CodeForces765B
B. Code obfuscation
Kostya likes Codeforces contests very much. However, he is very disappointed that his solutions are frequently hacked. That's why he decided to obfuscate (intentionally make less readable) his code before upcoming contest.
To obfuscate the code, Kostya first looks at the first variable name used in his program and replaces all its occurrences with a single symbol a, then he looks at the second variable name that has not been replaced yet, and replaces all its occurrences with b, and so on. Kostya is well-mannered, so he doesn't use any one-letter names before obfuscation. Moreover, there are at most 26 unique identifiers in his programs.
You are given a list of identifiers of some program with removed spaces and line breaks. Check if this program can be a result of Kostya's obfuscation.
Input
In the only line of input there is a string S of lowercase English letters (1 ≤ |S| ≤ 500) — the identifiers of a program with removed whitespace characters.
Output
If this program can be a result of Kostya's obfuscation, print "YES" (without quotes), otherwise print "NO".
Examples
input
abacaba
output
YES
input
jinotega
output
NO
Note
In the first sample case, one possible list of identifiers would be "number string number character number string number". Here how Kostya would obfuscate the program:
- replace all occurences of number with a, the result would be "a string a character a string a",
- replace all occurences of string with b, the result would be "a b a character a b a",
- replace all occurences of character with c, the result would be "a b a c a b a",
- all identifiers have been replaced, thus the obfuscation is finished.
//2017-02-14
#include <iostream>
#include <cstdio>
#include <cstring> using namespace std; int book[]; int main()
{
string S;
char cur_max;
bool fg;
while(cin>>S)
{
fg = true;
memset(book, , sizeof(book));
for(int i = ; i < S.length(); i++)
book[S[i]-'a']++;
for(int i = ; i < ; i++)
if(book[i]!= && book[i-]==)
{
fg = false;
break;
}
cur_max = S[];
if(cur_max != 'a')fg = false;
for(int i = ; i < S.length(); i++)
{
if(S[i]>cur_max && S[i]-cur_max==)cur_max = S[i];
else if(S[i]>cur_max && S[i]-cur_max>){
fg = false;
break;
}
}
if(fg)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
} return ;
}
CodeForces765B的更多相关文章
- WSDL协议简单介绍
WSDL – WebService Description Language – Web服务描述语言 通过XML形式说明服务在什么地方-地址. 通过XML形式说明服务提供什么样的方法 – 如何调用. ...
随机推荐
- poj3233 Matrix Power Series(矩阵快速幂)
题目要求的是 A+A2+...+Ak,而不是单个矩阵的幂. 那么可以构造一个分块的辅助矩阵 S,其中 A 为原矩阵,E 为单位矩阵,O 为0矩阵 将 S 取幂,会发现一个特性: Sk +1右上角 ...
- HDU 1710Binary Tree Traversals(已知前序中序,求后序的二叉树遍历)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1710 解题思路:可以由先序和中序的性质得到 : 先序的第一个借点肯定是当前子树的根结点, 那么在 中序 ...
- Android之常用类库
Android之常用类库 android.app :提供高层的程序模型.提供基本的运行环境android.content :包含各种的对设备上的数据进行访问和发布的类android.database ...
- 在android应用程序中启动其他apk程序
Android 开发有时需要在一个应用中启动另一个应用,比如Launcher加载所有的已安装的程序的列表,当点击图标时可以启动另一个应用. 一般我们知道了另一个应用的包名和MainActivity的名 ...
- 【NOIP2018】保卫王国 动态dp
此题场上打了一个正确的$44pts$,接着看错题疯狂$rush$“正确”的$44pts$,后来没$rush$完没将之前的代码$copy$回去,直接变零分了..... 这一题我们显然有一种$O(nm)$ ...
- 使用R进行分组统计
分组统计数据集是很常见的需求,R中也有相应的包支持数据集的分组统计.自己尝试了写了段R代码来完成分组统计数据集,支持公式,感觉用起来还算方便.代码分享在文章最后. 使用方式: step 1: sour ...
- C语言-apache mod(模块开发)-采用VS2017开发实战(windows篇)
C语言-apache mod(模块开发)-采用VS2017开发实战(windows篇) 名词解释:apxs apxs is a tool for building and installing ext ...
- Class与Style绑定
本文主要介绍如何使用Vue来绑定操作元素的class列表和内联样式(style属性). 因为class和style都是属性,所以通过v-bind命令来处理它们:只需要通过表达式计算出结果即可,不过字符 ...
- SQL查询排名第二名的信息
今天看见同学去面试的面试题,查询出某个字段排名第二名的信息,自己就看看 如果是Oracle ,这不就是考察Oracle分页么,以Oracle的emp表为例,根据薪水排名,查询排名第二的员工信息: se ...
- Postman—命令执行脚本及生成报告
前言 前面的应用中,都是在postman图形界面工具里面进行测试,但是有时候我们需要把测试脚本集成到CI平台,或者在非图形界面的系统环境下进行测试,那么我们该如果处理呢 通过newman来执行post ...