我的天，又FST在 long long piapiapia 扇死
题意：（我研究了半天才看懂）
第一行输入应用总数 n ，要一次打开应用的总数 ,每个屏幕最多的应用数。
第二行按位子（1~n)顺序输入应用的编号;
第三行输入要依次打开的应用。
问我们要求需要几次完成所有操作，打开所有应用。
注意：每次打开一个应用、该应用的位置就向前移动一位。最开始，屏幕显示在第一页，每次打开应用后，会回到第一页。
思路：稍微优化一下开两个数组换来换去，每换一次记录一下值就解决了。
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.
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.
Print a single number — the number of gestures that Anya needs to make to launch all the applications in the desired order.
8 3 3 1 2 3 4 5 6 7 8 7 8 1
7
5 4 2 3 1 5 2 4 4 4 4 4
8
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;
}