Codeforces Gym100952 C. Palindrome Again !!-回文字符串 (2015 HIAST Collegiate Programming Contest)

 

C. Palindrome Again !!

 

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

Given string with N characters, your task is to transform it to a palindrome string. It's not as easy as you may think because there is a cost for this transformation!!

First you have to start from character at given position P. From your position you always have 2 options:

- You can move one step to the right or to the left, the cost of each movement is 1. Assume that the string is cyclic, this means if you move one step to the left you will be at position P-1 if P > 1 or at the last character if P = 1, and if you move one step to the right you will be at position P+1 if P < N or at first character if P = N.

- You can change the letter at your current position by replacing it with the next or previous one in the English alphabet (assume that the alphabet is also cyclic so ‘a’ is after ‘z’). The cost of each replacement is also 1.

You should repeat that until the transformation is finished and the string is palindrome. What is the minimum cost to do that?

Input

The first line contains the number of test cases T ( 1  ≤  T  ≤  100 ). Each test case contains 2 lines, the first line contains two integers ( 1  ≤  N  ≤  100,000) the length of string and ( 1  ≤  P  ≤  N ) the initial position. While the second line contains a string with exactly N alphabetical characters.

Output

For each test case output one line contains the minimum cost that is needed to change the string into a palindrome one.

Examples

Input

1
8 3
aeabdaey

Output

8

Note

start with P = 3 ae(a)bdaey, move right => aea(b)daey, change to next => aea(c)daey, change to next => aea(d)deay, move left => ae(a)ddeay, move left => a(e)addeay, move left => (a)eaddeay, change to previous => (z)eaddeay, change to previous => (y)eaddeay. This costs 8 (4 movements and 4 replacements)

题意就是把一个串变成回文的，从指定的位置开始，然后就是可以向字母表左右变换，一个环，就是a->b,或者a->z都是变化1次。每次不变化走过的步数也算在内，问最少要多少次使得这个串变成回文的。

mdzz，写这个题的时候，因为是从0开始存的这个串，所以开始的位置也要-1，但是我是智障啊，忘了。。。

代码的意思就是只改变串的前一半，后一半就不变化了，然后再把走的步数加上就可以。

举个例子，aeabdaey，只看前4个，a和y不一样，要变化，是从a->b->c这样正着往下变呢，还是倒着a->z->y这样变呢，就需要找这两个的变化次数最小的，所以min比较一下。

然后第二个字母是e和倒数第二个字母e一样，不变化，再往下一步，第三个字母a和倒数第三个字母a一样，不变化，第四个字母b和倒数第四个字母d不一样，要变化。

接下来是走的步数怎么算呢。因为是拿前一半比较的，如果指定的位置在后一半，就把这个位置改成前一半，所以if(m>=n/2)m=n-m-1;

然后比较一下，这个变化之后的位置如果是在需要改变的第一个字母的前面，那么走的步数就是从最后一个需要改变的字母的位置-指定的位置就可以。

如果在之间，就是最后一个要改变的字母的位置-第一个要改变的字母的位置，再比较一下，指定的位置是先往左走还是先往右走使得走的步数最少，就需要min比较一下。

如果指定的位置在最后一个要改变的字母的右边，那就是指定的位置-第一个要改变的字母的位置。

其实就是一个区间的问题，很好想的，我这个智障都想明白了(托脸~)。

代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int INF=0x3f3f3f3f;
 4 const int N=1e5+10;
 5 char a[N];
 6 int main(){
 7     int t,n,m;
 8     int minn,maxx,ans;
 9     scanf("%d",&t);
10     while(t--){
11             scanf("%d%d",&n,&m);
12             scanf("%s",a);
13             m--;                                        //指定的位置-1。
14             minn=INF;maxx=-1;ans=0;
15         for(int i=0;i<n/2;i++){                         //只要前一半。
16             if(a[i]!=a[n-i-1]){                         
17               ans+=min(abs(a[i]-a[n-i-1]),26-abs(a[i]-a[n-i-1]));
18               minn=min(minn,i);maxx=max(maxx,i);
19             }
20         }
21         if(ans==0)printf("0
");
22         else{
23             if(m>=n/2)m=n-m-1;                          //吧啦吧啦，看上面的解释。
24             if(m<=minn)ans+=maxx-m;
25             else if(m>minn&&m<maxx)ans+=maxx-minn+min((maxx-m),(m-minn));
26             else if(m>=maxx)ans+=m-minn;
27             printf("%d
",ans);
28         }
29     }
30     return 0;
31 }