time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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).

Examples

input

3

2 2 3

intel

code

ch allenge

output

YES

input

4

1 2 3 1

a

bcdefghi

jklmnopqrstu

vwxyz

output

NO

input

4

13 11 15 15

to be or not to be that is the question

whether tis nobler in the mind to suffer

the slings and arrows of outrageous fortune

or to take arms against a sea of troubles

output

YES

Note

In the first sample, one can split words into syllables in the following way:

in-tel

co-de

ch al-len-ge

Since the word “ch” in the third line doesn’t contain vowels, we can ignore it. As the result we get 2 syllabels in first two lines and 3 syllables in the third one.

【题解】



a,e,i,o,u,y被认为是元音字母。

求一个字符串里面有x个元音字母。它就能被分成x个SMG;判断和所给的相不相同就可以了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string> using namespace std; int a[200];
string s; int main()
{
//freopen("F:\\rush.txt", "r", stdin);
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
getchar();
for (int i = 1; i <= n; i++)
{
getline(cin, s);
int len = s.size();
int num = 0;
for (int i = 0; i <= len - 1; i++)
if (s[i] == 'a' || s[i] == 'i' || s[i] == 'e' || s[i] == 'o' || s[i] == 'u' || s[i] == 'y')
num++;
if (num != a[i])
{
puts("NO");
return 0;
}
}
puts("YES");
return 0;
}

【57.14%】【codeforces 722B】Verse Pattern的更多相关文章

  1. 【dp入门题】【跟着14练dp吧...囧】

    A HDU_2048 数塔 dp入门题——数塔问题:求路径的最大和: 状态方程: dp[i][j] = max(dp[i+1][j], dp[i+1][j+1])+a[i][j];dp[n][j] = ...

  2. 【 BowWow and the Timetable CodeForces - 1204A 】【思维】

    题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制 ...

  3. 【57.97%】【codeforces Round #380A】Interview with Oleg

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  4. 【34.57%】【codeforces 557D】Vitaly and Cycle

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  5. 【中途相遇法】【STL】BAPC2014 K Key to Knowledge (Codeforces GYM 100526)

    题目链接: http://codeforces.com/gym/100526 http://acm.hunnu.edu.cn/online/?action=problem&type=show& ...

  6. 【77.78%】【codeforces 625C】K-special Tables

    time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...

  7. 【codeforces 754D】Fedor and coupons

    time limit per test4 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  8. 【codeforces 760A】Petr and a calendar

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  9. 【codeforces 750E】New Year and Old Subsequence

    time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

随机推荐

  1. 洛谷 P1852 奇怪的字符串

    P1852 奇怪的字符串 题目描述 输入两个01串,输出它们的最长公共子序列的长度 输入输出格式 输入格式: 一行,两个01串 输出格式: 最长公共子序列的长度 输入输出样例 输入样例#1: 复制 0 ...

  2. 前台技术--div的隐藏与显示

    怎样使用页面元素隐藏或显示. HTML为我们提供了两个变量visibility和display visibility:隐藏要元素可是元素所暂用的空间不予释放.也就是说元素隐藏了,可是页面上会流出一片空 ...

  3. css 兼容性前缀

    一.不同浏览器内核下的书写规则 二:transform  具体变性中心基点  transform-origin  默认情况下  rotate旋转.scale缩放.translate位移.矩阵matri ...

  4. Mahout项目开发环境搭建(Eclipse\MyEclipse + Maven)

    继续 http://www.tuicool.com/articles/rmiEz2 http://www.cnblogs.com/jchubby/p/4454888.html

  5. [ Tomcat ] [ startup ] Tomcat 無法在時限內開啟問題

    http://www.ewdna.com/2011/12/tomcat-server-in-eclipse-unable-to.html

  6. Tuple<int, int> Dictionary<string, object>妙用

    Tuple<int, int> Dictionary<string, object>妙用

  7. tomcat做成windows服务之后使用JMX监控的问题

    转载:http://blog.chinaunix.net/uid-20449851-id-2369842.html

  8. 利用jquery.fullPage.js 和 scrolloverflow.min.js 实现滚屏效果

    参考链接:https://blog.csdn.net/c11073138/article/details/79631036 /* 按着思路去search. */

  9. Python3的取余操作

    https://blog.csdn.net/u014647208/article/details/53368244 取余代码: 输入以下代码: >>>10%2 >>> ...

  10. [python]bug和debug

    bug:代码中存在的语法或者逻辑问题 debug:自查和解决代码中的问题 (coding五分钟,debug两小时) 一.出现bug原因的四大类型 1.粗心 1)错误案例 上面这个错误就是因为 if语句 ...