Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) B. Verse Pattern 水题
B. Verse Pattern
题目连接:
http://codeforces.com/contest/722/problem/B
Description
You are given a text consisting of n lines. Each line contains some space-separated words, consisting of lowercase English letters.
We define a syllable as a string that contains exactly one vowel any arbitrary number (possibly none) of consonants. In English alphabet following letters are considered to be vowels: 'a', 'e', 'i', 'o', 'u' and 'y'.
Each word of the text that contains at least one vowel can be divided into syllables. Each character should be a part of exactly one syllable. For example, the word "mamma" can be divided into syllables as "ma" and "mma", "mam" and "ma", and "mamm" and "a". Words that consist of only consonants should be ignored.
The verse patterns for the given text is a sequence of n integers p1, p2, ..., pn. Text matches the given verse pattern if for each i from 1 to n one can divide words of the i-th line in syllables in such a way that the total number of syllables is equal to pi.
You are given the text and the verse pattern. Check, if the given text matches the given verse pattern.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of lines in the text.
The second line contains integers p1, ..., pn (0 ≤ pi ≤ 100) — the verse pattern.
Next n lines contain the text itself. Text consists of lowercase English letters and spaces. It's guaranteed that all lines are non-empty, each line starts and ends with a letter and words are separated by exactly one space. The length of each line doesn't exceed 100 characters.
Output
If the given text matches the given verse pattern, then print "YES" (without quotes) in the only line of the output. Otherwise, print "NO" (without quotes).
Sample Input
3
2 2 3
intel
code
ch allenge
Sample Output
YES
Hint
题意
给你n个字符串,然后问你每个字符串里面是否恰好有p[i]个元音字母。
如果都符合的话,输出yes,否则输出no
题解:
暴力去判断就好了
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 105;
int p[maxn];
string s;
int dp[2][maxn];
int main()
{
int now=0,pre=1;
int n;scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&p[i]);
int flag = 1;getchar();
for(int i=1;i<=n;i++)
{
getline(cin,s);
int k = 0;
for(int j=0;j<s.size();j++)
{
if(s[j]=='a'||s[j]=='e'||s[j]=='i'||s[j]=='o'||s[j]=='u'||s[j]=='y')
k++;
}
if(k!=p[i])flag=0;
}
if(flag)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) B. Verse Pattern 水题的更多相关文章
- Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) A B C D 水 模拟 并查集 优先队列
A. Broken Clock time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- Intel Code Challenge Elimination Round (Div.1 + Div.2, combined)
A. Broken Clock time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- Intel Code Challenge Elimination Round (Div.1 + Div.2, combined)(set容器里count函数以及加强for循环)
题目链接:http://codeforces.com/contest/722/problem/D 1 #include <bits/stdc++.h> #include <iostr ...
- 二分 Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) D
http://codeforces.com/contest/722/problem/D 题目大意:给你一个没有重复元素的Y集合,再给你一个没有重复元素X集合,X集合有如下操作 ①挑选某个元素*2 ②某 ...
- 线段树 或者 并查集 Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) C
http://codeforces.com/contest/722/problem/C 题目大意:给你一个串,每次删除串中的一个pos,问剩下的串中,连续的最大和是多少. 思路一:正方向考虑问题,那么 ...
- Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) D. Generating Sets 贪心
D. Generating Sets 题目连接: http://codeforces.com/contest/722/problem/D Description You are given a set ...
- Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) C. Destroying Array 带权并查集
C. Destroying Array 题目连接: http://codeforces.com/contest/722/problem/C Description You are given an a ...
- Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) A. Broken Clock 水题
A. Broken Clock 题目连接: http://codeforces.com/contest/722/problem/A Description You are given a broken ...
- Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) C. Destroying Array
C. Destroying Array time limit per test 1 second memory limit per test 256 megabytes input standard ...
随机推荐
- 深入分析Java Web技术内幕
深入web请求过程 发起一个http请求的过程就是建立一个socket通信的过程 HTTPClient是一个开源的实现了http请求的工具包 深入分析java I/O的工作机制 深入分析java We ...
- CSS规范 - 命名规则--(来自网易)
使用类选择器,放弃ID选择器 ID在一个页面中的唯一性导致了如果以ID为选择器来写CSS,就无法重用. NEC特殊字符:"-"连字符 "-"在本规范中并不表示连 ...
- 用Canvas做动画
之前看过不少HTML5动画的书,讲解的是如何去做,对于其中的数学原理讲解的不详细,常有困惑.最近看的<HTML5+JavaScript 动画基础>这个是译本,Keith Peters曾写过 ...
- 【LibreOJ】#6354. 「CodePlus 2018 4 月赛」最短路 异或优化建图+Dijkstra
[题目]#6354. 「CodePlus 2018 4 月赛」最短路 [题意]给定n个点,m条带权有向边,任意两个点i和j还可以花费(i xor j)*C到达(C是给定的常数),求A到B的最短距离.\ ...
- iOS8 自定义navigationItem.titleView
navigationBar其实有三个子视图,leftBarButtonItem,rightBarButtonItem,以及titleView.前两种的自定义请参考http://www.cnblogs. ...
- Linux 网络操作
Linux 基础网路操作 ifconfig eth0 down # 禁用网卡 ifconfig eth0 up # 启用网卡 ifup eth0: # 启用网卡 mii-tool em1 # 查看网 ...
- [转]linux各文件夹介绍
本文来自linux各文件夹的作用的一个精简版,作为个人使用笔记. 下面简单看下linux下的文件结构,看看每个文件夹都是干吗用的? /bin 二进制可执行命令 /dev 设备特殊文件 /etc 系统管 ...
- Wordpress页脚
<?php /** * The template for displaying the footer */ ?> <?php if ( apply_filters( 'show_fl ...
- IL反编译的实用工具Ildasm.exe
初识Ildasm.exe——IL反编译的实用工具 https://www.cnblogs.com/yangmingming/archive/2010/02/03/1662307.html 学 ...
- malloc 函数详解
很多学过C的人对malloc都不是很了解,知道使用malloc要加头文件,知道malloc是分配一块连续的内存,知道和free函数是一起用的.但是但是: 一部分人还是将:malloc当作系统所提供的或 ...