Hey,
Here is something awesome i just made. “What’s up neighbors” will let you see what your neighbors are tweeting and where, based on your current location.
So, give it a try.
Have Fun.
Hey,
Here is something awesome i just made. “What’s up neighbors” will let you see what your neighbors are tweeting and where, based on your current location.
So, give it a try.
Have Fun.
Quora has given me a badge on my profile for most viewed writer in Social Psychology and Character & Personality.
It’s an achievement for me. The credits goes to this “Answer“. Though these are not so much upvotes, still matters to me a lot.
Have Fun.
Click here : https://www.quora.com/What-are-the-best-Python-scripts-youve-ever-written/answer/Karan-Dev
Have Fun.
Hi,
Get a visualization of how different algorithms sorts.
click here: Graphical representation sorting algorithms
Enjoy, Keep Coding 🙂
Hi,
If a=1, then how ++a + ++a=6?
I had been asked these type of questions many times by my teachers, but as I remember they all gave me different explanations that is why I could not remember what the correct explanation is ;).
But now I found a genuine answer for it. Below is the explanation’s link, as I can’t explain the concept better than Aparajita ji .
Thanks to her.
Enjoy, Keep Coding 🙂
Hi,
I wrote the code for Calendar in C. It will tell you the name of the day of any date after Jan 1st, 1901.
It is really interesting, you should try on your own before looking at the code.
#include<stdio.h> int main() { int i,s=0,r,n,d,m,y,j,k,l,b=0,kn=0; printf("enter day : "); scanf("%d",&d); printf("enter month : "); scanf("%d",&m); printf("enter year : "); scanf("%d",&y); i=1901; j=1; k=1; l=2; while(1) { while(1) { while(1) { if(i==y && j==m && k==d) { if(l==1) printf("\nMonday\n"); if(l==2) printf("\nTuesday\n"); if(l==3) printf("\nWednesday\n"); if(l==4) printf("\nThursday\n"); if(l==5) printf("\nFriday\n"); if(l==6) printf("\nSaturday\n"); if(l==7) printf("\nSunday\n"); kn=1; break; } l++; if(l==8) l=1; if((i%4==0 && i%100!=0) || i%400==0) { if(j==2) { if(k==29) { k=1; break; } } } else if(j==2) { if(k==28) { k=1; break; } } if(j==4 || j==6 || j==9 || j==11) { if(k==30) { k=1; break; } } else if(k==31) { k=1; break; } k++; } if(j==12) { j=1; break; } j++; if(kn==1) break; } i++; if(kn==1) break; } return 0; }
For any clarification, contact me.
Enjoy, Keep Coding 🙂
Hi,
Once my friend came to me and ask for a problem, which is to be implemented in the C/C++.
The problem:
You have to create a matrix, input element in the matrix using rand() function, the condition is that all element should be different(no duplicate) in the matrix.
I sugest you to first try to implement yourself before checking the solution.
C code for the above problem:
#include<stdio.h> #include<stdlib.h> int main() { int t,n,m,a[100][100],r,temp[200]={0},i=0,j=0,k=0,l,cnt=0,ex=1,rl; srand ( time(NULL) ); printf("enter size of the matrix : "); scanf("%d",&n); printf("enter maximum limit of random no. (should be greater than size*size) : "); scanf("%d",&rl); printf("\n"); while(i<n) { j=0;ex=1;cnt=0; while(j<n) { ex=1;cnt=0; r=rand()%rl+1; while(ex) { for(l=0;l<k;l++) { if(temp[l]==r) { cnt=1; break; } } if(cnt==0) ex=0; if(cnt==1) { r=rand()%rl+1; cnt=0; } } temp[k++]=r; a[i][j]=r; j++; } i++; } for(i=0;i<n;i++) { for(j=0;j<n;j++) { printf("%d ",a[i][j]); } printf("\n"); } return 0; }
Enjoy, Keep Coding 🙂
Hi, Please check my chrome extension “The Dictionary” that I made recently in JavaScript. Just select the word and click on the extension icon to get the meaning.
You can download The Dictionary extension in zip format from my github account. Also you can check for the instructions for how to install and how to use.
I will be happy to get the feedback. 🙂
Email – karandev43@gmail.com
Thank You.
Paste the below code in your chrome’s address bar, and make a temporary notepad to note anything.
data:text/html, <html contenteditable>
go enjoy.
#include<stdio.h>
int main()
{
int a,b,c;
int count;
count=1;
for (b=c=10;a=”- FIGURE?, UMKC,XYZHello Folks,\
TFy!QJu ROo TNn(ROo)SLq SLq ULo+\
UHs UJq TNn*RPn/QPbEWS_JSWQAIJO^\
NBELPeHBFHT}TnALVlBLOFAkHFOuFETp\
HCStHAUFAgcEAelclcn^r^r\\tZvYxXy\
T|S~Pn SPm SOn TNn ULo0ULo#ULo-W\
Hq!WFs XDt!” [b+++21]; )
for(; a– > 64 ; )
putchar ( ++c==’Z’ ? c = c/ 9:33^b&1);
return 0;
}