Educational Codeforces Round 12 C. Simple Strings 贪心
C. Simple Strings
题目连接:
http://www.codeforces.com/contest/665/problem/C
Description
zscoder loves simple strings! A string t is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string s. He wants to change a minimum number of characters so that the string s becomes simple. Help him with this task!
Input
The only line contains the string s (1 ≤ |s| ≤ 2·105) — the string given to zscoder. The string s consists of only lowercase English letters.
Output
Print the simple string s' — the string s after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string s' should also consist of only lowercase English letters.
Sample Input
aab
Sample Output
bab
Hint
题意
现在给你一个串,让你使得相邻的字符都不一样,要求修改的字符最少
问你最后的字符串长什么样
题解:
贪心,如果这个位置一样,那就变化就好了~
代码
#include<bits/stdc++.h>
using namespace std;
string s;
int main()
{
cin>>s;
s+='a';
for(int i=1;i<s.size()-1;i++)
{
if(s[i]==s[i-1])
{
for(int j=0;j<26;j++)
{
if(j+'a'==s[i-1]||(j+'a')==s[i+1])
continue;
s[i]=char(j+'a');
break;
}
}
}
for(int i=0;i<s.size()-1;i++)cout<<s[i];
cout<<endl;
}
Educational Codeforces Round 12 C. Simple Strings 贪心的更多相关文章
- Educational Codeforces Round 12 D. Simple Subset 最大团
D. Simple Subset 题目连接: http://www.codeforces.com/contest/665/problem/D Description A tuple of positi ...
- Educational Codeforces Round 12 B C题、
B. Shopping 题意:n个顾客,每个顾客要买m个物品,商场总共有k个物品,看hint就只知道pos(x)怎么算了,对于每一个Aij在k个物品中找到Aij的位置.然后加上这个位置对于的数值,然后 ...
- Codeforces Educational Codeforces Round 3 C. Load Balancing 贪心
C. Load Balancing 题目连接: http://www.codeforces.com/contest/609/problem/C Description In the school co ...
- Educational Codeforces Round 2 C. Make Palindrome 贪心
C. Make Palindrome Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/600/pr ...
- Educational Codeforces Round 12 F. Four Divisors 求小于x的素数个数(待解决)
F. Four Divisors 题目连接: http://www.codeforces.com/contest/665/problem/F Description If an integer a i ...
- Educational Codeforces Round 2 C. Make Palindrome —— 贪心 + 回文串
题目链接:http://codeforces.com/contest/600/problem/C C. Make Palindrome time limit per test 2 seconds me ...
- Educational Codeforces Round 12 E. Beautiful Subarrays 预处理+二叉树优化
链接:http://codeforces.com/contest/665/problem/E 题意:求规模为1e6数组中,连续子串xor值大于等于k值的子串数: 思路:xor为和模2的性质,所以先预处 ...
- Educational Codeforces Round 17 C. Two strings 打表二分
C. Two strings time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...
- Educational Codeforces Round 12 E. Beautiful Subarrays 字典树
E. Beautiful Subarrays 题目连接: http://www.codeforces.com/contest/665/problem/E Description One day, ZS ...
随机推荐
- 记一个logrotate的配置文件权限问题
问题描述 从git仓库更新了别人配置好的logrotate,发现不能正常运行.手工执行报错 error: Ignoring syslog because of bad file mode - must ...
- 谁说运维用ELK没用?我就说很有用,只是你之前不会用【转】
1.安装JDK 1)登陆ORACLE官网 (http://www.oracle.com/technetwork/java/javase/downloads/index-jsp-138363.html ...
- JVM的分区+查看GC对象是否存活+3种GC算法+7种垃圾收集器+如何减少GC次数
一.JVM的分区: 1.程序计数器(私有) 程序计数器是一块较小的内存分区,你可以把它看做当前线程所执行的字节码的指示器. 在虚拟机的概念模型里,字节码解释器工作时,就是通过改变计数器的值来选择下 ...
- Mac 终端自动补全忽略大小写
打开终端,输入:nano .inputrc 在里面粘贴上以下语句: set completion-ignore-case onset show-all-if-ambiguous onTAB: menu ...
- c++鼠标点点,获取坐标值,放入到txt文件中
// oj3.cpp : Defines the entry point for the console application.// #include "stdafx.h"#in ...
- 创建 dblink
目的:oracle中跨数据库查询 两台数据库服务器db_A(本地)和db_B(远程192.168.1.100),db_A下用户user_a 需要访问到db_B下user_b的数据解决:查询 ...
- 删除WP提示:自动升级WordPress失败
wordpress后台总有烦人的升级失败的提示,查了半天找不到怎么去掉:“自动升级WordPress失败--请再试一次”这个提示的方法,特意分享出来 方法/步骤 1 打开wordpress根目录找 ...
- phpmailer发送邮件出现错误:stream_socket_enable_crypto(): SSL operation failed with code 1.
如果开了调试,调试进去会看到错误提示: smtp_code:"stream_socket_enable_crypto(): SSL operation failed with code 1. ...
- nginx配置tomcat集群
显示nginx的核心配置 #user nobody;worker_processes 1; events { worker_connections 1024; #并发连接数} http { in ...
- Scrapy 增加随机请求头 user_agent
原文: 为什么要增加随机请求头:更好地伪装浏览器,防止被 Ban. 如何在每次请求时,更换不同的 user_agent,Scrapy 使用 Middleware 即可 Spider 中间件 (Midd ...