题目链接:Codeforces 486C Palindrome Transformation 题目大意:给定一个字符串,长度N.指针位置P,问说最少花多少步将字符串变成回文串. 解题思路:事实上仅仅要是对称位置不同样的.那么指针肯定要先移动到这里,改动字符仅仅须要考虑两种方向哪种更优即 可. 然后将全部须要到达的位置跳出来.贪心处理. #include <cstdio> #include <cstring> #include <cstdlib> #include <…

点击打开链接 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 m…

题目:戳我 题意:给定长度为n的字符串,给定初始光标位置p,支持4种操作,left,right移动光标指向,up,down,改变当前光标指向的字符,输出最少的操作使得字符串为回文. 分析:只关注字符串n/2长度,up,down操作是固定不变的,也就是不能优化了,剩下就是left,down的操作数,细想下中间不用管,只关注从左到中间第一个要改变的位置和最后一个要改变的位置即可,具体看代码. #include <iostream> #include <cstdio> #include…

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 meaningl…

题目链接:http://codeforces.com/problemset/problem/486/C 题目意思:给出一个含有 n 个小写字母的字符串 s 和指针初始化的位置(指向s的某个字符).可以对s进行四种操作:up,down,left,right.up/down是令到对称位置的字符相同所进行的操作次数.假设s[i] != s[j](i, j是对称的,假设分别是a, k),up: a(位置1) 根据字母表顺序变成 k(位置11) 需要 10 次(直接a->k).down:a 根据 字母表逆…

C. Palindrome Transformation     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 arro…

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 &…

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…

codeforces 704B - Ant Man 贪心 题意:n个点,每个点有5个值,每次从一个点跳到另一个点,向左跳:abs(b.x-a.x)+a.ll+b.rr 向右跳:abs(b.x-a.x)+a.lr+b.rl,遍历完所有的点,问你最后的花费是多少 思路:每次选一个点的时候,在当前确定的每个点比较一下,选最短的距离. 为什么可以贪心?应为答案唯一,那么路径必定是唯一的,每个点所在的位置也一定是最短的. #include <bits/stdc++.h> using namespace…

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 meaningl…