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

write the shortest code you can write to print all the odd no from 1 to 100?


Note:- Don't use google please

Replies

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

    I didn't execute it, by the way.
  • Manish Goyal
    Manish Goyal
    can you think something different

    i mean not using this and any modulus operator?
  • Reya
    Reya
    main()
    {
    for(int i=1;i<=100;i++)
    {
    if(i%2!=0)
    cout< }}
  • Hussanal Faroke
    Hussanal Faroke
    main()
    {
    pirntf("13579111315........99");
    }
  • Manish Goyal
    Manish Goyal
    Interesting
    Any other way friends
  • gaurav.bhorkar
    gaurav.bhorkar
    Hussanal Faroke
    main()
    {
    pirntf("13579111315........99");
    }
    Haha... but that will hurt your fingers ๐Ÿ˜€
  • rajesh_s
    rajesh_s
    main()
    {
    for(i=1;i<100;i++)
    {
    for(j=2;j {
    if(i%j!=0)
    cout< }
    }
    }
  • PraveenKumar Purushothaman
    PraveenKumar Purushothaman
    Heard of Structures? Using that we can do in this way... ๐Ÿ˜€

    struct ZeroEven {
        unsigned num : 1;
    }
    
    int main(){
        ZeroEven ze;
        for (i=0; i<100; i++)
        {
            ze.num = i;
            if (ze.num != 0)
                cout<Guys, do try this and say! ๐Ÿ˜€
                                        
  • Manish Goyal
    Manish Goyal
    This is what i was looking for Nice logic buddy

    Any other way
  • PraveenKumar Purushothaman
    PraveenKumar Purushothaman
    goyal420
    This is what i was looking for Nice logic buddy

    Any other way
    Thanx... ๐Ÿ˜€
  • rajesh_s
    rajesh_s
    is my logic not correct
  • gaurav.bhorkar
    gaurav.bhorkar
    praveenscience
    Heard of Structures? Using that we can do in this way... ๐Ÿ˜€

    struct ZeroEven {
        unsigned num : 1;
    }
    
    int main(){
        ZeroEven ze;
        for (i=0; i<100; i++)
        {
            ze.num = i;
            if (ze.num != 0)
                cout<Guys, do try this and say! ๐Ÿ˜€
    Nice solution. I never thought of using bit fields.
  • PraveenKumar Purushothaman
    PraveenKumar Purushothaman
    gaurav.bhorkar
    Nice solution. I never thought of using bit fields.
    He he ๐Ÿ˜› Thanx... ๐Ÿ˜
  • Sreekar B V
    Sreekar B V
    main()
    {
    int i=1;
    while((i<100)&&(i+=2))
    printf("%d",i);
    }
  • simplycoder
    simplycoder
    #include
    //simplycoder
    
    int main(int argc,char**argv)
    {
       int n=1;
      while(n<100)
      {
         if(n & 1 && printf("%d",n));
         n++;
      }
    return 0;
    }
    
  • Kaustubh Katdare
    Kaustubh Katdare
    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?
  • Tim Rossiter
    Tim Rossiter
    int main(int n)  //init n=-1
    {
         (n<99)?(printf("%d\n",main(n+2))):(n=n); //emote is colon left paren
         return(n);
    }
  • Kaustubh Katdare
    Kaustubh Katdare
    #-Link-Snipped-# - You may use [code] .... [/code] tags to enter code without having to deal with emoticons.
  • Tim Rossiter
    Tim Rossiter
    Thanks
  • Anoop Kumar
    Anoop Kumar
    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 ;
  • avii
    avii
    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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