#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

//分糖果的問題
/*
There are N children standing in a line. Each child is assigned a rating value.

You are giving candies to these children subjected to the following requirements:

Each child must have at least one candy.
Children with a higher rating get more candies than their neighbors.
*/
 


/*
解題思路：
遍歷兩邊，首先每個人得一塊糖，第一遍從左到右，若當前點比前一個點高就比前者多一塊。
這樣保證了在一個方向上滿足了要求。第二遍從右往左，若左右兩點，左側高於右側，但
左側的糖果數不多於右側，則左側糖果數等於右側糖果數+1，這就保證了另一個方向上滿足要求。

最後將各個位置的糖果數累加起來就可以了。
*/


int candyCount(vector<int>&rating) {

    int res = 0;
    //孩子總數
    int n = rating.size();

    //糖果集合
    vector<int> candy(n, 1
 
);
    //從左往右遍歷
    for (int i = 0;i < n - 1;i++) {
        if (rating[i + 1] > rating[i])candy[i + 1] = candy[i] + 1;
    }
    //從右往左
    for (int i = n - 1;i > 0;i--) {
        if (rating[i - 1] > rating[i] && candy[i - 1] <= candy[i])candy[i - 1] = candy[i] + 1;
    }

    //累加結果
    for
 
 (auto a : candy) {
        res += a;
    }

    return res;
}
//測試函式
int main() {

    vector<int> rating{1,3,2,1,4,5,2};
    cout << candyCount(rating) << endl;
    return 0;
}