coderforces 721b

题目描述：

B. Passwords

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.

Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.

Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.

Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.

The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.

The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.

Output

Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.

Examples

input

5 2
cba
abc
bb1
abC
ABC
abc

output

1 15

input

4 100
11
22
1
2
22

output

3 4

Note

Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2and only then the right one, spending 4 seconds at all.

 题目大意：一些列字符串，输入一次时间加一秒，当错误次数达到k时，必须等待5秒才能继续输入下一次;求输入正确密码最短时间和最长时间。

总结：本题卡了一阵主要是一下3个方面：

1.题目描述的是: some of his n passwords.也就是说正确的密码可能会出现好几次(没仔细看题目惹的祸，上次也有这种情况，应多注意)，而我认为只出现一次

2.if 之后的下一个判断没有加else，直接用的if -if导致第一个if可能改变了变量，使得满足第二个if的条件。应该if-else if(考虑不全面)

3.当出现正确密码时lcnt- 1;但是这样可能会导致负数，所以以后涉及减法时就应该考虑是否可能为负数(后来把lcnt设置成总数就不用考虑了)

    #include<stdio.h>
    #include<stdlib.h>
    #define _max 105
    char str[_max][_max];
    int main()
    {
    int len[_max];
    char r_pass[_max];
    int i = 0, n = 0, k = 0;
    int r_len = 0,scnt = 0,lcnt = 0,tt = 0;
    scanf("%d%d",&n,&k);
    for( i = 0; i < n; i++)
    {
    scanf("%s",str[i]);
    len[i] = strlen(str[i]);
    }
    scanf("%s",r_pass);
    r_len = strlen(r_pass);

    for( i = 0; i < n; i++)
    {
    if(len[i] < r_len)
    scnt++;
    if(len[i] == r_len)
    {
    lcnt++;
    if(strcmp(str[i],r_pass) == 0)
    tt++;
    }


    }
    lcnt+=scnt;//lcnt变成了总数
    lcnt-=(tt -1);//减去重复的正确密码次数
    if(scnt < k)
    scnt =scnt + 1;
    else if(scnt == k)
    scnt = scnt + 6;
    else if(scnt > k)
    {
    scnt = scnt + (scnt / k)*5  + 1;
    }
    if(lcnt <= k)
    lcnt = lcnt;

    else if(lcnt > k)
    {
    if(lcnt %k == 0)
    lcnt = lcnt + (lcnt/k - 1)*5;//当第n*k次时正好是正确密码，此时不用再等5秒了
    else
    lcnt = lcnt + (lcnt/k)*5;
    }
    printf("%d %d
",scnt,lcnt);

    }