C - Haiku
Problem description
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The i-th line contains the i-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Examples
Input
on codeforces
beta round is running
a rustling of keys
Output
YES
Input
how many gallons
of edo s rain did you drink
cuckoo
Output
NO
解题思路:只要符合俳句(由“五-七-五”,共十七字音组成,题目要求这些字音是必须都是元音字母)就输出"YES",否则输出"NO",java水过!
AC代码:
import java.util.Scanner;
public class Main {
final static char[] obk={'a','e','i','o','u'};
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String[] s1=scan.nextLine().split(" ");
String[] s2=scan.nextLine().split(" ");
String[] s3=scan.nextLine().split(" ");
int n1=r(s1,0),n2=r(s2,0),n3=r(s3,0);
if(n1==5&&n2==7&&n3==5)System.out.println("YES");
else System.out.println("NO");
}
public static int r(String[] s,int n){
for(int i=0;i<s.length;++i){
for(int j=0;j<s[i].length();++j){
for(int k=0;k<5;++k)
if(s[i].charAt(j)==obk[k])n++;
}
}
return n;
}
}
C - Haiku的更多相关文章
- 和風いろはちゃんイージー / Iroha and Haiku (ABC Edition) (水水)
题目链接:http://abc042.contest.atcoder.jp/tasks/abc042_a Time limit : 2sec / Memory limit : 256MB Score ...
- [Arc058E] Iroha and Haiku
[Arc058E] Iroha and Haiku 题目大意 问有多少\(n\)个数的正整数序列,每个数在\([1,10]\)之间,满足存在\(x,y,z,w\)使得\(x\to y-1,y\to z ...
- 水题 Codeforces Beta Round #70 (Div. 2) A. Haiku
题目传送门 /* 水题:三个字符串判断每个是否有相应的元音字母,YES/NO 下午网速巨慢:( */ #include <cstdio> #include <cstring> ...
- Solution -「ARC 058C」「AT 1975」Iroha and Haiku
\(\mathcal{Description}\) Link. 称一个正整数序列为"俳(pái)句",当且仅当序列中存在连续一段和为 \(x\),紧接着连续一段和为 \(y ...
- Libpci库的调用
这几天发现在Redhat AS6.5 X86_64下用outl(index, 0xcf8)和inl(0xcfc)下读取PCIe配置空间是系统有时性的会hang, 于是去寻找解决方案,首先想到的是用/d ...
- iOS 资源大全
这是个精心编排的列表,它包含了优秀的 iOS 框架.库.教程.XCode 插件.组件等等. 这个列表分为以下几个部分:框架( Frameworks ).组件( Components ).测试( Tes ...
- 盘点国内程序员不常用的热门iOS第三方库:看完,还敢自称”精通iOS开发”吗?【转载】
综合github上各个项目的关注度与具体使用情况,涵盖功能,UI,数据库,自动化测试,编程工具等类型,看完,还敢自称”精通iOS开发”吗? https://github.com/syedhali/EZ ...
- IOS中文版资源库
Swift 语言写成的项目会被标记为 ★ ,AppleWatch 的项目则会被标记为 ▲. [转自]https://github.com/jobbole/awesome-ios-cn#librari ...
- github上所有大于800 star OC框架
https://github.com/XCGit/awesome-objc-frameworks#awesome-objc-frameworks awesome-objc-frameworks ID ...
随机推荐
- vue移动端地址三级联动组件(二)
继续上一篇: 子组件css: <style scoped lang="less"> #city { width: 100%; height: 100%; positio ...
- hdu 4876
ZCC loves cards Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)T ...
- zk strom 本地环境启动命令
bin/zkServer.sh statusbin/zkServer.sh startbin/storm nimbus &bin/storm ui &bin/storm drpc &a ...
- hdu 2782 dfs(限定)
The Worm Turns Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)To ...
- 6.3.4 使用marshal 模块操作二进制文件
Python 标准库 marshal 也可以进行对象的序列化和反序列化,下面的代码进行了简单演示. import marshal x1 = 30 x2 = 5.0 x3 = [1,2,3] x4 = ...
- C#关键字详解第三节
byte:字节 字节是计算机信息技术用于计量存储容量的一种计量单位,通常情况下一字节等于八位,也在一些计算机编程 语言中表示数据类型和语言字符.这是百度百科给出的解释,在C#语言中byte也可以是一种 ...
- Spark源码值提交任务
/** * Return the number of elements in the RDD. */ def count(): Long = sc.runJob(this, Utils.getIt ...
- 08springMVC拦截器
u 概述 u 拦截器接口 u 拦截器适配器 u 运行流程图 u 拦截器HelloWorld u 常见应用之性能监控 1 概述 1.1 简介 Spring Web M ...
- 造成segment fault,产生core dump的可能原因
1.内存访问越界 a) 由于使用错误的下标,导致数组访问越界 b) 搜索字符串时,依靠字符串结束符来判断字符串是否结束,但是字符串没有正常的使用结束符 c) 使用strcpy, strcat, spr ...
- Go语言基础介绍
Go是一个开源的编程语言.Go语言被设计成一门应用于搭载Web服务器,存储集群或类似用途的巨型中央服务器的系统编程语言.目前,Go最新发布版本为1.10. Go语言可以运行在Linux.FreeBSD ...