find out the record-breaking days.

输入格式
第一行包含整数 T,表示共有 T 组测试数据.

每组数据第一行包含整数 N.

第二行包含 N 个整数,其中第 i 个表示 Vi.

输出格式
每组数据输出一个结果,每个结果占一行.

结果表示为 Case #x: y,其中 x 为组别编号（从 1 开始）,y for record-breaking days.

数据范围
1≤T≤100,
0≤Vi≤2×105,
对于每个测试点,满足 1≤N≤2×105 的数据一定不超过 10 组,其余数据则满足 1≤N≤1000.

输入样例：
4
8
1 2 0 7 2 0 2 0
6
4 8 15 16 23 42
9
3 1 4 1 5 9 2 6 5
6
9 9 9 9 9 9
输出样例：
Case #1: 2
Case #2: 1
Case #3: 3
Case #4: 0
样例解释
For test samples 1,Records were broken on the second and fourth days.

For test samples 2,Only the last day broke the record.

For test samples 3,第一天、The third and sixth days broke records.

For test samples 4,Not a single day broke the record.

难度：简单
时/空限制：1s / 64MB
总通过数：26
总尝试数：36
来源：Google Kickstart2020 Round D Problem A
算法标签
None

2：代码实现

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 200010;
int n, st[N];

int main()
{
    
    int T;
    cin >> T;
    for(int t = 1; t <= T; t++){
    
        cin >> n;
        for(int i = 0; i < n; i++) cin >> st[i];

        int res = 0, mav = 0;
        for(int i = 0; i < n; i++)
            if(mav < st[i]){
    
                mav = st[i];
                if(st[i] > st[i + 1]) res++;
            }

        cout << "Case #" << t << ": " << res << endl;
    }

    return 0;
}