CF665B Shopping
CF665B Shopping
题目描述
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains $ k $ items. $ n $ customers have already used the above service. Each user paid for $ m $ items. Let $ a_{ij} $ denote the $ j $ -th item in the $ i $ -th person’s order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the $ i $ -th order he will find one by one all the items $ a_{ij} $ ( $ 1<=j<=m $ ) in the row. Let $ pos(x) $ denote the position of the item $ x $ in the row at the moment of its collection. Then Ayush takes time equal to $ pos(a_{i1})+pos(a_{i2})+…+pos(a_{im}) $ for the $ i $ -th customer.
When Ayush accesses the $ x $ -th element he keeps a new stock in the front of the row and takes away the $ x $ -th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
输入格式
The first line contains three integers
n
n
n ,
m
m
m and
k
k
k (
1
≤
n
,
k
≤
100
,
1
≤
m
≤
k
1\le n,k\le 100,1\le m\le k
1≤n,k≤100,1≤m≤k ) — the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains
k
k
k distinct integers
p
l
p_{l}
pl (
1
<
=
p
l
<
=
k
1<=p_{l}<=k
1<=pl<=k ) denoting the initial positions of the items in the store. The items are numbered with integers from
1
1
1 to
k
k
k .
Each of the next
n
n
n lines contains
m
m
m distinct integers
a
i
j
a_{ij}
aij (
1
<
=
a
i
j
<
=
k
1<=a_{ij}<=k
1<=aij<=k ) — the order of the
i
i
i -th person.
输出格式
Print the only integer $ t $ — the total time needed for Ayush to process all the orders.
输入输出样例
样例输入1
2 2 5
3 4 1 2 5
1 5
3 1
样例输出1
14
说明/提示
Customer
1
1
1 wants the items
1
1
1 and
5
5
5 .
p
o
s
(
1
)
=
3
pos(1)=3
pos(1)=3 , so the new positions are:
[
1
,
3
,
4
,
2
,
5
]
[1,3,4,2,5]
[1,3,4,2,5] .
p
o
s
(
5
)
=
5
pos(5)=5
pos(5)=5 , so the new positions are:
[
5
,
1
,
3
,
4
,
2
]
[5,1,3,4,2]
[5,1,3,4,2] .
Time taken for the first customer is
3
+
5
=
8
3+5=8
3+5=8 .
Customer
2
2
2 wants the items
3
3
3 and
1
1
1 .
p
o
s
(
3
)
=
3
pos(3)=3
pos(3)=3 , so the new positions are:
[
3
,
5
,
1
,
4
,
2
]
[3,5,1,4,2]
[3,5,1,4,2] .
p
o
s
(
1
)
=
3
pos(1)=3
pos(1)=3 , so the new positions are:
[
1
,
3
,
5
,
4
,
2
]
[1,3,5,4,2]
[1,3,5,4,2] .
Time taken for the second customer is
3
+
3
=
6
3+3=6
3+3=6 .
Total time is
8
+
6
=
14
8+6=14
8+6=14 .
Formally
p
o
s
(
x
)
pos(x)
pos(x) is the index of
x
x
x in the current row.
思路
题上首先给了三个数
n
,
m
,
k
n,m,k
n,m,k,然后在第二行里给出了
k
k
k 个数,然后又
n
×
m
n\times m
n×m 个操作,代表每一次把这个操作的数移动到第一位,其他的往后推,每次移动的时候有一个权值(从第一个数到第这个数的相差的位置数),问的是所有权值的和,数据量不大,直接模拟就好。
注
看到楼上用的 vector
,我这里用数组。
Code
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n, m, k, x, ans = 0, a[105];
scanf("%d %d %d", &n, &m, &k);
for (register int i(1); i <= k; ++i)
scanf("%d", &a[i]);
for (register int i(1); i <= n * m; ++i)
{
scanf("%d", &x);
register int j, l;
for (j = 1; j <= k; ++j)
if (a[j] == x)
break;
ans += j;
for (l = j; l >= 2; --l)
a[l] = a[l - 1];
a[1] = x;
}
printf("%d\n", ans);
return 0;
}
广告
绿树公司 - 官方网站:https://wangping-lvshu.github.io/LvshuNew/
绿树智能 - 官方网站:https://wangping-lvshu.github.io/LvshuZhineng/
(现在使用,人人均可获得300元大奖)
CF665B Shopping的更多相关文章
- Shopping(山东省第一届ACM省赛)
Shopping Time Limit: 1000MS Memory limit: 65536K 题目描述 Saya and Kudo go shopping together.You can ass ...
- sdutoj 2154 Shopping
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2154 Shopping Time Limit: ...
- Shopping(SPFA+DFS HDU3768)
Shopping Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Sub ...
- eclipse中 起动tomcat时报Multiple Contexts have a path of "/shopping"
eclipse中 启动tomcat时报Multiple Contexts have a path of "/shopping". 这个是由于你的server服务器中的server. ...
- 洛谷P2732 商店购物 Shopping Offers
P2732 商店购物 Shopping Offers 23通过 41提交 题目提供者该用户不存在 标签USACO 难度提高+/省选- 提交 讨论 题解 最新讨论 暂时没有讨论 题目背景 在商店中, ...
- UVALive - 6572 Shopping Malls floyd
题目链接: http://acm.hust.edu.cn/vjudge/problem/48416 Shopping Malls Time Limit: 3000MS 问题描述 We want to ...
- Codeforces Gym 100803C Shopping 贪心
Shopping 题目连接: http://codeforces.com/gym/100803/attachments Description Your friend will enjoy shopp ...
- Codeforces Round #332 (Div. 2) A. Patrick and Shopping 水题
A. Patrick and Shopping Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...
- poj 1170 Shopping Offers
Shopping Offers Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4696 Accepted: 1967 D ...
随机推荐
- IDEA通用配置
文件的自动author注释
- JS 的立即执行函数
JS 的立即执行函数 本文写于 2019 年 12 月 7 日 其实 ES6 之后有了之后,很多之前的用法都没必要了,立即执行函数就是其一. 今天看到一道面试题: 请「用自己的语言」简述 立即执行函数 ...
- Kafka 万亿级消息实践之资源组流量掉零故障排查分析
作者:vivo 互联网服务器团队-Luo Mingbo 一.Kafka 集群部署架构 为了让读者能与小编在后续的问题分析中有更好的共鸣,小编先与各位读者朋友对齐一下我们 Kafka 集群的部署架构及服 ...
- Fail2ban 命令详解 fail2ban-server
Fail2ban的服务端操作命令,用于启动一个Fail2ban服务. root@local:~# fail2ban-server --help Usage: /usr/bin/fail2ban-ser ...
- CoaXPress 时间戳 Time Stamping
背景 在CXP2.0之前,CXP没有定义Time Stamping时间戳的概念,但是用户对Time Stamping是有实际需求的,比如我们要对比多台设备拍摄同一个物体不同角度的照片,或者记录触发完成 ...
- 【SpringCloud原理】万字剖析OpenFeign之FeignClient动态代理生成源码
年前的时候我发布两篇关于nacos源码的文章,一篇是聊一聊nacos是如何进行服务注册的,另一篇是一文带你看懂nacos是如何整合springcloud -- 注册中心篇.今天就继续接着剖析Sprin ...
- Navicat可视化MySQL数据库
Navicat可视化MySQL数据库 Navicat内部封装了所有的操作数据库的命令,用户只需要点击操作即可,无需书写sql语句. navicat能够充当多个数据库的客户端. 具体操作参考百度. py ...
- 1.3温度转换(中国大学Mooc-Python 语言程序设计)
温度转换 温度刻画的两种不同体系 1.摄氏度:(中国等世界大多数国家使用) 以1标准大气压下水的结冰点为0度,沸点为100度,将温度进行等分刻画 2.华氏度:(美国.英国等国家使用) 以1标准大气压 ...
- nvm安装与使用及乱码问题
前端开发工作中经常负责多个项目(新项目.多年的老项目及团队合作项目),经常会遇到npm install安装依赖包或者启动本地服务时依赖报错的情况,大多数是因为NodeJS和npm与依赖之间版本的问题, ...
- 五分钟搞懂POM设计模式
转载请注明出处️ 作者:IT小学生蔡坨坨 原文链接:五分钟搞懂POM设计模式 大家好,我是IT小学生蔡坨坨. 今天,我们来聊聊Web UI自动化测试中的POM设计模式. 为什么要用POM设计模式 前期 ...