CodeForces 632A(反向模拟)

Grandma Laura and Apples

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 632A

Description

Grandma Laura came to the market to sell some apples. During the day she sold all the apples she had. But grandma is old, so she forgot how many apples she had brought to the market.

She precisely remembers she had n buyers and each of them bought exactly half of the apples she had at the moment of the purchase and also she gave a half of an apple to some of them as a gift (if the number of apples at the moment of purchase was odd), until she sold all the apples she had.

So each buyer took some integral positive number of apples, but maybe he didn't pay for a half of an apple (if the number of apples at the moment of the purchase was odd).

For each buyer grandma remembers if she gave a half of an apple as a gift or not. The cost of an apple is p (the number p is even).

Print the total money grandma should have at the end of the day to check if some buyers cheated her.

Input

The first line contains two integers n and p (1 ≤ n ≤ 40, 2 ≤ p ≤ 1000) — the number of the buyers and the cost of one apple. It is guaranteed that the number p is even.

The next n lines contains the description of buyers. Each buyer is described with the string half if he simply bought half of the apples and with the string halfplus if grandma also gave him a half of an apple as a gift.

It is guaranteed that grandma has at least one apple at the start of the day and she has no apples at the end of the day.

Output

Print the only integer a — the total money grandma should have at the end of the day.

Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Sample Input

Input

2 10halfhalfplus

Output

15

Input

3 10halfplushalfplushalfplus

Output

55

题目大意:有n个人买苹果,当苹果剩余偶数时买走一半,当苹果剩余奇数时,先买走一半,再用半价买走一个苹果,最终苹果恰好卖完.农民收入为多少.

解体思路:因为不知道起始状态,但知到终止状态,所以我们可以从尾向前推,反向模拟一遍.

代码如下:

#include<iostream>#include<string>using namespace std;string s[45];int main(){int n,p,i;long long sum,now;sum=now=0;cin>>n>>p;for(i=n-1;i>=0;i--)cin>>s[i];for(i=0;i<=n-1;i++){if(s[i]=="half"){sum+=now*p;now=2*now;}else{sum+=now*p+p/2;now=now*2+1;}}cout<<sum<<endl;return 0;}