Wednesday, February 26, 2014

Graduate Aptitude Test in Engineering for Electronics and Communication Enigneering Study Materials Ebooks

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Hai Engineers,

Here We have Uploaded Useful study materials for Gate preparation(ECE Stream).
The Following Concepts are Uploaded  :)
  • Amplitude Modulation
  • Circuit Theorems
  • Complex Variables
  • Digital Electronics Question papers
  • Discrete Time Systems
  • Number Systems and Boolean Algebra
  • Operational Amplifiers
  • Semiconductor Physics
  • Signals and Systems
  • Transfer Functions
  • Transmission Lines

Cambridge Business English Certificates(BEC) Preliminary and Vantage Examination Study Materials

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Hai Engineers, 
Cambridge University offers a Certification program in Business English Language(Professional English) through ESOL Examination Centre for all Student Engineers and other peoples in the name of BEC(Business English Certificates).
BEC has three levels,
  • BEC Preliminary (Suitable for All College Students)
  • BEC Vantage (Suitable for Final year Students and job seeking candidates)
  • BEC Higher (Needed for Working Professionals)
All Three levels focus on the Candidates Skills in READING,WRITING,SPEAKING and LISTENING in various different level of standard.

This Cambridge English Business Certificates(BEC) is recognized by many leading International,National and MNC companies.


This will be useful for Students who want to get higher grades in BEC Examination.These materials are got from the Examination training centre and Google Search.

                                                                   Download

Links for References:


Cisco CCNA Study Guide Ebooks

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Hai Engineers,here is the study guide for Cisco CCNA { Cisco Certified Network Associate (CCNA) } certification Program.



MBA Entrance Exam Common Admission Test (CAT) Complete Study materials and Ebooks

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Hai Engineers,For getting into IIM and starting your dream career with MBA course,its important to take up the MBA entrance exams and securing a good percentage. There are quite many MBA Entrance Exams in India for doing Master of Business administration in Post graduation. The main and notable exam is The CAT (common admission test) for MBA.It is important to bell the cat for getting into IIMs. So,here are providing you with various best resources for preparation phase in CAT. 

"Make use of it and Succeed your dream career in IIMs"

The Study Resources for CAT includes all A-Z materials required for Knowing and Preparing CAT.
  • All Previous year CAT Question Papers
  • 3 Model Mock Full length Question papers
  • Quantitative Aptitude all Syllabus 30+
  • Logical Reasoning Ebooks 15+
  • Math basic Formulas to Advanced Formulas over 3000+
  • Non IIMs accepting CAT scores
  • General knowledge Ebooks
  • Puzzles Ebooks
  • Mental Agility Quizzes Ebooks
  • Reading Comprehension Ebooks
                                                                       Download

Green Computing and Linux Power management Ebooks and Video Lectures

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What is Green Computing?

Green computinggreen IT or ICT Sustainability, refers to environmentally sustainable computing or IT. In the article Harnessing Green IT: Principles and Practices, San Murugesan defines the field of green computing as "the study and practice of designing, manufacturing, using, and disposing of computers, servers, and associated subsystems—such as monitors, printers, storage devices, and networking and communications systems — efficiently and effectively with minimal or no impact on the environment.

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https://drive.google.com/folderview?id=0Bz4q5gvytHVmRW1fZFk1R0lfT00&usp=sharing


               http://fortunelao.blogspot.in/


CISCO CCNA PPTs Study Materials for All 4 Modules

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CISCO CCNA PPTs Study Materials for All 4 Modules for Cisco CCNA Certification Program.These PPTs are prepared from the Trainers of CISCO CCNA Training center.These materials covers the all the modules of CISCO CCNA syllabus including
  • Accessing WAN
  • LAN Switching
  • Network Accessing
  • Routing Accessing

Placement Papers of 65 Leading Companies in India

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Huge Collection of Placement Papers for 65 Leading Companies in India:

  • AAI
  • ABB
  • Accenture
  • Aditi
  • Adobe
  • Agreeya
  • Alcatel Lucent
  • Alumnus
  • Amazon
  • Amdocs
  • American MegaTrends
  • Andhra bank
  • Aricent
  • Ashok Leyland
  • Aspire
  • Bajaj
  • Bank of Maharashtra
  • BEL
  • BHEL
  • Birlasoft
  • Blue Star
  • BOB
  • BPL
  • Brakes India
  • BSNL
  • Cadence
  • Calsoft
  • Canara Bank
  • Canarys
  • Cisco
  • Citicorp
  • CMC
  • Crompton
  • Dell
  • Deloitte
  • Deutsche Bank
  • Ericsson
  • Essar
  • Godrej
  • Google
  • HCL
  • Honeywell
  • IBM
  • ICICI
  • Indian Air Force
  • Infosys
  • InfoTech
  • ISRO
  • L and T
  • Mahindra Satyam
  • Mastek
  • MBT
  • Microsoft
  • Motorola
  • NTPC
  • Oracle
  • Patni
  • Polaris
  • Reliance
  • Robert Bosch
  • Satyam
  • State bank of India
  • TATA
  • TCS
  • Tech Mahindra

    Wednesday, February 19, 2014

    C Program to Perform Copying Content of One File Into Another File

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main(int argc,char *argv[])
    {
    FILE *fp,*fp1;
    int i;
    char oout[40];
    fp=fopen(argv[1],"w");
    for(i=3;i<argc;i++)
    {
    fprintf(fp,"%s\n",argv[i]);
    }
    fclose(fp);
    fp=fopen(argv[1],"r");
    fp1=fopen(argv[2],"w");
    for(i=3;i<argc;i++)
    {
    fscanf(fp,"%s\n",out);
    fprintf(fp1,"%s\n",out);
    }
    fclose(fp);
    fclose(fp1);
    }

    C Program to Display Student Details Using File

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    FILE *fp;
    char name[10][10],gr[10][10],res[10][10];
    int m1[10],m2[10],m3[10],tot[10],n,rno[10],i;
    float avg[10];
    clrscr();
    fp=fopen("stud.txt","w");
    printf("\nStudent Detail Using File:\n");
    printf("\n*********************\n");
    printf("\nEnter the Number of Students:\n");
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
    printf("\nEnter the Student Name:\n");
    scanf("%s",name[i]);
    printf("\nEnter the Student Number:\n");
    scanf("%d",&rno[i]);
    printf("\nEnter the Student Marks:\n");
    scanf("%d%d%d",&m1[i],&m2[i],&m3[i]);
    fprintf(fp,"%s\t%d\t%d\t%d\t%d\n",name[i],rno[i],m1[i],m2[i],m3[i]);
    }
    fclose(fp);
    fp=fopen("stud.txt","r");
    printf("\n\t\tStudent Marklist:\n");
    printf("\n\t\t*********************\n");
    printf("\---------------------------------");
    printf("Name\tNO\tTotal\tAverage\tResult\tGrade");
    printf("\n-------------------------------------------");
    i=0;
    for(i=1;i<=n;i++)
    {
    fscanf(fp,"%s\t%d\t%d\t%d\t%d\n",name[i],&rno[i],&m1[i],&m2[i],&m3[i]);
    tot[i]=m1[i]+m2[i]+m3[i];
    avg[i]=tot[i]/3;
    if(m1[i]>=35 && m2[i]>=35 && m3[i]>=35)
    {
    strcpy(res[i],"PASS");
    if(avg [i]>=80)
    strcpy(gr[i],"Distinction");
    else if(avg[i]>=60 && avg[i]<80)
    strcpy(gr[i],"I Class");
    else if(avg[i]>=50 && avg[i]<60)
    strcpy(gr[i],"II Class");
    else
    strcpy(gr[i],"III Class");
    }
    else
    {
    strcpy(res[i],"Fail");
    strcpy(gr[i],"_");
    }
    printf("\n%s\t%d\t%d\t%f\t%s\t%s\t",name[i],rno[i],tot[i],avg[i],res[i],gr[i]);
    printf("\n");
    }
    fclose(fp);
    getch();
    }

    C Program to Perform Swapping of Two Numbers Using Pointers

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int a,b,c;
    int *x,*y;
    clrscr();
    printf("\t\n Swapping of two numbers using Pointers\n");
    printf("\t\n********************************\n");
    printf("\n 1.Using Temporary Variable\n");
    printf("\n 2.Without Using Temporary Variable\n");
    printf("\n 3.Exit\n");
    printf("\n Enter your choice:\n");
    scanf("%d",&c);
    switch(c)
    {
    case 1:
    printf("\n Input:\n");
    printf("\n*****\n");
    printf("\nBefore Swapping\n");
    printf("\n*************\n");
    printf("\nEnter the a value:\n");
    scanf("%d",&a);
    printf("\nEnter the b value:\n");
    scanf("%d",&b);
    x=&a;
    y=&b;
    c=*x;
    *x=*y;
    *y=c;
    printf("\nOutput:\n");
    printf("\n*******\n");
    printf("\nAfter Swapping\n");
    printf("\n*************\n");
    printf("\nA value is:%d\n",a);
    printf("\nB value is:%d\n",b);
    break;
    case 2:
    printf("\n Input:\n");
    printf("\n******\n");
    printf("\nBefore Swapping\n");
    printf("\n***************\n");
    printf("\nEnter the a value:\n");
    scanf("%d",&a);
    printf("\nEnter the b value:\n");
    scanf("%d",&b);
    x=&a;
    y=&b;
    *x=*x+*y;
    *y=*x-*y;
    *x=*x-*y;
    printf("\nOutput:\n");
    printf("\n*******\n");
    printf("\nAfter Swapping\n");
    printf("\n*************\n");
    printf("\nA value is:%d\n",a);
    printf("\nB value is:%d\n",b);
    break;
    case 3:
    exit(0);
    break;
    }
    getch();
    }

    C Program to Calculate Electricity Bill Using Structure

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    struct
    {
    char name[20][20];
    int unit,curr,prev,amt;
    int eno;
    }ebill[10];
    void main()
    {
    int i,n;
    clrscr();
    printf("\tElectricity bill\n");
    printf("\t************\n");
    printf("Enter the limit to customer\n");
    scanf("%d",&n);
    printf("Enter the customers details\n");
    printf("*********************\n");
    for(i=1;i<=n;i++)
    {
    printf("\nEnter %d customer details\n",i);
    printf("Enter the customer number\n");
    scanf("%d",&ebill[i].eno);
    printf("Enter the customer name\n");
    scanf("%s",&ebill[i].name);
    printf("Enter the previous reading\n");
    scanf("%d",&ebill[i].prev);
    printf("Enter the current reading\n\n");
    scanf("%d",&ebill[i].curr);
    }
    for(i=1;i<=n;i++)
    {
    if(ebill[i].prev>ebill[i].curr)
    {
    printf("Previous reading is must less than the current reading");
    }
    else
    {
    ebill[i].unit=ebill[i].curr-ebill[i].prev;
    if(ebill[i].unit<100)
    {
    ebill[i].amt=ebill[i].unit*0.50;
    }
    else if(ebill[i].unit>100&&ebill[i].unit<200)
    {
    ebill[i].amt=ebill[i].unit*1;
    }
    else
    {
    ebill[i].amt=ebill[i].unit*2;
    }
    }
    }
    printf("\-----------------------------------------------------------------\n");
    printf("\tEno\tName\tPrevious Reading\tCurrent Reading\tUnit\tAmount");
    printf("\n----------------------------------------------------------------\n");
    for(i=1;i<=n;i++)
    {
    printf("\t%d\t%s\t\t%d\t\t%d\t\t%d\t%d\n",ebill[i].eno,ebill[i].name,ebill[i].prev,ebill[i].curr,ebill[i].unit,ebill[i].amt");
    }
    getch();
    }

    C Program to Display Students Mark Details Using Structure

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    struct student
    {
    char name[50],grade[20],res[20];
    int rollno,m1,m2,m3,tot;
    float avg;
    }s[50];
    int n,i;
    clrscr();
    printf("\n\tStudent Details");
    printf("\n\t************\n");
    printf("\nEner the Number of Students:\n");
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
    printf("\nEnter the Student Name:");
    scanf("%s",&s[i].name);
    printf("\nEnter the Roll no:");
    scanf("%d",&s[i].rollno);
    printf("\nEnter the mark1:");
    scanf("%d",&s[i].m1);
    printf("\nEnter the mark2:");
    scanf("%d",&s[i].m2);
    printf("\nEnter the mark3:");
    scanf("%d",&s[i].m3);
    }
    for(i=0;i<n;i++)
    {
    s[i].tot=s[i].m1+s[i].m2+s[i].m3;
    s[i].avg=s[i].tot/3;
    if((s[i].m1>=35)&&(s[i].m2>=35)&&(s[i].m3>=35))
    {
    strcpy(s[i].res,"Pass");
    if(s[i].avg>=80)
    strcpy(s[i].grade,"Distinction");
    else if(s[i].avg>=60)
    strcpy(s[i].grade,"First Class");
    else if(s[i].avg>=50)
    strcpy(s[i].grade,"Second Class");
    else
    strcpy(s[i].grade,"Third Class");
    }
    else
    {
    strcpy(s[i].res,"Fail");
    strcpy(s[i].grade,"-");
    }
    }
    printf("\n\t\t Student Details");
    printf("\n--------------------");
    printf("\nRollNo\tName\tMark1\tMark2\tMark3\tTotal\tAverage\tResult\tGrade");
    printf("\n-------------------------------------------------------------------------");
    for(i=0;i<n;i++)
    {
    printf("\n%d\t%s\t%d\t%d\t%d\t%d\t%.2f\t%s\t%s\t",s[i].rollno,s[i].name,s[i].m1,s[i].m2,s[i].m3,s[i].tot,s[i].avg,s[i].res,s[i].grade);
    }
    getch();
    }

    C Program to Find NCR and NPR Using Recursive Function

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    int fact(int n);
    void main()
    {
    int n,r,ncr,npr,n1,n2,n3,n4;
    clrscr();
    printf("\n NCR & NPR Using Recursive Function:\n");
    printf("\n-------------------------------------------\n");
    printf("\nInput:\n");
    printf("\n********\n");
    printf("\nEnter the n value:\n");
    scanf("%d",&n);
    printf("\nEnter the r value:\n");
    scanf("%d",&r);
    n3=n-r;
    n1=fact(n);
    n2=fact(r);
    n3=n-r;
    n4=fact(n3);
    ncr=n1/(n2*n4);
    npr=n1/n4;
    printf("\nOutput:\n");
    printf("\n********\n");
    printf("\nNCR Value is:%d",ncr);
    printf("\nNPR Value is:%d",npr);
    getch();
    }

    int fact(int n)
    {
    int f;
    if(n==0)
    return 1;
    else
    f=n*fact(n-1);
    return f;
    }

    C Program to Find Largest Element in an Array

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    int big(int n,int a[]);
    void main()
    {
    int n,i,a[50],c;
    clrscr();
    printf("\t Largest Element in an Array:\n");
    printf("\t **********************\n");
    printf("\nInput:\n");
    printf("\n--------\n");
    printf("\nEnter the Number:\n");
    scanf("%d",&n);
    printf("\nEnter the Number one by one:\n");
    for(i=0;i<n;i++)
    {
    scanf("%d",&a[i]);
    }
    c=big(n,a);
    printf("\nOutput:\n");
    printf("\n---------\n");
    printf("\nThe Biggest Value is:%d",c);
    getch();
    }

    int big(int n1,int a[20])
    {
    int b=a[0],i;
    for(i=0;i<n1;i++)
    {
    if(b<a[i])
    {
    b=a[i];
    }
    }
    return(b);
    }

    C Program to Find Fibonacci Series Using Function

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    int fib(int);
    void main()
    {
    int n,i;
    clrescr();
    printf("\t Fibonacci Series Using Function:\n");
    printf("\t **************************\n");
    printf("\nInput:\n");
    printf("\n------\n");
    printf("\nEnter the Number:\n");
    scanf("%d",&n);
    printf("\n Output:\n");
    printf("\n---------\n");
    printf("\nFibonacci Series is:");
    fib(n);
    getch();
    }
    int fib(int n1)
    {
    int i,f1=-1,f2=1,f3;
    for(i=1;i<=n1;i++)
    {
    f3=f1+f2;
    f1=f2;
    f2=f3;
    printf("\n \n %d",f3);
    }
    return(f3);
    }

    C Program to Find Factorial Value Using Function

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    int fact(int);
    void main()
    {
    int n,i,f;
    clrscr();
    printf("\t Factorial Value Using Function:\n");
    printf("\t *************************\n");
    printf("\nInput:\n");
    printf("\n---------\n");
    printf("\nEnter the Number:\n");
    scanf("%d",&n);
    f=fact(n);
    printf("\n Output:\n");
    printf("\n --------\n");
    printf("\n Factorial Value is:%d",f);
    getch();
    }
    int fact(int n1)
    {
    int i,s=1;
    for(i=1;i<=n;i++)
    {
    s=s*i;
    }
    return(s);
    }

    C Program for Sorting the Names in Alphabetical Order

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    #include<string.h>
    void main()
    {
    char name[20][15],t[20][15];
    int i,j,n;
    clrscr();
    printf("\n Sorting the names in Alphabetical Order\n");
    printf("\n *******************************\n");
    printf("\n Enter the Number of Names \n");
    scanf("%d",&n);
    printf("\n Enter the Name one by one\n");
    for(i=0;i<n;i++)
    {
    scanf("%s",name[i]);
    }
    for(i=0;i<n;i++)
    {
    for(j=i+1;j<n;j++)
    {
    if(strcmp(name[i],name[j])>0)
    {
    strcpy(t[i],name[i]);
    strcpy(name[i],name[j]);
    strcpy(name[j],t[i]);
    }
    }
    }
    printf("\n OUTPUT\n");
    printf("\n Sorting the Names \n");
    printf("\n****************\n");
    for(i=0;i<n;i++)
    {
    printf("\n%s",name[i]);
    }
    getch();
    }

    C Program for Sorting the Characters in Alphabetical Order

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    #include<string.h>
    void main()
    {
    char name[10],t;
    int i,j,n;
    clrscr();
    printf("\n Sorting the Characters in Alphabetical Order \n");
    printf("\n Enter the Name\n");
    scanf("%s",name);
    n=strlen(name);
    printf("\nLength of Characters: %d",n);
    for(i=0;i<n;i++)
    {
    for(j=i+1;j<n;j++)
    {
    if(name[i]>name[j])
    {
    t=name[i];
    name[i]=name[j];
    name[j]=t;
    }
    }
    }
    printf("\n OUTPUT:\n");
    printf("\n Sorting the Characters \n");
    for(i=0;i<n;i++)
    {
    printf("\n %c",name[i]);
    }
    getch();
    }

    C Program to Perform String Manipulation Using Built in Function

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    #include<string.h>
    void main()
    {
    char str1[20],str2[20];
    int l,s,y,ch;
    clrscr();
    do
    {
    printf("\n Menu\n");
    printf("\n 1.Concatenation\n");
    printf("\n 2.Length\n");
    printf("\n 3.Copy\n");
    printf("\n 4.Comparison\n");
    printf("\n 5.Reverse\n");
    printf("\n 6.Exit\n");
    printf("\n Enter our choice\n");
    scanf("%d",&ch);
    switch(ch)
    {
    case 1:
    printf("\n Concatenation of the String\n");
    printf("\n************************************\n");
    printf("\n Enter the String1\n");
    scanf("%s",str1);
    printf("\n Enter the String2\n");
    scanf("%s",str2);
    strcat(str1,str2);
    printf("\n Concatenation is: %s",str1);
    break;
    }
    case 2:
    {
    printf("\n Length of the String\n");
    printf("\n****************************\n");
    printf("\n Enter the String\n");
    scanf("%d",str1);
    l=strlen(str1);
    printf("\n Length of the String is: %d",l);
    break;
    }

    case 3:
    {
    printf("\n Copy the String\n");
    printf("\n**************************\n");
    printf("\n Enter the String1 \n");
    scanf("%s",str2);
    strcpy(str1,str2);
    printf("\n Original String: %s",str2);
    printf("\n Copied String:%s",str1);
    break;
    }

    case 4:
    {
    printf("\n Comparision of the String \n");
    printf("\n **********************************\n");
    printf("\n Enter the String1 \n");
    scanf("%s",str1);
    printf("\n Enter the String2 \n");
    scanf("%s",str2);
    s=strcmp(str1,str2);
    if(s==0)
    {
    printf("\n The Given Strings are Equal \n");
    }
    else
    {
    printf("\n The Given Strings are not Equal\n");
    }
    break;
    }

    case 5:
    {
    printf("\n Reverse the String\n");
    printf("\n ********************************\n");
    printf("\n Enter the String\n");
    scanf("%s",str1);
    strrev(str1);
    printf("Reverse the String is: %s",str1);
    break;
    }

    case 6:
    exit();
    break;
    }
    printf("\n Do you want to continue(0/1) \n");
    scanf("%d",&y);
    }
    while(y!=0);
    getch();
    }

    C Program to Display Students Mark Details Using Array

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    char name[20][20],result[20][20],grade[20][20];
    int sno[20],m1[20],m2[20],m3[20],total[20],n,I,s=1;
    float avg[20];
    clrscr();
    printf("Student Mark Details\n");
    printf("*****************\n");
    printf("Enter the number of students\n");
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
    printf("Student:%d\n",i);
    printf("Enter the Student number\n");
    scanf("%d",&sno[i]);
    printf("Enter the Student name\n");
    scanf("%s",name[i]);
    printf("Enter the Mark1\n");
    scanf("%d",&m1[i]);
    printf("Enter the Mark2\n");
    scanf("%d",&m2[i]);
    printf("Enter the Mark3\n");
    scanf("%d",&m3[i]);
    s++;
    }
    for(i=1;i<=n;i++)
    {
    total[i]=m1[i]+m2[i]+m3[i];
    avg[i]=total[i]/3;
    if(m1[i]>=35&&m2[i]>=35&&m3[i]>=35)
    {
    strcpy(result[i],"Pass");
    if(avg>=80)
    strcpy(grade[i],"Distinction");
    else if(avg[i]>=60&&avg[i]<=80)
    strcpy(grade[i],"First Class");
    if(avg[i]>=50&&avg[i]<=60)
    strcpy(grade[i],"Second Class");
    else
    strcpy(grade[i],"Third Class");
    }
    else
    {
    strcpy(result[i],"Fail");
    strcpy(grade[i],"***********");
    }
    }
    printf("-----------------------------------------------------------------\n");
    printf("SNo\tName\tMark1\tMark2\tMark3\tTotal\tAverage\tResult\tGrade\n");
    for(i=1;i<=n;i++)
    {
    printf("%d\t%s\t%d\t%d\t%d\t%d\t%2f\t%s\t%s\n",sno[i],name[i],m1[i],m2[i],m3[i],total[i],avg[i],result[i].grade[i]);
    }
    getch();
    }

    C Program to Perform Transpose of a Matrix

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int a[50][50],b[50][50],i,j,r,c;
    clrscr();
    printf("\t Trtanspose Matrix:\n");
    printf("\t ***************\n");
    printf("Input:\n");
    printf("****\n");
    printf("\nEnter the order of matrix:\n");
    scanf("%d %d",&r,&c);
    printf("\nEnter the element of the matrix:");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c;j++)
    {
    scanf("%d",&a[i][j]);
    }
    }
    printf("\n Given Matrix is:\n");
    printf("\n*************\n");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c;j++)
    {
    printf("%d\t",&a[i][j]);
    }
    printf("\n");
    }
    printf("\nOutput:\n");
    printf("\n*****\n");
    printf("\nTranspose of a matrix is:\n");
    printf("\n*******************\n");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c;j++)
    {
    printf("%d\t",a[j][i]);
    }
    printf("\n");
    }
    getch();
    }

    C Program to Perform Matrix Multiplication

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int a[50][50],b[50][50],x[50][50];
    int i,j,r,r1,c,c1,k;
    clrscr();
    printf("\t Matrix Multiplication:\n");
    printf("\t ***********\n");
    printf("Input:\n");
    printf("****\n");
    printf("\nEnter the order of first matrix:\n");
    scanf("%d %d",&r,&c);
    printf("\nEnter the order of second matrix:\n");
    scanf("%d %d",&r1,&c1);
    if(c==r1)
    {
    printf("\nEnter the element for the first matrix:\n");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c;j++)
    {
    scanf("%d",&a[i][j]);
    }
    }
    printf("\nFirst Matrix is:\n");
    printf("\n*********\n");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c;j++)
    {
    printf("%d\t",a[i][j]);
    }
    printf("\n");
    }
    printf("\nEnter the element for the first matrix:\n");
    for(i=1;i<=r1;i++)
    {
    for(j=1;j<=c1;j++)
    {
    scanf("%d",&b[i][j]);
    }
    }
    printf("\n Second Matrix is:\n");
    printf("\n***********\n");
    for(i=1;i<=r1;i++)
    {
    for(j=1;j<=c1;j++)
    {
    printf("%d\t",b[i][j]);
    }
    printf("\n");
    }
    printf("\nOutput:\n");
    printf("\n*****\n");
    printf("\nMultiplication of Two Matrix:\n");
    printf("\n***************\n");

    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c1;j++)
    {
    x[i][j]=0;
    for(k=1;k<=c;k++)
    {
    x[i][j]=x[i][j]+a[i][k]*b[k][j];
    }
    }
    }
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c1;j++)
    {
    printf("%d\t",x[i][j]);
    }
    printf("\n");
    }
    }
    else
    {
    printf("\nOutput:\n");
    printf("\n*****\n");
    printf("Matrix Multiplication is Not possible:");
    }
    getch();
    }

    C Program to Perform Matrix Subtraction

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int a[50][50],b[50][50],x[50][50];
    int i,j,r,r1,c,c1;
    clrscr();
    printf("\t Matrix Subtraction:\n");
    printf("\t ***************\n");
    printf("Input:\n");
    printf("****\n");
    printf("\nEnter the order of first matrix:\n");
    scanf("%d %d",&r,&c);
    printf("\nEnter the order of second matrix:\n");
    scanf("%d %d",&r1,&c1);
    if(r==r1 && c==c1)
    {
    printf("\nEnter the element for the first matrix:\n");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c;j++)
    {
    scanf("%d",&a[i][j]);
    }
    }
    printf("\nFirst Matrix is:\n");
    printf("\n***********\n");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c;j++)
    {
    printf("%d\t",a[i][j]);
    }
    printf("\n");
    }
    printf("\nEnter the element for the first matrix:\n");
    for(i=1;i<=r1;i++)
    {
    for(j=1;j<=c1;j++)
    {
    scanf("%d",&b[i][j]);
    }
    }
    printf("\n Second Matrix is:\n");
    printf("\n**************\n");
    for(i=1;i<=r1;i++)
    {
    for(j=1;j<=c1;j++)
    {
    printf("%d\t",b[i][j]);
    }
    printf("\n");
    }
    printf("\nOutput:\n");
    printf("\n******\n");
    printf("\nSubtraction of Two Matrix:\n");
    printf("\n*********************\n");

    for(i=1;i<=r;i++)
    {
    for(j=1;j<=r1;j++)
    {
    x[i][j]=a[i][j]-b[i][j];
    }
    }
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=r1;j++)
    {
    printf("%d\t",x[i][j]);
    }
    printf("\n");
    }
    }
    else
    {
    printf("Matrix Subtraction is not possible:");
    }
    getch();
    }

    C Program to Perform Matrix Addition

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int a[50][50],b[50][50],x[50][50];
    int i,j,r,r1,c,c1;
    clrscr();
    printf("\t Matrix Addition:\n");
    printf("\t**********\n");
    printf("Input:\n");
    printf("****\n");
    printf("\nEnter the order of first matrix:\n");
    scanf("%d %d",&r,&c);
    printf("\nEnter the order of second matrix:\n");
    scanf("%d %d",&r1,&c1);
    if(r==r1 && c==c1)
    {
    printf("\nEnter the element for the first matrix:\n");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c;j++)
    {
    scanf("%d",&a[i][j]);
    }
    }
    printf("\nFirst matrix is:\n");
    printf("\n********\n");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=c;j++)
    {                
    printf(%d\t",a[i][j]);
    }
    printf("\n");
    }
    printf("\nEnter the element for the second matrix:\n");
    for(i=1;i<=r1;i++)
    {
    for(j=1;j<=c1;j++)
    {
    scanf("%d",&b[i][j]);
    }
    }
    printf("\nSecond matrix is:\n");
    printf("\n**********\n");
    for(i=1;i<=r1;i++)
    {
    for(j=1;j<=c1;j++)
    {
    printf("%d\t",b[i][j]);
    }
    printf("\n");
    }
    printf("\n Output:\n");
    printf("\n*****\n");
    printf("\nAddition of Two matrix:\n");
    printf("\n*************\n");
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=r1;j++)
    {
    x[i][j]=a[i][j]+b[i][j];
    }
    }
    for(i=1;i<=r;i++)
    {
    for(j=1;j<=r1;j++)
    {
    printf("%d\t",x[i][j]);
    }
    printf("\n");
    }
    }
    else
    {
    printf("\nOutput:\n");
    printf("\n*****\n");
    printf("Matrix Addition is not possible:");
    }
    getch();
    }

    C Program to Demonstrate Descending Order

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int a[50],i,j,n,t;
    clrscr();
    printf("\n\t Descending Order:\n");
    printf("\n\t***********\n");
    printf("Input:\n");
    printf("****\n");
    printf("\nEnter the no of numbers:\n");
    scanf("%d",&n);
    printf("\nEnter the Array of Elements:\n");
    for(i=1;i<=n;i++)
    {
    scanf("%d",&a[i]);
    }
    for(i=1;i<=n;i++)
    {
    for(j=i+1;j<=n;j++)
    {
    if(a[i]<a[j])
    {
    t=a[i];
    a[i]=a[j];
    a[j]=t;
    }
    }
    }
    printf("\nOutput:\n");
    printf("\n****\n");
    printf("\nDescending order is:\n");
    for(i=1;i<=n;i++)
    {
    printf("%d\n",a[i]);
    }
    getch();
    }

    C Program to Demonstrate Ascending Order

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int a[50],i,j,n,t;
    clrscr();
    printf("\n\t Ascending Order:\n");
    printf("\n\t**************\n");
    printf("Input:\n");
    printf("*****\n");
    printf("\nEnter the no of numbers:\n");
    scanf("%d",&n);
    printf("\nEnter the Array of Elements:\n");
    for(i=1;i<=n;i++)
    {
    scanf("%d",&a[i]);
    }
    for(i=1;i<=n;i++)
    {
    for(j=i+1;j<=n;j++)
    {
    if(a[i]>a[j])
    {
    t=a[i];
    a[i]=a[j];
    a[j]=t;
    }
    }
    }
    printf("\nOutput:\n");
    printf("\n*******\n");
    printf("\nAscending order is:\n");
    for(i=1;i<=n;i++)
    {
    printf("%d\n",a[i]);
    }
    getch();
    }

    C Program for Searching an Element

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int a[50],i,j=0,n,x;
    clrscr();
    printf("\n\t Search an Element:\n");
    printf("\n\t ************\n");
    printf("Input:\n");
    printf("****\n");
    printf("\nEnter the no of numbers:\n");
    scanf("%d",&n);
    printf("\nEnter the Array of Elements:\n");
    for(i=1;i<=n;i++)
    {
    scanf("%d",&a[i]);
    }
    printf("\nEnter the no to be search:\n");
    scanf("%d",&x);
    for(i=1;i<=n;i++)
    {
    if(a[i]==x)
    j++;
    }
    printf("\nOutput:\n");
    printf("\n****\n");
    if(j>0)
    printf("The No %d is present %d times",x,j);
    else
    printf("The No %d is not present",x);
    getch();
    }

    C Program for Swapping of Two Numbers

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int a,b,c;
    clrscr();
    printf("\t\t\n Swapping of Two Number\n");
    printf("\t\t\n ***************\n");
    begin:
    printf("\n 1.Using Temporary Variable\n");
    printf("\n 2.Without using Temporary variable\n");
    printf("\n 3.Exit \n");
    printf("\n Enter your choice:\n");
    scanf("%d",&c);
    switch(c)
    {
    case 1:
    printf("\n Input:\n");
    printf("\n****\n");
    printf("\nBefore Swapping\n");
    printf("\n**********\n");
    printf("\nEnter the a value:\n");
    scanf("%d",&a);
    printf("\nEnter the b value:\n");
    scanf("%d",&b);
        c=a;
        a=b;
        b=c;
    printf("\n Output:\n");
    printf("\n*****\n");
    printf("\nAfter Swapping\n");
    printf("\n*********\n");
    printf("\nA value is:%d\n",a);
    printf("\nB value is:%d\n",b);
    break;

    case 2:
    printf("\nInput:\n");
    printf("\n***\n");
    printf("\nBefore Swapping\n");
    printf("\n**********\n");
    printf("\nEnter the a value:\n");
    scanf("%d",&a);
    printf("\nEnter the b value:\n");
    scanf("%d",&b);
    a=a+b;
    b=a-b;
    a=a-b;
    printf("\nOutput:\n");
    printf("\n****\n");
    printf("\nAfter Swapping\n");
    printf("\n*********\n");
    printf("\nA Value is:%d\n",a);
    printf("\nB Value is:%d\n",b);
    break;
    case 3:
    exit(0);
    break;
    }
    getch();
    }

    C Program for Number Checking - Sum Reverse Palindrome Prime and Armstrong Number

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int n,s,r,n1,c,p,i;
    clrscr();
    printf("\n 1.Sum of Digit \n");
    printf("\n 2.Reverse Number \n");
    printf("\n 3.Palindrome \n");
    printf("\n 4.Prime Number \n");
    printf("\n 5.Armstrong Number \n");
    printf("\n 6.Exit \n");
    printf("\n Enter your Choice\n");
    scanf("%d",&c);
    switch(c)
    {
    case 1:
    printf("\n Sum of Digit:\n");
    printf("\n -----------\n");
    printf("\n Input:\n");
    printf("******\n");
    printf("\n Enter the Number:\n");
    scanf("%d",&n);
    read:
    s=0;
    while(n>0)
    {
    r=n%10;
    s=s+r;
    n=n/10;
    }
    if(s>9)
    {
    n=s;
    goto read;
    }
    printf("\nOutput:\n");
    printf("\n*****\n");
    printf("\nSum of Digit is:\n%d\n",s);
    break;

    case 2:
    s=0;
    printf("\n Reverse of Digit:\n");
    printf("\n --------------\n");
    printf("Input: \n");
    printf("****\n");
    printf("\nEnter the Number:\n");
    scanf("%d",&n);
    while(n>0)
    {
    r=n%10;
    s=s*10+r;
    n=n/10;
    }
    printf("\nOutput:\n");
    printf("\n*****\n");
    printf("\nReverse Number is:%d",s);
    break;

    case 3:
    printf("\n Palindrome Number:\n");
    printf("\n ------------------\n");
    printf("Input: \n");
    printf("****\n");
    printf("\nEnter the Number:\n");
    scanf("%d",&n);
    n1=n;
    while(n>0)
    {
    r=n%10;
    s=s*10+r;
    n=n/10;
    }
    printf("\nOutput:\n");
    printf("\n*****\n");
    if(s==n1)
       printf("\n The Given no is Palindrome\n");
    else
       printf("\n The Given no is Not Palindrome\n");
    break;

    case 4:
    printf("\n Prime Number:\n");
    printf("\n -------------\n");
    printf("Input: \n");
    printf("****\n");
    printf("\nEnter the Number:\n");
    scanf("%d",&n);
    for(i=2;i<=n/2;i++)
    {
     if(n%i==0)
    p=0;
    }
    printf("\nOutput:\n");
    printf("\n*****\n");
    if(p==0)
       printf("\nThe Given No is not prime\n");
    else
       printf("\nThe Given No is prime\n");
    break;

    case 5:
    printf("\n Armstrong Number:\n");
    printf("\n -----------------\n");
    printf("Input: \n");
    printf("****\n");
    printf("\nEnter the Number:\n");
    scanf("%d",&n);
    n1=n;
    while(n>0)
    {
    r=n%10;
    s=s+(r*r*r);
    n=n/10;
    }
    printf("\nOutput:\n");
    printf("\n*****\n");
    if(s==n1)
          printf("\n The Given no is armstrong no\n");
    else
          printf("\n The Given no is not armstrong no\n");
    break;
    case 6:
    exit(0);
    break;
    }
    getch();
    }

    C Program to Demonstrate Adam Number Checking

    with 0 Comment
    #include<stdio.h>
    #include<conio.h>
    void main()
    {
    int n,x1,x,r,r1,s=0,s1,s2=0,a;
    clrscr();
    Begin:
    {
    printf("\t\t\n Adam Number \n");
    printf("\t\t\n ************\n");
    printf("\n Input: \n");
    printf("\n****\n");
    printf("\n Enter the Number:\n");
    scanf("%d",&n);
    x=n;
    while(n>0)
    {
    r=n%10;
    s=s*10+r;
    n=n/10;
    }
    s1=s*s;
    x1=x*x;
    while(x1>0)
    {
    r1=x1%10;
    s2=s2*10+r1;
    x1=x1/10;
    }
    printf("\n Output:\n");
    printf("\n ****\n");
    if(s1==s2)
    printf("\n The Given no is adam\n");
    else
    printf("\n The Given no is not adam\n");
    pritnf("\n Do you want to continue(0/1) \n");
    scanf("%d",a);
    if(a==1)
    {
    goto Begin;
    }
    }
    getch();
    }

    ------------
    OUTPUT:
    ------------
    INPUT:
    --------
    Enter the Number:
    12

    OUTPUT:
    ----------
    The Given no is adam
    Do you want to continue(0/1)

    INPUT:
    --------
    Enter the Number:
    15

    OUTPUT:
    ----------
    The Given no is not adam
    Do you want to continue(0/1)

    CISCO CCNA Module 1,2,3,4 All Chapters Model Questions and Answers

    with 0 Comment
    CCNA Module Sample Questions:
    Good News for all CCNA Aspirants,here are the collections of all model questions and answers for all chapter exams in module 1,2,3 and 4.This Would be very useful while preparing for CISCO CCNA certification exams.
     
    Module I:

    Chapter 1, Chapter 2, Chapter 3, Chapter 4, Chapter 5, Chapter 6, Chapter 7, Chapter 8, Chapter 9, Chapter 10, Chapter 11

    Module II:

    Chapter 1, Chapter 2, Chapter 3, Chapter 4, Chapter 5, Chapter 6, Chapter 7, Chapter 8, Chapter 9, Chapter 10, Chapter 11

    Module III:

    Chapter 1, Chapter 2, Chapter 3, Chapter 4, Chapter 5, Chapter 6 ,Chapter 7

    Module IV:

    Chapter 1, Chapter 2, Chapter 3, Chapter 4, Chapter 5, Chapter 6 ,Chapter 7

    Dear Engineers,
    If you have any model or study materials related to CISCO certification program.please do share with us for the benefits of all ccna aspirants.

    Tuesday, February 18, 2014

    with 0 Comment
    AnnaUniversity Engineers Related Sites:

    IndiaEngineer.Net , Rejinpaul , Knowafest , TechnicalSymposium , Studentblog100 , mykalvi , grabthenew , Mechanical , AnnaUnivOnline , AnnaUnivquestionpapers , QuestionBank4u , ursarjun , IndiaStudyChannel , Paprix , Chennaikalvi , EEways , EnggEdu , ResultSaifi , TnSchools  , HTcampus , StudyChaCha , VisionStudents ,

    Technology Blogs:
    Labnol , ShoutMetech , Mashable , Gizmodo , GigaOM , ZDNet , TechCrunch , ComputerWorld , GoogleEnterPrise , Apple Weblog , HowtoGeek , ITBusinessEdge , Techankit , Gupable , LatestTechTips , Avgjoegeek , Techbeat ,

    Hacking blogs:
    HackingArticles , Hackw0rm , Secrethacker-akshay , OfficialNet4hack , AkshayHacker7

    Thesis theme Related sites:
    diythemes , ThesisHooks , ThesisCustomizationGuide ,

    Wordpress Theme Sites:

    Blogger Template sites:

    Wednesday, February 5, 2014

    Why Engineering Students Should blog?

    with 0 Comment

    Engineering Course:

    • Engineering is one of the professional course among the medical and law studies (or) branches.The 'Number' of engineering students passing out has been increasing every year,ever since engineering has become the favourite of most of the above average students.In 2014,a whole number of 12lakh engineers are graduated from engineering colleges in India.

    Knowledge Sharing Concept:

    • The only term that revolves around all the engineering buddies is 'Knowledge'.The concept 'Knowledge Sharing' is not very common among engineering students and engineering colleges.In current trend,the 'knowledge sharing' is achieved through symposiums,workshops and conferences only.Have they been effective?If yes! How much it is been useful and how much it is used by the engineering students.

    Symposium,Workshops,Conferences???

    • To enrich the knowledge sharing concept and eradicate the above featured issue,every engineering student should start a blog and write about his own interest and gained knowledge in day to day life during engineering or even outside. It should be viewed by other engineering students to update their knowledge in the respective field.Surely,there is no doubt that blogging helps others to enrich their knowledge and encourages the blogger to study,learn more and blog more.

    Start at Blogger:

    • Starting a blog does not gonna cost even a single penny.The only thing you need is to have a good source of knowledge and producing it in a quality content manner.You can create a blog in blogger.com or wordpress.com.Blogger is very user friendly and THE recommended content mangement system(CMS) platform for begineers.So,Don't wait to start a blog.Start Early ! Stay Ahead ! 

    Saturday, February 1, 2014

    Wireless ARM based Automatic Meter Reading System Using RFID

    with 0 Comment
    • This Project implements a wireless ARM-based automatic meter reading control system(WAMRCS) for distribution automation.It is designed based on a 32-bit ARM microprocessor to deal with power data processing and relay control.In order to provide a cost -effective,wireless,always-connected,two way data link between utility company and WAMRCS. 
      • The WAMRCS sends information of utility usage,power quality and outage alarm to utility company via wireless network.Compared with analog utility meter reading by man power,WAMRCS is more accurate ,reliable,cost-effective,quick and free from man-made errors.It can provide extra capabilities such as distribution automation,load management and time-of-use rate.

      Improved Algorithm for Quadruple Deblocking Filter in the Removal of Block Artifacts

      with 0 Comment
      • The Recent researches showed that the existing quardruple deblocking algorithm fails in selecting the artifact factors and increases the time computation complexity.A Proposed novel algorithm for selecting the block artifacts and to justify the blocked pixels from the decompressed video and the blocks can be removed by involving three different stage by stage approach,which are selected through each pixel,which improves the visual perspective of the video. 
        • This innovation is achieved by considering both the luminance and chrominance data arranging in the filtering schedule .This gives higher real-time performance and also provides a good trade-off between enhancing the performance and it gives the improvement in both subjective and objective image quality.The proposed algorithm is justified in the output video by observing  the visual perspective and the processing time is been much reduced compared to the quadruple deblocking filter.This improved algorithm is implemented using MATLAB GUIDE.

        Secured System Access Using RFID

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        • The Purpose of our project is to provide a secured access to the highly secured area.The security given for the user should be flexible to use,so inorder to provide high security and flexibility,we go for RFID(Radio Frequency Identification) Technology.It is a generic term for technologies that uses radio waves to automatically identify people or objects. 
          • RFID system in general consists of a RFID tag and reader.The tag consist of a chip and antenna,the antenna enables the chip to transmit the ID information to the reader and the reader gets the information and passes the information to the system through parallel port and checks with the database maintained and displays whether the access is granted or denied.Thus this system limits the time taken for accessing and provides a flexible access.This is the challenging criteria to be solved by our project.

          Drowsy Driver Detection and Warning System

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          • In this project our proposed system is used for finding the status of the driver and to alert him in case of drowsiness.DDDS is a non-invasive method to detect the drowsiness in the transportation industry and it would allow avoidance of accidents due to loss of alertness.In DDDS,a web camera is used for capturing the video,which is then splitted into frames.Then the face is detected through skin tone detection method using YCbCr color space. 
            • Now the eye region is segmented from the face area,which is then used for template matching technique.In order to detect whether the driver is drowsy or not,the status of the eye has to be tracked.For this the centroid of the eye has to be calculated.If the centroid cannot be found for ten consecutive frames,the system draws the conclusion that the driver is falling asleep.Thus the driver fatigue is confirmed and issuing a warning signal can alert the driver.The experimental results show validity of our proposed method.

            Design of AutoGuard System Based on RFID and GSM Network

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            • The Project features on reducing the theft rate of the car and meets the auto-guard demand of people.The auto-guard system was designed by combining the technology of Radio Frequency Identification(RFID) and Global System for Mobile(GSM)Communication.The Embedded microcontroller used here is LPC2129 microcontroller.
            • RFID Technology is used to identify the user using the electronic product code.The Electronic product code is transferred to the embedded microcontroller.The Microcontroller is programmed to check whether the number corresponds to authorized person or not.After identifying the person as authorized one,the user is allowed to enter the password.If the entered password is correct,the relay circuit will activate the car engine.If the user is unauthorized,the message will be sent to the car owner immediately through GSM modem.The user can change the password instantly.The Thieves willnot be able to restore the password and cannot be able to switch ON the car once again.
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