Given a character as input, check if it is a vowel. Print YES if it is so, otherwise NO.
Input
A character c which takes a value in the range 'a', ...., 'z'.
Output
YES if c is a vowel, otherwise NO.
Example
Input
e
Output
YES
Input
u
Output
YES
Input
g
Output
NO
This is straightforward.
None
Read and check if the char is either 'a', 'e', 'i', 'o' or 'u'. This is implemented by a function isvowel. Print YES or NO accordingly.
// vowel.c
#include <stdio.h>
int isvowel(char ch) {
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u')
return 1;
else
return 0;
}
int main() {
char ch;
scanf("%c", &ch);
if ( isvowel(ch) )
printf("YES\n");
else
printf("NO\n");
return 0;
}Given a sequence of characters as input, check if every character is a vowel. Print YES if it is so, otherwise NO.
Input
A sequence s which every character takes a value in the range 'a', ...., 'z'.
Output
YES if each character is a vowel, otherwise NO.
Example
Input
aeiou
Output
YES
Input
eeeee
Output
YES
Input
abeioou
Output
NO
Note: In the third example, the second char is not a vowel.
Read through the sequence of chars, one at a time, check if it is a vowel. Print NO upon encountering the first non-vowel char. Otherwise print YES after the end of the sequence is reached.
Although implementing the loop is straightforward, identifying the end the sequence is not. The exit criteria for the loop should be properly defined. since the size of the sequence is arbitrary and not known. We note that the sequence could be followed by (i) a blank space ' ', (ii) a tab space '\t' or (iii) a new line '\n'. There is a 4th possibility, that is, end-of-file denoted by a macro EOF. Pressing Ctrl+D produces this char.
Read and check if each char is either 'a', 'e', 'i', 'o' or 'u' until one of the following chars: ' ', '\t', '\n' and EOF is met. This is implemented by the function isend. Print YES or NO accordingly.
// vowelsequence.c
#include <stdio.h>
int isvowel(char ch) {
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u')
return 1;
else
return 0;
}
int isend(char ch) {
if (ch == ' ' || ch == '\t' || ch == '\n' || ch == EOF)
return 1;
else
return 0;
}
int main() {
char ch;
scanf("%c", &ch); // Read a char
while ( !isend(ch) ) { // while it not the end char
if ( isvowel(ch) ) // if it is vowel
scanf("%c", &ch); // read the next char
else { // criteria broken
printf("NO\n");
return 0; // No need to scan the rest
}
}
// We did not find a single consonant char all through
printf("YES\n");
return 0;
}Given a sequence of characters as input, determine if there are two consecutive consonants.
Input
A sequence s which every character takes a value in the range 'a', ...., 'z'.
Output
Print "NO" if there are two consecutive consonants. Otherwise "YES".
Example
Input
axeyizowu
Output
YES
Input
aeiou
Output
YES
Input
abeicdoou
Output
NO
Maintain the last two chars. Check if both of them are consonants.
None
Define variables ch and last. For the first time, read the first 2 chars into last and ch respectively. Check if both are consonants. If so, print "NO" and exit. If not, copy ch to last and read the next char into ch. Repeat this till the end of the sequence. Finally print "YES".
// no2consonants.c
#include <stdio.h>
int isvowel(char ch) {
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u')
return 1;
else
return 0;
}
int isend(char ch) {
if (ch == ' ' || ch == '\t' || ch == '\n' || ch == EOF)
return 1;
else
return 0;
}
int main() {
char ch, last;
scanf("%c", &last); // read the first char
scanf("%c", &ch); // read the next char
while ( !isend(ch) ) { // while the next char is not the end char
if ( !isvowel(ch) && !isvowel(last) ) { // if last 2 chars are consonants
printf("NO\n"); // no need to scan the rest
return 0;
}
else {
last = ch;
scanf("%c", &ch);
}
}
// We did not find a two consecutive consonants all through
printf("YES\n");
return 0;
}Given a sequence of characters as input, determine if there are two consecutive consonants. Also, the sequence should not end with a consonant.
Input
A sequence s which every character takes a value in the range 'a', ...., 'z'.
Output
Print "NO" if there are two consecutive consonants or the sequence ends with a consonant. Otherwise "YES".
Example
Input
axeyizowu
Output
YES
Input
aeixyou
Output
NO
Input
abeicooud
Output
NO
Maintain the last two chars. Check if both of them are consonants. Finally, check if the last char of the sequence is a consonant.
None
Define variables ch and last. For the first time, read the first 2 chars into last and ch respectively. Check if both are consonants. If so, print "NO" and exit. If not, copy ch to last and read the next char into ch. Repeat this till the end of the sequence. At the end, check if last holds a consonant. If so, print "NO" and exit. Finally, print "YES".
// nolastconsonant.c
#include <stdio.h>
int isvowel(char ch) {
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'n')
return 1;
else
return 0;
}
int isend(char ch) {
if (ch == ' ' || ch == '\t' || ch == '\n' || ch == EOF)
return 1;
else
return 0;
}
int main() {
char ch, last;
scanf("%c", &last); // read the first char
scanf("%c", &ch); // read the next char
while ( !isend(ch) ) { // while the next char is not the end char
if ( !isvowel(ch) && !isvowel(last) ) { // if last 2 chars are consonants
printf("NO\n"); // no need to scan the rest
return 0;
}
else {
last = ch;
scanf("%c", &ch);
}
}
if ( !isvowel(last) ) { // Note: ch now holds end char. last holds the last char of the input sequence
printf("NO\n");
return 0;
}
// We did not find a two consecutive consonants all through and last char is not a consonant
printf("YES\n");
return 0;
}Given a sequence of characters as input, determine if there are two consecutive consonants. The sequence should not end with a consonant. The only exception is for the character 'n'. 'n' can be followed by any character vowel or constant. Also, the sequence can end with 'n'.
Input
A sequence s which every character takes a value in the range 'a', ...., 'z'.
Output
Print "NO" if there are two consecutive consonants or the sequence ends with a consonant. Otherwise "YES".
Example
Input
axenyizowu
Output
YES
Input
aeixnyou
Output
NO
Input
abeicooud
Output
NO
Maintain the last two chars. Check if both of them are consonants. Finally, check if the last char of the sequence is a consonant.
None
Define variables ch and last. For the first time, read the first 2 chars into last and ch respectively. Check if both are consonants. If so, print "NO" and exit. If not, copy ch to last and read the next char into ch. Repeat this till the end of the sequence. At the end, check if last holds a consonant. If so, print "NO" and exit. Finally, print "YES".
// notwithn.c
#include <stdio.h>
int isvowel(char ch) {
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u')
return 1;
else
return 0;
}
int isend(char ch) {
if (ch == ' ' || ch == '\t' || ch == '\n' || ch == EOF)
return 1;
else
return 0;
}
int main() {
char ch, last;
scanf("%c", &last); // read the first char
scanf("%c", &ch); // read the next char
while ( !isend(ch) ) { // while the next char is not the end char
if ( !isvowel(ch) && (!isvowel(last) && last != 'n') ) { // if last 2 chars are consonants
printf("NO\n"); // no need to scan the rest
return 0;
}
else {
last = ch;
scanf("%c", &ch);
}
}
if ( !isvowel(last) && last != 'n') {
printf("NO\n"); // no need to scan the rest
return 0;
}
// We did not find a two consecutive consonants all through with the exception of 'n'
printf("YES\n");
return 0;
}- Check out the problem 1008A Romaji in https://codeforces.com/problemset/problem/1008/A
- After testing your program thoroughly, submit your solution and get an Accepted message.