HackerRank - Stock Maximize

最新推荐文章于 2017-12-07 19:49:03 发布

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最新推荐文章于 2017-12-07 19:49:03 发布

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分类专栏： C++ 文章标签： HackerRank

本文链接：https://blog.youkuaiyun.com/lilong_dream/article/details/39616767

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本文探讨了如何利用预测市场趋势的技术，通过动态规划方法优化股票买卖策略，以实现最大利润。具体步骤包括分析每日股价预测，决定买卖时机，确保在股价波动中获取最高收益。

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题目： https://www.hackerrank.com/challenges/stockmax

Your algorithms have become so good at predicting the market that you now know what the share price of Wooden Orange Toothpicks Inc. (WOT) will be for the next N days.

Each day, you can either buy one share of WOT, sell any number of shares of WOT that you own, or not make any transaction at all. What is the maximum profit you can obtain with an optimum trading strategy?

Input

The first line contains the number of test cases T. T test cases follow:

The first line of each test case contains a number N. The next line contains N integers, denoting the predicted price of WOT shares for the next N days.

Output

Output T lines, containing the maximum profit which can be obtained for the corresponding test case.

Constraints

1 <= T <= 10
1 <= N <= 50000

All share prices are between 1 and 100000

Sample Input

Sample Output

0
197
3

Explanation

For the first case, you cannot obtain any profit because the share price never rises.
For the second case, you can buy one share on the first two days, and sell both of them on the third day.
For the third case, you can buy one share on day 1, sell one on day 2, buy one share on day 3, and sell one share on day 4.

分析：

动态规划的思想，从后往前处理，如果后面的最大值大于当前值，则将差值加到结果上，否则更新最大值。

代码：

#include <iostream>
#include <vector>

int main() {
    int T;
    std::cin >> T;
    
    int N;
    
    while(T--) {
        std::cin >> N;
        
        std::vector<int> nums(N, 0);
        for(int i = 0; i < N; ++i) {
            std::cin >> nums[i];
        }
        
        long long result = 0;  // !
        int max = nums[N - 1];
        for(int i = N - 2; i >= 0; --i) {
            if(max > nums[i]) {
                result += (max - nums[i]);
            } else {
                max = nums[i];
            }
        }
        std::cout << result << std::endl;
    }
    
    return 0;
}