Codeforces 486 C Palindrome Transformation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.

There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n, the cursor appears at the beginning of the string).

When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.

Initially, the text cursor is at position p.

Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 105) and p (1 ≤ p ≤ n), the length of Nam's string and the initial position of the text cursor.

The next line contains n lowercase characters of Nam's string.

Output

Print the minimum number of presses needed to change string into a palindrome.

Sample test(s)

input

8 3aeabcaez

output

6

Note

A string is a palindrome if it reads the same forward or reversed.

In the sample test, initial Nam's string is:  (cursor position is shown bold).

In optimal solution, Nam may do 6 following steps:

The result, , is now a palindrome.

分情况考虑，也就是分治了

虽然一发过了，但是感觉想清楚还是有些难度

一边写才一边想清楚的

最后的结论：答案就是从p走过一个区间花费的的最小代价，加上改变字符的代价

#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<climits>#include<list>#include<iomanip>#include<stack>#include<set>using namespace std;int n,p;int cnt(int l,int r){int ans=INT_MAX;if(p<l){ans=min(ans,p+n-l);ans=min(ans,r-p);}else if(p>r){ans=min(ans,p-l);ans=min(ans,n-p+r);}else{ans=min(ans,p-l+r-l);ans=min(ans,r-p+r-l);}return ans;}int main(){cin>>n>>p;string s;cin>>s;s="a"+s;int l=0,m=n/2;for(int i=1;i<=m;i++)if(s[i]!=s[n-i+1]){l=i;break;}if(l==0){cout<<0;return 0;}int r;for(int i=m;i>=l;i--)if(s[i]!=s[n-i+1]||i==l){r=i;break;}int ans=min(cnt(l,r),cnt(n-r+1,n-l+1));for(int i=l;i<=r;i++)if(s[i]!=s[n-i+1]){int t=max(s[i],s[n-i+1])-min(s[i],s[n-i+1]);ans+=min(t,26-t);}cout<<ans;}