Palindrome Transformation
2015-07-16 21:54
323 查看
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
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.
题意:使原文成为回文串的最小操作数。
题解:使原文成为回文串,因为是对称的,我们都在左边操作。然后根据要操作的最左端和要操作的最右端,哪边小就走哪边。
代码:
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 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:使原文成为回文串的最小操作数。
题解:使原文成为回文串,因为是对称的,我们都在左边操作。然后根据要操作的最左端和要操作的最右端,哪边小就走哪边。
代码:
#include <stdio.h> #include <string.h> #define maxn 100009 int abs(int x) { return x >= 0 ? x : -x; } int main() { int n, p, i, j; char str[maxn]; int visited[maxn]; scanf("%d%d", &n, &p); scanf("%s", str+1); for(int i = 1; i <= n;i++) visited[i] = 0; int count = 0; int mid; if(n%2) { mid = n / 2 + 1; for(i = mid - 1,j = mid + 1;i!=0;i--,j++) { if(str[i]!=str[j]) { visited[i] = 1; int temp = abs(str[i] - str[j]); if(temp >= 13) temp = 26 - temp; count += temp; } } } else { mid = n / 2; for(i = mid,j = mid + 1;i!=0;i--,j++) { if(str[i]!=str[j]) { visited[i] = 1; int temp = abs(str[i] - str[j]); if(temp >= 13) temp = 26 - temp; count += temp; } } } //printf("%d\n",count); if(p>n/2) p = 1 + n - p; int left = 0; int right = 0; for(i = 1;i<p;i++) { if(visited[i] == 1) { left = p - i ; break; } } for(i = mid;i>p;i--) { if(visited[i] == 1) { right = i - p; break; } } if(left < right) count += 2 * left + right; else count += 2 * right + left; printf("%d\n",count); //for(int i =1;i<=n;i++) // printf("%d",visited[i]); }
相关文章推荐
- VIM Tutorial
- (七十二)自定义通知NSNotification实现消息传递
- oracle用户和表空间
- linux之配置IP地址
- ThreadPoolExecutor运转机制详解
- 怎样更健康解决一个人时的性需求
- leetcode83 Remove Duplicates from Sorted List
- poj1236 强连通分量+缩点
- Objective-C编码规范:26个方面解决iOS开发问题
- mysql或则oracle的declare可以放在操作语句的后面吗?
- Mac新手操作指南(五)
- java EE之jsp的7个动作指令 复习
- vimrc for windows
- AFNetworking2.0源码解析AFURLResponseSerialization
- 架构师速成6.6-知识的收集整理学习
- AIX PASE 和unix
- 架构师速成6.6-知识的收集整理学习 分类: 架构师速成 2015-07-16 21:50 254人阅读 评论(0) 收藏
- 指针注意
- Android Studio使用Git@SC分享项目
- 开发基于 IBM Lotus Domino 的 Web 2.0 应用的最佳实践