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• Problem installing virtualbox on ubuntu 9.04

  - by debanjan

  I'm using ubuntu 9.04. From last few days I'm trying to install a virtualbox,but i can't. I went to this link for virtualbox. https://www.virtualbox.org/wiki/Download_Old_Builds_3_0 Found the virtualbox for my version. After downloading it in xp i restarted my pc and opened ubuntu. While installing by package installer there is a problem. But the status giving a massage everytime I'm trying to install.it saying "Error: Dependency is not satisfiable: libcurl3 (=7.16.2-1)". Here is an icon named "install package" but its hidden. I don't know what's the problem.I'm new to ubuntu,pls try to help me.

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• Correct algorith for checking leap year

  - by Debanjan

  What will be the exact definition of leap year ? AFAIK "A year which is divisible by 4 is a leap year.But for century years' the years which are divisible by 400 is a leap year." But that definition makes 100, 200, 300, 400.... upto 1700 NOT LEAP years! But in Gregorian calender all of them are all leap year,check this out. You can also try "call 1700" in Linux to verify. So the correct algorithm foe leap year would be if ( (year % 4 == 0) && ( year <= 1700 || year % 100 != 0 || year % 400 == 0 )) printf("%d is a leap year.\n", year); else printf("%d is not a leap year.\n", year); But is this specific to Gregorian calender ? Even if that is the case why is it not mentioned here ? Regards, PS:The history of Gregorian callender seems interesting check out the September month of 1752.

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• Is this mingw bug ?

  - by Debanjan

  Hi, I have been trying to execute this program on my migw ,through code::blocks, #include <string.h> #include <math.h> #include <stdio.h> #define N 100 int p[N / 64]; int pr[N]; int cnt; void sieve() { int i,j; for(i=0;i<N;i++) pr[i]=1; pr[0]=pr[1]=0; for(i=2;i<N;i++) if(pr[i]) { p[cnt]=i; cnt++; for(j=i+i;j<=N;j+=i) pr[j]=0; } } int main(){ sieve(); int i; for(i=0;i<cnt;i++) printf("%d ",p[i]); puts(""); printf("Total number of prime numbers : %d",cnt); return 0; } In My system the output is : 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 Total number of prime numbers : 22 Which is completely insane,since I am completely sure about the implementation of my algorithm. So I decided to try it in Ideone where it gives correct output.Can anybody point out the reason ?

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• Crash the program with cmd line args

  - by Debanjan

  Lets us consider the following program : #include <stdlib.h> int main(int argc, char **argv){ int a,b; if (argc != 3) return -1; a = atoi(argv[1]); b = atoi(argv[2]); a = b ? a/b : 0; return a; } The task is to crash the program by providing arguments in command-line.

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• Correct expression for checking leap years

  - by Debanjan

  What will be the exact definition of leap year? AFAIK "A year which is divisible by 4 is a leap year. But for century years' the years which are divisible by 400 is a leap year." But that definition makes 100, 200, 300, 400.... upto 1700 NOT LEAP years! But in Gregorian calendar all of them are all leap year, check this out. You can also try "call 1700" in Linux to verify. So the correct algorithm for leap years would be: if ( (year % 4 == 0) && ( year <= 1700 || year % 100 != 0 || year % 400 == 0 )) printf("%d is a leap year.\n", year); else printf("%d is not a leap year.\n", year); But is this specific to Gregorian calendar? Even if that is the case why is it not mentioned here? Regards, PS:The history of Gregorian callender seems interesting check out the September month of 1752.

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