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

Manish Goyal · 2011-02-28T05:13:35+00:00

write the shortest code you can write to print all the odd no from 1 to 100? Note:- Don't use google please

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

Manish Goyal
@manish-r2Hoep

Updated: Feb 28, 2011

Views: 1.7K

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

Note:- Don't use google please

0

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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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