Codeforces 18D Seller Bob （贪心）

D. Seller Bob

time limit per test

2 seconds

memory limit per test

128 megabytes

input

standard input

output

standard output

Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:

• A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
• Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.

Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally.

Input

The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000).

Output

Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time.

Examples

input

7win 10win 5win 3sell 5sell 3win 10sell 10

output

1056

input

3win 5sell 6sell 4

output

0

题意：题目告诉你你能够按顺序获得一些价值为2^x的硬盘，你可以保留一个并且在之后的几天卖掉，但是你在卖掉当前这个之前是不能保留其他的硬盘的，

在已经知道什么时候会有人买什么类型的硬盘的前提下，问你能获得的最大收益是多少

思路：从大到小枚举x， 然后遇到一个sell x， 就往前找第一个win这个x的， 然后这一段全都做上标记。。

import java.math.BigInteger;import java.util.Scanner;import java.util.*;public class Main {public static void main(String[] args) {boolean[] flag = new boolean[5005];boolean[] book = new boolean[5005];BigInteger ans = BigInteger.valueOf(0);BigInteger ER = BigInteger.valueOf(2);Scanner sc = new Scanner(System.in);String s = new String();int n = sc.nextInt();int[] x = new int[5005];for(int i = 0; i < 5005; i++)x[i] = -1;for(int i = 0; i < n; i++){s = sc.next();x[i] = sc.nextInt();if(s.equals("win")) flag[i] = true;}for(int i = 2000; i >= 0; i--){for(int j = 0; j < n; j++){if(flag[j] == false && x[j] == i && book[j] == false){int k; for(k = j-1; k >= 0; k--){if(book[k] == true || (flag[k] == true && x[k] == i)) break;}if(k == -1 || book[k] == true) continue;for(int l = k; l <= j; l++) book[l] = true;ans = ans.add(ER.pow(i));}}}System.out.println(ans);  }}