CodeForces 632A(역방향 시뮬레이션)

Grandma Laura and Apples
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
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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 10
half
halfplus

Output

15

Input

3 10
halfplus
halfplus
halfplus

Output
55
제목의 대의: n명이 사과를 사면 짝수가 남을 때 반을 사고, 사과가 홀수가 남을 때 먼저 반을 사고 반값에 사과를 하나 사면 결국 사과가 마침 다 팔린다.농민 수입은 얼마입니까?
해체 사고방식: 시작 상태를 모르지만 종료 상태를 알기 때문에 우리는 뒤에서 앞으로 밀고 반대로 시뮬레이션을 할 수 있다.
코드는 다음과 같습니다.

#include
#include
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<