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  1. Practice
  2. C++
  3. Introduction
  4. For Loop
  5. Discussions

For Loop

  • Khushboo_Watwani + 1 comment

    Thank you so much for explaining this question... :)

    • akhii + 10 comments
      int a,b,i;
    cin>>a;
    cin>>b;
    for(i=a;i<=b;i++)
    {
        switch(i)
            {
            case 1: cout<<"one \n";break;
            case 2: cout<<"two \n";break;
            case 3: cout<<"three \n";break;
            case 4: cout<<"four \n";break;
            case 5: cout<<"five \n";break;
            case 6: cout<<"six \n";break;
            case 7: cout<<"seven \n";break;
            case 8: cout<<"eight \n";break;
            case 9: cout<<"nine \n";break;
        }
    }
        for(i=10;i<=b;i++)
        {
            if((i%2)==0)
            cout<<"even \n";
            else
                cout<<"odd \n";
        }
      • decoder_ + 1 comment

        Ain't this wrong?

        • akhii + 1 comment

          no its correct

          • decoder_ + 1 comment

            what if the value I give is a=8 and b=5?

            • akhii + 1 comment

              You will be given two positive integers, a and b (a<=b), separated by a newline. read the question first .....AH

              • decoder_ + 1 comment

                calm down okay! It was just a mistake. Sorry that I asked you.

                • drkato + 1 comment

                  Good practice would be to not assume such things, so your intuition is spot on. A simple check at the beginning of the function hurts nothing, as so...

                  int a, b;
cin >> a;
cin >> b;

if((b - a) < 0)
    return 0; //or whatever other error handling is needed
                  • RA1511008010245 + 0 comments

                    you are right .

      • shahashish991 + 2 comments
        [deleted]
        • akhii + 0 comments

          its coreect u mother fucker

        • andriesuaktiwa + 1 comment

          lmao the comment and reply.. hahaha you guys, made my day

          • b6032682 + 0 comments

            very funny stuff

      • deekshith565 + 0 comments
        [deleted]
      • shajol484 + 0 comments

        thanks

      • samplenhold + 0 comments

        Pretty poor solution. Unnecessary use of two for loops. You can do it in one for loop. The code would be easier to read if it was written using one loop and in the loop had an initial if/else structure that determines if the current number in the iteration is less than or equal to 9. If the count is less than 10, you have if statements that print the english word out, and if the count is greater than or equal to 10, then you do modulo to determine even or odd.

      • omarf7120 + 1 comment

        it is totally worng answer

        • drkato + 0 comments

          It is kind of poorly written, but should give the right results. It can certainly be done all in one for loop though, and in my opinion, using a string array is much cleaner than using a switch statement.

      • alice_ferrazzi + 0 comments

        using a double for loop is not necessary

      • bartlomiejpopek + 0 comments

        This is wrong. When I Try imput more then 10 eg. 23 and 25 I get to many even and odds.

      • shuklarituraj9 + 1 comment

        Try this

        #include <iostream>
#include <cstdio>
using namespace std;

int main() {
    int i,c,b;
    string a[11]={"even","odd","one","two","three","four","five","six","seven","eight","nine"};
    cin>>c>>b;
    for(i=c;i<=b;i++)
    {
        if((i>9) && (i%2==0))  cout<<a[0]<<endl;
        else if((i>9) && (i%2!=0)) cout<<a[1]<<endl;
        else    cout<<a[i+1]<<endl;
    }
    return 0;
}
        • haberpeter + 0 comments

          I see this kind of solution in many cases. But in fact this is wrong. This works only in case the first input is less than the second one. If the test case would be 12 and 11 then this failed the test case because the code does not enter into the for loop, because c is greater than b.

      • vvyy346 + 0 comments

        bro just skill the break statement in switch case and the you will be needing only second loop