BUY LOW, BUY LOWER
| Time Limit: 1000MS |
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Memory Limit: 30000K |
| Total Submissions: 8327 |
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Accepted: 2888 |
Description The advice to "buy low" is half the formula to success in the bovine stock market.To be considered a great investor you must also follow this problems' advice:
"Buy low; buy lower"
Each time you buy a stock, you must purchase it at a lower price than the previous time you bought it. The more times you buy at a lower price than before, the better! Your goal is to see how many times you can continue purchasing at ever lower prices.
You will be given the daily selling prices of a stock (positive 16-bit integers) over a period of time. You can choose to buy stock on any of the days. Each time you choose to buy, the price must be strictly lower than the previous time you bought stock. Write a program which identifies which days you should buy stock in order to maximize the number of times you buy.
Here is a list of stock prices:
Day 1 2 3 4 5 6 7 8 9 10 11 12
Price 68 69 54 64 68 64 70 67 78 62 98 87
The best investor (by this problem, anyway) can buy at most four times if each purchase is lower then the previous purchase. One four day sequence (there might be others) of acceptable buys is:
Day 2 5 6 10
Price 69 68 64 62 Input * Line 1: N (1 <= N <= 5000), the number of days for which stock prices are given
* Lines 2..etc: A series of N space-separated integers, ten per line except the final line which might have fewer integers.
Output Two integers on a single line:
* The length of the longest sequence of decreasing prices
* The number of sequences that have this length (guaranteed to fit in 31 bits)
In counting the number of solutions, two potential solutions are considered the same (and would only count as one solution) if they repeat the same string of decreasing prices, that is, if they "look the same" when the successive prices are compared. Thus, two different sequence of "buy" days could produce the same string of decreasing prices and be counted as only a single solution.
Sample Input 12
68 69 54 64 68 64 70 67 78 62
98 87
Sample Output 4 2
Source USACO 2002 February |
题意：
最长递减子序列+最长的不同子序列计数。
思路：（贴别人的）
注意：重复的只算一次
如何去掉一些重复的是本题的关键
去重思路：
7
5 3 7 6 3 2 1
6
5 3 7 3 1
5
5 3 2 1 3
第一组在推到 数字 2 的时候有 3会出现重复， 显然前面一个3是可有可无的。
第二组也一样，前面一个3是可有可无的。
1.如果最长下降序列中有后面一个3， 如第一二组数据，那么前面一个3是无用的，在推好后面一个3之后把之前的所用重复的计数数组清零
2.如果最长下降序列中只有前面一个3，如第三组数据， 做情况1的操作也是不会影响结果的，因为第三组的前面一个3的状态已经推到2后才被清零的。
代码：
#include
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