我的天，又FST在  long long piapiapia 扇死 

題意：（我研究了半天才看懂）

第一行輸入應用總數 n ，要一次開啟應用的總數 ,每個螢幕最多的應用數。

第二行按位子（1~n)順序輸入應用的編號;

第三行輸入要依次開啟的應用。

問我們要求需要幾次完成所有操作，開啟所有應用。

注意：每次開啟一個應用、該應用的位置就向前移動一位。最開始，螢幕顯示在第一頁，每次開啟應用後，會回到第一頁。

思路：稍微優化一下開兩個陣列換來換去，每換一次記錄一下值就解決了。

C. Anya and Smartphone time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Anya has bought a new smartphone that uses Berdroid

 operating system. The smartphone menu has exactly n applications, each application has its own icon. The icons are located on different screens, one screen contains k icons. The icons from the first to the k-th one are located on the first screen, from the (k + 1)-th to the 2k-th ones are on the second screen and so on (the last screen may be partially empty).

Initially the smartphone menu is showing the screen number 1. To launch the application with the icon located on the screen t, Anya needs to make the following gestures: first she scrolls to the required screen number t

, by making t - 1 gestures (if the icon is on the screen t), and then make another gesture — press the icon of the required application exactly once to launch it.

After the application is launched, the menu returns to the first screen. That is, to launch the next application you need to scroll through the menu again starting from the screen number 1.

All applications are numbered from 1 to n. We know a certain order in which the icons of the applications are located in the menu at the beginning, but it changes as long as you use the operating system. Berdroid is intelligent system, so it changes the order of the icons by moving the more frequently used icons to the beginning of the list. Formally, right after an application is launched, Berdroid swaps the application icon and the icon of a preceding application (that is, the icon of an application on the position that is smaller by one in the order of menu). The preceding icon may possibly be located on the adjacent screen. The only exception is when the icon of the launched application already occupies the first place, in this case the icon arrangement doesn't change.

Anya has planned the order in which she will launch applications. How many gestures should Anya make to launch the applications in the planned order?

Note that one application may be launched multiple times.

Input

The first line of the input contains three numbers n, m, k (1 ≤ n, m, k ≤ 105) — the number of applications that Anya has on her smartphone, the number of applications that will be launched and the number of icons that are located on the same screen.

The next line contains n integers, permutation a1, a2, ..., an — the initial order of icons from left to right in the menu (from the first to the last one), ai —  is the id of the application, whose icon goes i-th in the menu. Each integer from 1 to n occurs exactly once among ai.

The third line contains m integers b1, b2, ..., bm(1 ≤ bi ≤ n) — the ids of the launched applications in the planned order. One application may be launched multiple times.

Output

Print a single number — the number of gestures that Anya needs to make to launch all the applications in the desired order.

Sample test(s) input

8 3 3
1 2 3 4 5 6 7 8
7 8 1

output

7

input

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

output

8

Note

In the first test the initial configuration looks like (123)(456)(78), that is, the first screen contains icons of applications 1, 2, 3, the second screen contains icons 4, 5, 6, the third screen contains icons 7, 8.

After application 7 is launched, we get the new arrangement of the icons — (123)(457)(68). To launch it Anya makes 3 gestures.

After application 8 is launched, we get configuration (123)(457)(86). To launch it Anya makes 3 gestures.

After application 1 is launched, the arrangement of icons in the menu doesn't change. To launch it Anya makes 1 gesture.

In total, Anya makes 7 gestures.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;

int main()
{
     int pos[100005] , app[100005] , n , m , k , launch;
     long long ans = 0 ;
     cin >> n >> m >> k;
     for(int i = 1 ; i <= n ; i++)
     {
          cin >> pos[i];
          app[ pos[i] ] = i;
     }
     
     for(int i = 0 ; i < m ; i++)
     {
          cin >> launch;
          int pa = app[launch];
          ans += ceil(1.0 * pa / k);
          if(pa != 1)
          {
               swap(pos[pa] , pos[pa - 1]);
               swap(app[ pos[pa] ] , app[ pos[pa - 1] ] ) ;
          }
     }
     cout << ans;
     return 0;
}