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  • Shortest C++ code to print odd numbers between 1 and 100

    Manish Goyal

    Member

    Updated: Oct 26, 2024
    Views: 1.6K
    write the shortest code you can write to print all the odd no from 1 to 100?


    Note:- Don't use google please
    0
    Replies
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Replies

  • gaurav.bhorkar

    Member • Feb 28, 2011

    main( )
    {
    for (int i = 1; i <= 100; i+=2)
    cout << i;
    }

    I didn't execute it, by the way.
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  • Manish Goyal

    Member • Feb 28, 2011

    can you think something different

    i mean not using this and any modulus operator?
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  • Reya

    Member • Feb 28, 2011

    main()
    {
    for(int i=1;i<=100;i++)
    {
    if(i%2!=0)
    cout<<i;
    }}
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  • Hussanal Faroke

    Member • Feb 28, 2011

    main()
    {
    pirntf("13579111315........99");
    }
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  • Manish Goyal

    Member • Feb 28, 2011

    Interesting
    Any other way friends
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  • gaurav.bhorkar

    Member • Mar 1, 2011

    Hussanal Faroke
    main()
    {
    pirntf("13579111315........99");
    }
    Haha... but that will hurt your fingers 😀
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  • rajesh_s

    Member • Mar 1, 2011

    main()
    {
    for(i=1;i<100;i++)
    {
    for(j=2;j<i;j++)
    {
    if(i%j!=0)
    cout<<i;
    }
    }
    }
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  • PraveenKumar Purushothaman

    Member • Mar 2, 2011

    Heard of Structures? Using that we can do in this way... 😀

    struct ZeroEven {
    unsigned num : 1;
}

int main(){
    ZeroEven ze;
    for (i=0; i<100; i++)
    {
        ze.num = i;
        if (ze.num != 0)
            cout<<i;
    }
    return 0;
}
    Guys, do try this and say! 😀
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  • Manish Goyal

    Member • Mar 3, 2011

    This is what i was looking for Nice logic buddy

    Any other way
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  • PraveenKumar Purushothaman

    Member • Mar 3, 2011

    goyal420
    This is what i was looking for Nice logic buddy

    Any other way
    Thanx... 😀
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  • rajesh_s

    Member • Mar 3, 2011

    is my logic not correct
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  • gaurav.bhorkar

    Member • Mar 3, 2011

    praveenscience
    Heard of Structures? Using that we can do in this way... 😀

    struct ZeroEven {
    unsigned num : 1;
}

int main(){
    ZeroEven ze;
    for (i=0; i<100; i++)
    {
        ze.num = i;
        if (ze.num != 0)
            cout<<i;
    }
    return 0;
}
    Guys, do try this and say! 😀
    Nice solution. I never thought of using bit fields.
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  • PraveenKumar Purushothaman

    Member • Mar 4, 2011

    gaurav.bhorkar
    Nice solution. I never thought of using bit fields.
    He he 😛 Thanx... 😁
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  • Sreekar B V

    Member • Sep 18, 2014

    main()
    {
    int i=1;
    while((i<100)&&(i+=2))
    printf("%d",i);
    }
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  • simplycoder

    Member • Sep 28, 2014 

    #include<stdio.h>
//simplycoder

int main(int argc,char**argv)
{
   int n=1;
  while(n<100)
  {
     if(n & 1 && printf("%d",n));
     n++;
  }
return 0;
}
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  • Kaustubh Katdare

    Administrator • Sep 28, 2014

    Without wanting to distract this discussion; I think the guys who used print got it all right. Someone gotta beat that!

    Or we need to rephrase the problem statement?
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  • Tim Rossiter

    Member • Sep 29, 2014 

    int main(int n)  //init n=-1
{
     (n<99)?(printf("%d\n",main(n+2))):(n=n); //emote is colon left paren
     return(n);
}
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  • Kaustubh Katdare

    Administrator • Sep 29, 2014

    #-Link-Snipped-# - You may use [code] .... [/code] tags to enter code without having to deal with emoticons.
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  • Tim Rossiter

    Member • Sep 29, 2014

    Thanks
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  • Anoop Kumar

    Member • Sep 29, 2014

    I don't know C++ but here is java logic

    void printOddNumbers(int maxLimit) {
       for(int i=1; i <= maxLimit; i+=2){
            System.out.println("Odd : " +i);
    }
}
    Number of iterations: maxLimit / 2 ;
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  • avii

    Member • Oct 2, 2014

    well here's in Go, if that interests anyone:

    package main

import "fmt"

func main() {
    for i := 1; i < 100; i+=2 {
        fmt.Println(i)
    }
}
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