Codeforces Round #256 (Div. 2) B
Bizon the Champion isn't just a bison. He also is a favorite of the "Bizons" team.
At a competition the "Bizons" got the following problem: "You are given two distinct words (strings of English letters), s and t.
You need to transform word s into word t". The task
looked simple to the guys because they know the suffix data structures well. Bizon Senior loves suffix automaton. By applying it once to a string, he can remove from this string any single character. Bizon Middle knows suffix array well. By applying it once
to a string, he can swap any two characters of this string. The guys do not know anything about the suffix tree, but it can help them do much more.
Bizon the Champion wonders whether the "Bizons" can solve the problem. Perhaps, the solution do not require both data structures. Find out whether the guys can solve the problem and if they can, how do they do it? Can they solve it either only with use of suffix
automaton or only with use of suffix array or they need both structures? Note that any structure may be used an unlimited number of times, the structures may be used in any order.
The first line contains a non-empty word s. The second line contains a non-empty word t.
Words s and t are different. Each word consists
only of lowercase English letters. Each word contains at most 100 letters.
In the single line print the answer to the problem. Print "need tree" (without the quotes) if word s cannot
be transformed into word teven with use of both suffix array and suffix automaton. Print "automaton"
(without the quotes) if you need only the suffix automaton to solve the problem. Print "array" (without the quotes) if you need only the suffix array to solve
the problem. Print "both" (without the quotes), if you need both data structures to solve the problem.
It's guaranteed that if you can solve the problem only with use of suffix array, then it is impossible to solve it only with use of suffix automaton. This is also true for suffix automaton.
automaton
tomat
automaton
array
arary
array
both
hot
both
need
tree
need tree
AC代码:
#include<cstdio>
#include<cstring>
int n,i,j,l1,l2,a[26],b[26];
char s[101],t[101];
int main()
{
scanf("%s",s);
scanf("%s",t);
l1=strlen(s);
l2=strlen(t);
for(i=0;i<l1;++i)++a[s[i]-'a'];
for(i=0;i<l2;++i)++b[t[i]-'a'];
for(i=0;i<26;++i)if(a[i]<b[i])break;
if(i<26)
printf("need tree\n");
else{
for(i=j=0;i<l2&&j<l1;++i,++j)while(j<l1&&s[j]!=t[i])++j;
printf("%s\n",i<l2?(l1==l2?"array":"both"):"automaton");
}
return 0;
}
Codeforces Round #256 (Div. 2) B的更多相关文章
- Codeforces Round #256 (Div. 2) D. Multiplication Table(二进制搜索)
转载请注明出处:viewmode=contents" target="_blank">http://blog.csdn.net/u012860063?viewmod ...
- Codeforces Round #256 (Div. 2) B. Suffix Structures(模拟)
题目链接:http://codeforces.com/contest/448/problem/B --------------------------------------------------- ...
- Codeforces Round #256 (Div. 2/B)/Codeforces448B_Suffix Structures(字符串处理)
解题报告 四种情况相应以下四组数据. 给两字符串,推断第一个字符串是怎么变到第二个字符串. automaton 去掉随意字符后成功转换 array 改变随意两字符后成功转换 再者是两个都有和两个都没有 ...
- Codeforces Round #256 (Div. 2) 题解
Problem A: A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standar ...
- Codeforces Round #256 (Div. 2) A. Rewards
A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standard input out ...
- Codeforces Round #256 (Div. 2) D. Multiplication Table 二分法
D. Multiplication Table time limit per test 1 second memory limit per test 256 megabytes input st ...
- Codeforces Round #256 (Div. 2) D. Multiplication Table
主题链接:http://codeforces.com/contest/448/problem/D 思路:用二分法 code: #include<cstdio> #include<cm ...
- Codeforces Round #256 (Div. 2/A)/Codeforces448A_Rewards(水题)
解题报告 意思就是说有n行柜子,放奖杯和奖牌.要求每行柜子要么全是奖杯要么全是奖牌,并且奖杯每行最多5个,奖牌最多10个. 直接把奖杯奖牌各自累加,分别出5和10,向上取整和N比較 #include ...
- Codeforces Round #256 (Div. 2) D. Multiplication Table 很有想法的一个二分
D. Multiplication Table time limit per test 1 second memory limit per test 256 megabytes input stand ...
- Codeforces Round #256 (Div. 2)A-D
题目连接:http://codeforces.com/contest/448 A:给你一些奖杯与奖牌让你推断能不能合法的放在给定的架子上.假设能够就是YES否则就是NO. <span style ...
随机推荐
- 基于Angular4+ server render(服务端渲染)开发教程
目标: 1.更好的 SEO,方便搜索爬虫抓取页面内容 2.更快的内容到达时间(time-to-content) 影响: 1.用户:比原来更快的看到渲染的页面,提升用户体验 2.开发人员:某些代码可能需 ...
- [Apple开发者帐户帮助]六、配置应用服务(2)创建DeviceCheck私钥
要验证与DeviceCheck服务的通信,您将使用启用了DeviceCheck的私钥. 首先创建并下载启用了DeviceCheck 的私钥.然后获取密钥标识符(kid)以创建JSON Web令牌(JW ...
- bitmap实现背景透明
近日在项目中,一直被一个问题搞得头大的很,美工要把按钮图片弄成不规则的,但是在winform里实现又不仅仅是使用简单的png图片而已.在网上找到一些方法,稍微改了一点加工成项目所需. 贴出解决方案,以 ...
- SQLServer2008 关于数据转换
全进位 select cast(ceiling(2.1111) as dec(18,0)) 结果:3
- 通过getSystemServices获取手机管理大全
getSystemService是Android很重要的一个API,它是Activity的一个方法,根据传入的NAME来取得对应的Object,然后转换成相应的服务对象.以下介绍系统相应的服务. 传入 ...
- vsftp服务器搭建
1.FTP的主动模式和被动模式的区别: 最大的区别是数据端口并不总是20, 主动模式和被动模式的优缺点: 主动FTP对FTP服务器的管理和安全很有利,但对客户端的管理不利.因为FTP服务器企图与客户端 ...
- 相机标定:PNP基于单应面解决多点透视问题
利用二维视野内的图像,求出三维图像在场景中的位姿,这是一个三维透视投影的反向求解问题.常用方法是PNP方法,需要已知三维点集的原始模型. 本文做了大量修改,如有不适,请移步原文: ...
- 时序分析:KMP算法用于序列识别
考研基础资料之一的<算法与数据结构>,KMP算法作为串匹配的基本算法,为必考题目之一.对于算法入门来说,也是复杂度稍高的一个基本算法. KMP算法作为串匹配的非暴力算法,是为了减少回溯而设 ...
- Memcached 之缓存雪崩现象、实际案例和缓存无底洞现象
一.缓存雪崩现象 由于集群中某个memcached服务器宕机的原因,造成集群中的服务器命中率下降.只能通过访问数据库得到数据,是的数据库的压力倍增,造成数据库服务器崩溃.重启数据库还是会崩溃,但是数据 ...
- css image-set 让浏览器自动切换1x,2x图片
方法一: <img src="img.png" srcset="path/img.png 2x,path/img.png.png 3x"/> 方法二 ...