# CodeForces 665B Shopping

B. Shopping

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 aij 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 aij (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(ai1) + pos(ai2) + ... + pos(aim) 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.

Input

The first line contains three integers nm and 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 distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.

Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.

Output

Print the only integer t — the total time needed for Ayush to process all the orders.

Example

input

```2 2 5
3 4 1 2 5
1 5
3 1```

output

14

```#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
int n,m,k;
int tag[105];
int a[105];
int main()
{
int x;
scanf("%d%d%d",&n,&m,&k);
for(int i=1;i<=k;i++)
{
scanf("%d",&a[i]);
tag[a[i]]=i;
}
int ans=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
scanf("%d",&x);
ans+=tag[x];
for(int p=1;p<=k;p++)
{
if(tag[a[p]]<tag[x])
tag[a[p]]++;
}
tag[x]=1;
}
}
printf("%d\n",ans);
return 0;

}```

0 条评论

• ### HOJ-2056 Bookshelf（线性动态规划）

L is a rather sluttish guy. He almost never clean up his surroundings or regulat...

• ### POJ-1953 World Cup Noise（线性动规）

World Cup Noise Time Limit: 1000MS Memory Limit: 30000K Total Submissio...

• ### ZOJ 3202 Second-price Auction

Time Limit: 1 Second      Memory Limit: 32768 KB

• ### POJ-1953 World Cup Noise（线性动规）

World Cup Noise Time Limit: 1000MS Memory Limit: 30000K Total Submissio...

• ### ZOJ 3623 Battle Ships

Battle Ships Time Limit: 2 Seconds Memory Limit: 65536 KB Battle Ships is a ne...

• ### Gazebo機器人仿真學習探索筆記（一）安裝與使用

Gazebo提供了多平臺的安裝和使用支持，大部分主流的linux，Mac以及Windows，這裏結合ROS以Ubuntu爲例進行介紹。

• ### 什么是SAP物料主数据里的Batch

Materials are produced and theoretically have the same properties. Nevertheless ...

• ### 如何启用SAP Business by design里的Correction Invoice功能

Subject: [Tip] How to enable the function Correction Invoice for customer invoic...

• ### How to find “hidden” remote jobs using Google Search.

By using a special search operator with Google search, you can find remote jobs ...