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 <= 50000All share prices are between 1 and 100000
Sample Input
3 3 5 3 2 3 1 2 100 4 1 3 1 2Sample Output
0 197 3Explanation
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.
经典的dp题目，每天可以买进一份股票，或者卖出任意份的股票（当然你要有这么多份股票），或者什么也不做。
首先，如果你在某一天买一份股票，那么这一天的股票价格一定不是从当天起到最后一天的最大值，想一想。
其次，如果选择在某一天卖出若干份股票，那么一定是将所有的股票都卖出才是最优策略。
最后，如果选择在某一天卖出股票，那么当天的股票价格一定是从当天起到最后一天的最大值。
Accepted Code:
1 #include <iostream>
2 using namespace std;
3
4 typedef long long LL;
5 const int maxn = 50002;
6 int a[maxn];
7
8 int main(void) {
9 ios::sync_with_stdio(false);
10 int T;
11 cin >> T;
12 while (T--) {
13 int n;
14 cin >> n;
15 for (int i = 1; i <= n; i++) cin >> a[i];
16 int max_price = -1;
17 LL ans = 0;
18 for (int i = n; i >= 1; i--) {
19 max_price = max(max_price, a[i]);
20 ans += max_price - a[i];
21 }
22 cout << ans << endl;
23 }
24
25 return 0;
26 }
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