Educational Codeforces Round 9 A. Grandma Laura and Apples【水题】

解决祖母卖苹果问题,通过逆向计算确定初始苹果数量及赠品,最终得出总收入。
A. Grandma Laura and Apples
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

Examples
input
2 10
half
halfplus
output
15
input
3 10
halfplus
halfplus
halfplus
output
55
Note

In the first sample at the start of the day the grandma had two apples. First she sold one apple and then she sold a half of the second apple and gave a half of the second apple as a present to the second buyer.

题意:老太太带着苹果去街上卖,刚好卖完,卖了多少个已经忘记了,输入n次操作和苹果的单价m,每次操作只有两种字符串,halfplus表示卖出苹果总数的一半,并且送出半个苹果。half表示卖出总数的一半苹果。问老太太最好能赚多少钱?


水题,从后往前计算出苹果总量和送出苹果的数目,再计算出赚的钱。


#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<iomanip>
#include<cmath>
long long sum=0,c[100],cnum=0;
int n,m;
using namespace std;
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        string s;
        cin>>s;
        if(s.size()>4)c[i]=1;
    }
    for(int i=n;i>0;i--){
        if(c[i]==1){
            sum=2*sum+1;
            cnum++;
        }
        else sum=sum*2;
    }
    cout<<sum*m-cnum*m/2<<endl;

    return 0;
}


### Codeforces Round 927 Div. 3 比赛详情 Codeforces是一个面向全球程序员的比赛平台,定期举办不同级别的编程竞赛。Div. 3系列比赛专为评级较低的选手设计,旨在提供更简单的问让新手能够参与并提升技能[^1]。 #### 参赛规则概述 这类赛事通常允许单人参加,在规定时间内解决尽可能多的问来获得分数。评分机制基于解决问的速度以及提交答案的成功率。比赛中可能会有预测试案例用于即时反馈,而最终得分取决于系统测试的结果。此外,还存在反作弊措施以确保公平竞争环境。 ### 目解析:Moving Platforms (G) 在这道中,给定一系列移动平台的位置和速度向量,询问某时刻这些平台是否会形成一条连续路径使得可以从最左端到达最右端。此问涉及到几何学中的线段交集判断和平面直角坐标系内的相对运动分析。 为了处理这个问,可以采用如下方法: - **输入数据结构化**:读取所有平台的数据,并将其存储在一个合适的数据结构里以便后续操作。 - **时间轴离散化**:考虑到浮点数精度误差可能导致计算错误,应该把整个过程划分成若干个小的时间间隔来进行模拟仿真。 - **碰撞检测算法实现**:编写函数用来判定任意两个矩形之间是否存在重叠区域;当发现新的连接关系时更新可达性矩阵。 - **连通分量查找技术应用**:利用图论知识快速求解当前状态下哪些节点属于同一个集合内——即能否通过其他成员间接相连。 最后输出结果前记得考虑边界条件! ```cpp // 假设已经定义好了必要的类和辅助功能... bool canReachEnd(vector<Platform>& platforms, double endTime){ // 初始化工作... for(double currentTime = startTime; currentTime <= endTime ;currentTime += deltaT){ updatePositions(platforms, currentTime); buildAdjacencyMatrix(platforms); if(isConnected(startNode,endNode)){ return true; } } return false; } ```
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