Intermediate.
Put your skills to the test.

Problem summary

Classify each sentence: one language never has four consecutive consonants in a word, while the other has at least one such word. Count the two languages, treating a,e,i,o,u,y as vowels.

Constraints

At most 1000 sentences; at most 1000 characters per sentence.

Worked example

Input and output for the worked example discussed below.

3
tststsbreh kvassssi hubvlka.
miyyy takvuz sum si em ubava em gizdava pa i umna i fruclavu.
tststs shte si htshtschup glvta s tttttteb.

1 2

Solution analysis

1. Read whole lines because a sentence contains multiple words. Maintain the current run of consonant letters; increment it on a consonant and reset it on a vowel or a word separator.
2. Once the run reaches 4, the sentence belongs to the consonant-heavy language. Continue scanning or remember a flag; later vowels cannot undo the existence of that earlier run.
3. The first sample sentence begins with a long tststs consonant sequence, and the third does as well. The second has no four-consonant block, so the output is 1 2. The letter y explicitly counts as a vowel here.

Why it works

The running counter is exactly the number of consecutive consonants ending at the current character, because every non-consonant resets it. A flag becomes true if and only if some such run has length at least four, exactly the stated classification criterion.

Watch out. Do not forget y in the vowel set, join consonants across spaces, or use operator>> to read only the first word. Consume the newline after the initial count.

An independently written complete program with standard headers.

#include<iostream>
#include<string>
using namespace std;
int main(){
    int n;
    cin>>n;
    string sentence;
    getline(cin,sentence);
    int light=0;
    int heavy=0;
    const string vowels="aeiouy";
    for(int i=0;i<n;i++){
        getline(cin,sentence);
        int run=0;
        bool found=false;
        for(char letter:sentence){
            if(letter>='a'&&letter<='z'&&vowels.find(letter)==string::npos){
                run++;
                // A later separator cannot cancel a consonant run already found.
                found=found||run>=4;
            }
            else{
                run=0;
            }
        }
        if(found){
            heavy++;
        }
        else{
            light++;
        }
    }
    cout<<light<<' '<<heavy<<'\n';
}