wideIT

Advance Sparse Matrix Operations in C languages by DIBYENDU SWAR (rajamax1987@yahoo.com) ----------------------------------------------------------------------------------------- #include<stdio.h> #include<conio.h> int isSparse(int [3][3]); void getSparse(int [3][3]); void dispSparse(int [3][3]); void sumSparse(int [3][3],int [3][3],int [3][3]); void multiSparse(int [3][3],int [3][3],int [3][3]); void main() { int mat[3][3],mat1[3][3],rslt[3][3]; int i; do{ clrscr(); printf("\n\t\tDATA STRUCTUTE OVER 2D ARRAY [DIBYENDU]\n\t\t------------------------------\n"); printf("\n\t\tSPARSE MATRIX OPERATIONS\n\n\t\t1. LOAD SPARSE [GET DATA]\n\n\t\t2. SUM SPARSE\n\nt\t\t3. MULTIPLY SPARSE\n\n\t\t4. DISPLAY MATRIXes\n\n\t\t5. DISPLAY SPARSEes\n\n\t\t6. EXIT\n\n\t\tENTER CHOICE :: "); scanf("%d",&i); switch(i) { case 1: getSparse(mat); getSparse(mat1); break; case 2: su

As the figures of Q4 2010 smartphone shipment is out, Google’s Android platform has emerged as the leading smartphone platform with 32.9 million shipment leaving Nokia’s Symbian platform with 31 million shipment behind in 2010 across the globe. Though Nokia did managed to retain its position as the leading global smart phone vendor, with 28% market share, it was Android which has helped the global smartphone market to witness 101.2 million units shipment in 2010 with 89% growth compare to 2009., speaks a latest report from Canalys. In Q4 2010, volumes of Google OS-based smart phones (Android, OMS and Tapas) were again boosted by strong performances from a number of vendors, notably LG, Samsung, Acer and HTC, whose volumes across these platforms grew 4,127%, 1,474%, 709% and 371% respectively year-on-year. HTC and Samsung together accounted for nearly 45% of Google OS-based handset shipments. However, it will be difficult to say whether Android will be stealing the show again in 20

Stay connect with ORACLE server whereever you are... Friend cnn As String = "Password=fusion;Persist Security Info=True;User ID=fms;Data Source=(DESCRIPTION =(ADDRESS_LIST =(ADDRESS = (PROTOCOL = TCP)(HOST =<HOST_IP>)(PORT = 1521)))(CONNECT_DATA =(SID =<G DB NAME>) ));"

Unicode is a character encoding standard that has widespread acceptance. Microsoft software uses Unicode at its core. Whether you realize it or not, you are using Unicode already! Basically, “computers just deal with numbers. They store letters and other characters by assigning a number for each one. Before Unicode was invented, there were hundreds of different encoding systems for assigning these numbers. No single encoding could contain enough characters.” This has been the problem we, in SIL, have often run into. If you are using a legacy encoding your font conflicts with the font someone in another area of the world uses. You might have an in your font while someplace else someone used a at the same codepoint. Your files are incompatible. Unicode provides a unique number for every character and so you do not have this problem if you use Unicode. If your document calls for U+0289  it will be clear to any computer program what the character should be.

class test_math { int sum(int a,int b) { System.out.print("\nHits on Integer function\n"); return(a+b); } float sum(float a,float b) { System.out.print("\nHits on Float function\n"); return(a+b); } } public class Test { public static void main(String args[]) { test_math t=new test_math(); System.out.println(t.sum(2.5f,3.5f)); System.out.println(t.sum(2,3)); System.out.println(t.sum(2,3.9f)); } }

#include<stdio.h> #include<conio.h> int x[50],ic=0,fg=0; main() { int c=0,f=0; clrscr(); for(c=0;c<=49;c++) { if(x[c]!=0) { f++; } } printf("\n\t\tEXAMPLE::1 DIMENTIONAL ARRAY- AS A DATA STRUCTURE\n\t\t\tAL/22/09-Dibyendu Swar (Sonu)\n\n"); printf("\n\t\tCHOICES\n"); printf("\n\t\t1.INSERT [%d]\n\t\t2.MODIFY\n\t\t3.DISPLAY\n\t\t4.QUIT\n\t\t5.DELETE\n\t\t6.ARRANGE DATA",f); printf("\n\n\t\tYOUR CHOICE::");    scanf("%d",&c);    if(c==1)    { ins(); getch();    }    else if(c==3)    { dispc(); getch();    }    else if(c==2)    { modify(); getch();    }    else if(c==4)    { exit();    }    else if(c==5)    { dlete(); getch();    }    else if(c==6)    {   arng(); getch();    }    else    { main();  

public class whilechk1 {     public static void main(String args[])     {         int row,column,y;         System.out.print("\nMultiplication Table\n");         row=1;         do         {             column=1;             do             {                 y=row*column;                 System.out.print("\t"+y);                 column++;             }while(column<=10);             System.out.println("\n");             row++;         }while(row<=10);     } }