贪心+构造 Codeforces Round #277 (Div. 2) C. Palindrome Transformation

题目传送门

 /*
     贪心+构造：因为是对称的，可以全都左一半考虑，过程很简单，但是能想到就很难了
 */
 /************************************************
 Author        :Running_Time
 Created Time  :2015-8-3 9:14:02
 File Name     :B.cpp
  *************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 char s[MAXN];
 int a[MAXN];
 int n, p;

 int main(void)    {       //Codeforces Round #277 (Div. 2) C. Palindrome Transformation
     scanf ("%d%d", &n, &p); scanf ("%s", s + );
     if (p > n / )  p = n - p + ;
     int l = n + , r = ;   int ans = ;
     for (int i=; i<=n/; ++i)    {
         if (s[i] != s[n-i+]) {
             l = min (l, i); r = max (r, i);
             ans += min (abs (s[i] - s[n-i+]),  - abs (s[i] - s[n-i+]));
         }
     }
     if (ans == )   puts ("");
     else    {
         printf ("%d\n", ans + r - l + min (abs (p - l), abs (r - p)));
     }

     return ;
 }

下面的图片更形象点。。。