Tyndex.Brome

题意翻译

### 题目描述 每个测试字符串$S$都有一个**独立**的$ans$。 对于测试字符串$S$的每个字母$S_{i}$，可以找到标准字符串$C$中字母$S_{i}$的离$i$的最近位置$j$。将位置的绝对差$\left|i-j \right|$加到$ans$中。也就是说对于每个$i$，取位置$j$，使得 $S_i = C_j$ 且$\left|i-j \right|$最小。 如果标准字符串中不存在字母$C_i$，则将测试字符串$S$的长度添加到$ans$。 ### 输入格式 第$1$行包含两个整数$n$和$k$$\left( n,k\leq 10^5 \right)$。$n$是测试字符串的数目，$k$是标准字符串的长度。 第$2$行是长度为$k$且字符均为小写字母的标准字符串。 第$3$至$n+3$行每行输入一个测试字符串。保证所有字符串的总长度不超过$2\times10^5$。 ### 输出格式 $n$行，每行输出一个数字代表$ans$。

题目描述

Tyndex is again well ahead of the rivals! The reaction to the release of Zoozle Chrome browser was the release of a new browser Tyndex.Brome! The popularity of the new browser is growing daily. And the secret is not even the Tyndex.Bar installed (the Tyndex.Bar automatically fills the glass with the finest 1664 cognac after you buy Tyndex.Bottles and insert in into a USB port). It is highly popular due to the well-thought interaction with the user. Let us take, for example, the system of automatic address correction. Have you entered codehorses instead of codeforces? The gloomy Zoozle Chrome will sadly say that the address does not exist. Tyndex.Brome at the same time will automatically find the closest address and sent you there. That's brilliant! How does this splendid function work? That's simple! For each potential address a function of the $ F $ error is calculated by the following rules: - for every letter $ c_{i} $ from the potential address $ c $ the closest position $ j $ of the letter $ c_{i} $ in the address ( $ s $ ) entered by the user is found. The absolute difference $ |i-j| $ of these positions is added to $ F $ . So for every $ i $ ( $ 1<=i<=|c| $ ) the position $ j $ is chosen such, that $ c_{i}=s_{j} $ , and $ |i-j| $ is minimal possible. - if no such letter $ c_{i} $ exists in the address entered by the user, then the length of the potential address $ |c| $ is added to $ F $ . After the values of the error function have been calculated for all the potential addresses the most suitable one is found. To understand the special features of the above described method better, it is recommended to realize the algorithm of calculating the $ F $ function for an address given by the user and some set of potential addresses. Good luck!

输入输出格式

输入格式

The first line contains two integers $ n $ and $ k $ ( $ 1<=n<=10^{5},1<=k<=10^{5} $ ). They are the number of potential addresses and the length of the address entered by the user. The next line contains $ k $ lowercase Latin letters. They are the address entered by the user ( $ s $ ). Each next $ i $ -th ( $ 1<=i<=n $ ) line contains a non-empty sequence of lowercase Latin letters. They are the potential address. It is guaranteed that the total length of all the lines does not exceed $ 2·10^{5} $ .

输出格式

On each $ n $ line of the output file print a single number: the value of the error function when the current potential address is chosen. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

输入输出样例

输入样例 #1

2 10
codeforces
codeforces
codehorses

输出样例 #1

0
12

输入样例 #2

9 9
vkontakte
vcontacte
vkontrakte
vkollapse
vkrokodile
vtopke
vkapuste
vpechke
vk
vcodeforcese

输出样例 #2

18
14
36
47
14
29
30
0
84

题意翻译

### 题目描述 每个测试字符串$S$都有一个**独立**的$ans$。 对于测试字符串$S$的每个字母$S_{i}$，可以找到标准字符串$C$中字母$S_{i}$的离$i$的最近位置$j$。将位置的绝对差$\left|i-j \right|$加到$ans$中。也就是说对于每个$i$，取位置$j$，使得 $S_i = C_j$ 且$\left|i-j \right|$最小。 如果标准字符串中不存在字母$C_i$，则将测试字符串$S$的长度添加到$ans$。 ### 输入格式 第$1$行包含两个整数$n$和$k$$\left( n,k\leq 10^5 \right)$。$n$是测试字符串的数目，$k$是标准字符串的长度。 第$2$行是长度为$k$且字符均为小写字母的标准字符串。 第$3$至$n+3$行每行输入一个测试字符串。保证所有字符串的总长度不超过$2\times10^5$。 ### 输出格式 $n$行，每行输出一个数字代表$ans$。