Output has one extra line c++

Question

I am typing a program in c++ where it takes a number and does the following.

• If the number is a multiple of 3 but not 5, the output is Fizz.

• If the number is a multiple of 5 but not 3, the output is Buzz.

• If the number is a multiple of both 5 and 3, the output is FizzBuzz.

• If the number is neither a multiple of 3 or 5, it just prints out the a number.

Whenever I compile the code with 15 as the test input, instead of getting the expected output:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

I instead get this

FizzBuzz
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

My code is as follows:

#include <bits/stdc++.h>
#include <iostream>

using namespace std;

string ltrim(const string &);
string rtrim(const string &);




void fizzBuzz(int n) {
    
    for(int i = 0; i <= n; i++)
    {
        if(i % 3 == 0 && i % 5 != 0)
        {
            cout << "Fizz" << endl;
        }
        
        if (i % 3 != 0 && i % 5 == 0)
        {
            cout << "Buzz" << endl;
        }
        
        if (i % 3 != 0 && i % 5 != 0)
        {
            cout << i << endl;
        }
        
        if (i % 3 == 0 && i % 5 == 0)
        {
            cout << "FizzBuzz" << endl;
        }
    }
     
}

int main()
{
    string n_temp;
    getline(cin, n_temp);

    int n = stoi(ltrim(rtrim(n_temp)));

    fizzBuzz(n);

    return 0;
}

string ltrim(const string &str) {
    string s(str);

    s.erase(
        s.begin(),
        find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
    );

    return s;
}

string rtrim(const string &str) {
    string s(str);

    s.erase(
        find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
        s.end()
    );

    return s;
}

Would like to know what I'm doing wrong that's making the extra line in the output show up, any advice will be greatly appreciated!

Answer 1

For loop should start from 1.

void fizzBuzz(int n) {
    for(int i = 1; i <= n; i++){
        if(i % 3 == 0 && i % 5 != 0) cout << "Fizz" << endl;
        if (i % 3 != 0 && i % 5 == 0) cout << "Buzz" << endl;
        if (i % 3 != 0 && i % 5 != 0) cout << i << endl;
        if (i % 3 == 0 && i % 5 == 0) cout << "FizzBuzz" << endl;
    }  
}

Answer 2

@JIN's answer is correct, because 0 is a multiple of both 3 and 5. However, I'd like to further suggest you change the order with which you check the conditions.

Currently you check if it's a multiple of 3, then a multiple of 5, then a multiple of neither, and then if it's a multiple of 15. This works, but we can be more efficient without coding if we check like so:

if (i % 3 == 0 && i % 5 == 0) {
    // Multiple of 15.
}
else if (i % 3 == 0) {
    // If we reach this stage and we know that i is divisible by 3, 
    // then we *know* the only reason it failed the first condition
    // is that it's *not* divisible by 5, so we don't need to check
    // that.
    // Multiple of 3.
}
else if (i % 5 == 0) {
    // As in the previous condition, if we're here, we know that
    // i is not a multiple of 3.
    // Multiple of 5.
}
else {
    // Must be a multiple of neither 3 nor 5.
}