codeforces 665B B. Shopping(水题)

题目链接：

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 n, m 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

Note

Customer 1 wants the items 1 and 5.

pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].

pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].

Time taken for the first customer is 3 + 5 = 8.

Customer 2 wants the items 3 and 1.

pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].

pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].

Time taken for the second customer is 3 + 3 = 6.

Total time is 8 + 6 = 14.

Formally pos(x) is the index of x in the current row.

题意：

找到所有顾客的所有物品的位置的和,每找到一件物品就把这件物品拿到最前面;

思路：

暴力求解;

AC代码：

/*2014300227    665B - 11    GNU C++11    Accepted    15 ms    2176 KB*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
const ll mod=1e9+7;
int n,m,fk,a[200],x;
int main()
{
    scanf("%d%d%d",&n,&m,&fk);
    for(int i=1;i<=fk;i++)
    {
        scanf("%d",&a[i]);
    }
    int ans=0;
    while(n--)
    {
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&x);
            for(int j=1;j<=fk;j++)
            {
                if(a[j]==x)
                {
                    ans+=j;
                    for(int k=j;k>1;k--)
                    {
                        a[k]=a[k-1];
                    }
                    a[1]=x;
                    break;
                }
            }
        }
    }
    printf("%d
",ans);
    return 0;
}