Codeforces Round #277 (Div. 2)-C. Palindrome Transformation （贪心）_html/css_WEB-ITnose

Palindrome Transformation

time limit per test

memory limit per test

input

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

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.

解题思路：贪心。只考虑一边的情况即可，以左半为例，不断贪心考虑pos离哪边的不匹配边界近，如此往复即可。

AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cmath>using namespace std;int main(){    int n, pos;    string s;//    freopen("in.txt", "r", stdin);    while(scanf("%d%d", &n, &pos)!=EOF){        pos --;        cin>>s;        int l = 0, r = n/2 - 1;        if(n/2 <= pos)  pos = n - 1 - pos;        while(l < n/2 && s[l] == s[n - 1 - l])  l++;      //找左边不匹配边界        while(r >= 0 && s[r] == s[n - 1 - r])  r--;       //<span style="font-family: Arial, Helvetica, sans-serif;">找左边不匹配边界</span>        int ans = 0;        for(int i=l; i<=r; i++){            int temp = abs(s[i] - s[n-1-i]);            ans += min(temp, 26 - temp);        }        if(l < r)            ans += min(abs(pos - l), abs(r - pos)) + r - l;        else if(l == r)            ans += abs(pos - r);        cout<<ans<<endl;    }    return 0;}