codeforces 486c-Palindrome Transformation【贪心】

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.

Examples

input

8 3
aeabcaez

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.

题意：
给你一个字符串，通过right、left（指针移动）、up、down（字符变化）将它改成一个回文字符串。现在让你求最小的移动次数。
思路：
因为每次操作都是移动光标或者改变字符串，所以只需要字符前半部分和后半部分进行比较。首先把需要翻转字符的次数记录下来。如果光标不需要来回，一次就可以完成，否则，光标一定是先移动完一边在去另一边，求出最小值就可以。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
//typedef long long LL;
typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 10;
char str[MAXN];
int main() {
    int n, p;
    while (scanf("%d%d", &n, &p) != EOF) {
        scanf("%s", str + 1);
        int psum = 0, las, fir;
        bool flag = true;
        //把p换到左边
        if(p > n/2)  p = n + 1 - p;
        for (int i = 1; i <= n/2; i++) {
            int op = abs(str[i] - str[n + 1 - i]);
            if (op) {
                if (flag) {
                    fir = i; flag = false;
                }
                psum += min(op, 26 - op);
                las = i;
            }
        }
        if (psum == 0) printf("0
");
        else if (fir >= p) printf("%d
", las - p + psum);
        else if (las <= p) printf("%d
", p - fir + psum);
        else{
            printf("%d
", min(2*(p - fir) + las - p, 2*(las - p) + p - fir) + psum);
        }
    }
    return 0;
}