CF Educational Round 12, B

B. Shopping
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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：
给你一组数字（从1到n，乱序），每次让你找到某一个数字，记录下它的位置i，然后将i加到ans上，然后将这个数字提到队首，于是在新的数字排序中找下一个数，最后输出ans
题解：
模拟整个过程，暴力解法，注意每次初始化小数组储存用户的order

#include<iostream>#include<stdlib.h>#include<string.h>using namespace std;int main(){    int str[103];    int numofcus;    int numoforder;    int numofitem;    int ans=0;    cin>>numofcus>>numoforder>>numofitem;    for(int i=1;i<=numofitem;i++){cin>>str[i];}    int turn=numofcus;    while(turn){        int cusorder[numoforder];        memset(cusorder,0,sizeof(cusorder));        for(int i=0;i<numoforder;i++){            cin>>cusorder[i];        }        for(int i=0;i<numoforder;i++){            for(int j=1;j<=numofitem;j++){                if(cusorder[i]==str[j]){                    ans+=j;                   int temp=str[j];                   for(int k=j;k>1;k--){//重组数组                    str[k]=str[k-1];                   }                    str[1]=temp;                    j=1;                    break;                }            }        }        turn--;    }cout<<ans<<endl;}