Name-Class-XII

Computer SciencePractical

Simple

Programs

Q1. Write a function that takes time as three integer arguments(hours,minutes,seconds),and returns the number of seconds since the clock last struck "12". Use thisfunction to write a program to calculate the amount of time in seconds betweentwo times, both of which are ithin one 12 hour cycle of clock.

#include<iostream.h>#include<conio.h>#include<math.h>int seconds(int hh,int mm,int ss){int z,s;z=abs(hh-12);s=(z*60*60)+(mm*60)+ss;return s;}

int main(){ clrscr();int h1,h2,m1,m2,s1,s2,t1,t2;cout<<"Enter the first time:"<<endl;cout<<"\tHours:";cin>>h1;cout<<"\tMinutes:";cin>>m1;cout<<"\tSeconds:";cin>>s1;t1=seconds(h1,m1,s1);cout<<"Enter the second time:"<<endl;cout<<"\tHours:";cin>>h2;cout<<"\tMinutes:";cin>>m2;cout<<"\tSeconds:";cin>>s2;t2=seconds(h2,m2,s2);cout<<"The difference in time is:"<<abs(t1-t2);return 0;}

OUTPUT:

Q2. Write a program to find the sum of the sequence#include<iostream.h>#include<conio.h>#include<math.h>void sum(int x,int n){ double sum=0,s=0;int k=1;long fac=1;for(int i=1;i<=n-1;i++){ for(int j=(2*k);j>=1;j--) {fac*=j;}sum+=(pow(x,i)/fac);}s=1+sum;cout<<"\nThe sum is:\n";cout<<s;}void main(){ clrscr();int x,n;cout<<"\nEnter the limiting value\n";cin>>n;cout<<"\nEnter the value \n";cin>>x;sum(x,n);getch();}

OUTPUT:

Q3. Write a c++ program to sum the sequence: -x -(x2/2!) + (x4/4!) -(x6/6!) + .......

#include<iostream.h>#include<conio.h>#include<math.h>int main(){int x,p,i,j;double fact=1.0,ans=0;cout<<"Enter the value of x:";cin>>x;cout<<"Enter till what power you want:";cin>>p;ans=x;for(i=2,j=1;i<=p;i++,j++){fact=fact*i;if(i%2==0)ans+=(pow(-1,j))*((pow(x,i))/(fact));}cout<<"The sum of the series is:"<<ans;return 0;}

OUTPUT:

Q4. Write a program to accept three digits and print all possible combinations from these digits.

#include<iostream.h>#include<conio.h>void main(){int a[3];clrscr();cout<<"Enter three digits :"<<endl;cin>>a[0]>>a[1]>>a[2];for (int i=0;i<3;i++){for(int j=0;j<3;j++){for (int k=0;k<3;k++){if (i!=j&&i!=k&&j!=k)cout<<endl<<endl<<a[i]<<a[j]<<a[k];}}}getch();}

OUTPUT:

Q5. Computers are playing an increasing role in education. Write a program that will help elementary school students learn multiplication. Use rand function to produce two positive one digit integers.

#include<iostream.h>#include<conio.h>#include<stdlib.h>#include<time.h>void main(){ clrscr();int x[2],ans;time_t t;char c;do{ srand(time(&t));x[0]=rand()%10;srand(x[0]);x[1]=rand()%10;cout<<"\n\nWhat is "<<x[0]<<"times "<<x[1]<<" ?\nANS: ";do{cin>>ans;if(ans!=(x[0]*x[1]))cout<<"\nWRONG! TRY AGAIN\n";}while(ans!=(x[0]*x[1]));if(ans==(x[0]*x[1]));{cout<<"correct!\n\n\nDO YOU WANT TO CONTINUE?";cin>>c;}if(c=='N'||c=='n')

{break;}}while(1);}

OUTPUT:

Q6. Write a program to find the LCM and GCD of two numbers.

#include<iostream.h>#include<conio.h>void main(){ clrscr();int x,y,gcd=1;cout<< "ENTER 1st NO : ";cin>>x;cout<<"\n\nEnter 2nd NO. :";cin>>y;for(int i=1;i<1000;++i){if((x%i==0)&&(y%i==0))gcd=i;}cout<<"\n\n\nGCD :"<<gcd;cout<<"\n\n\nLCM :"<<(x*y)/gcd;getch();}

OUTPUT:

Q7. Write a program to find the sum of sequence: - 1 + 1/1! + 1/2!+ .........

#include<iostream.h>#include<conio.h>void main(){ clrscr();int n,i;float sum=1.00;float fact(int a);cout<<"Enter n:";cin>>n;for(i=1;i<n;i++) {sum+=(1/fact(i));}cout<<"Sum of series ="<<sum;getch();}

float fact (int a){ int f=1,i;for (i=1;i<=a;i++)f*=i;return f;}

OUTPUT:

Q8. Write a program in c++ to check if the entered number is prime or not.

#include<iostream.h>#include<conio.h>void main(){clrscr();int p=0,i,n;cout<<"Enter the no :"<<endl;cin>>n;for (i=2;i<=(n/2);i++){if(n%i==0)p=1;}if(p==0)cout<<"Prime";else if (p==1)cout<<"Not Prime";getch();}

OUTPUT:

Q9. Write a program to find the sum of series x2/1 + x3/1 + x3/2 + x4/2 + x4/4 + x4/12 + ………………………………………….

#include<iostream.h>#include<conio.h>#include<math.h>void main(){clrscr();int j,n,x,f,i;float s=1;cout<<"Enter numerator & limit"<<endl;cin>>x>>n;for(i=1;i<n;i++){for(j=1;j<i;j++){f=f*j;}s=s+(pow(x,i)/f);}cout<<"Answer +"<<s;getch();}

OUTPUT:

Q10.Write a program to find the sum of the series

#include<iostream.h>#include<conio.h>#include<math.h>void main(){clrscr();float b=0,x,c,n,s=0,e,r,f=1;int i;cout<<"Enter x & limit"<<endl;cin>>x>>n;for(i=2;i<=n;i++){f=f*i;if(i%2==0)s=s+(pow(x,i)/f);elses=(s-(pow(x,i)/f));}cout<<"Result ="<<s;getch();}

OUTPUT:

Classand

Structure

Based

Programs

Q1. Write a program in c++ to store and display the data of audio tape and book.#include<iostream.h>#include<conio.h>#include<process.h>#include<stdio.h>class publication {char title[20];float price;public:void getdata(){cout<<"Enter Title"<<endl;gets(title);cout<<"Enter Price"<<endl;cin>>price;

}void putdata(){cout<<"Title"<<endl;puts(title);cout<<"Price"<<endl;cout<<price;}};class book : protected publication{int pagecount;public:void putdata(){publication :: putdata();cout<<endl<<"Pg Count"<<endl;cout<<pagecount;}void getdata(){publication::getdata();cout<<"Pg Count"<<endl;cin>>pagecount;}};class tape : protected publication{float time;public:void getdata(){publication :: getdata();cout<<"Enter Time"<<endl;cin>>time;}void putdata(){publication :: putdata();cout<<endl<<"Time "<<endl;cout<<time;}};void main(){clrscr();book b;tape t;

int ch;x:{cout<<endl<<"1. BOOK 2:Tape 3:Exit "<<endl;cin>>ch;switch(ch){case 1:b.getdata();b.putdata();goto x;case 2:t.getdata();t.putdata();goto x;case 3:exit(0);}}

getch();}

OUTPUT:

Q2. Write a c++ program to illustrate a calculator. Perform calculations on two operands using a class calculator. Calculator should add, subtract, multiply and divide operands.

#include<iostream.h>#include<conio.h>#include<process.h>class calculator{float result;int o1,o2;public:void enter();void showresult();void add();void sub();void mul();void div();void clear();};void calculator::enter(){cout<<"Enter a operant:";cin>>o1;cout<<"Enter the other operant:";cin>>o2;}void calculator::showresult(){

cout<<"The result of the operation is:"<<result<<endl;}void calculator::add(){result=o2+o1;}void calculator::sub(){result=o1-o2;}void calculator::mul(){result=o1*o2;}void calculator::div(){result=o1/o2;}void calculator::clear(){result=0;}int main(){char ch='y';calculator c1;while(ch=='y'){c1.enter();cout<<"Which operation do you want to perform:";cin>>ch;switch(ch){case '+':c1.add();break;case '-':c1.sub();break;case '*':c1.mul();break;case '/':c1.div();break;default :cout<<"Wrong choice:";exit(0);}c1.showresult();c1.clear();cout<<"Do you want to continue:";cin>>ch;

}return 0;}OUTPUT:

Q3. Imagine a ticket selling booth at a fair. People passing by are requested to purchase a ticket. A ticket is priced at Rs. 2.50. The booth keeps track of the number of people that have visited the fair and of the total amount of money collected. Model this ticket selling booth with a class tick. Include a program to rest this class.

#include<iostream.h>#include<conio.h>#include<process.h>class tick{int nop;float total;public:tick(){nop=0;

total=0;}void inc();void display();void displaynop();};void tick::inc(){nop++;total+=2.50;cout<<"\nThe no. of people and the total have beem incremented";}void tick::display(){cout<<"\nThe number of people who have entered the fair are:"<<nop<<endl;cout<<"The total amount collected till now is:"<<total<<endl;}void tick::displaynop(){cout<<"The no. of people who have visited so far are:"<<nop<<endl;}int main(){char ch='y';int choice;tick t1;l1:cout<<"\n\n\n\n1.Increment person and total"<<endl;cout<<"2.Display no. of people and the amount collected till now"<<endl;cout<<"3.Display no. of people who entered"<<endl;cout<<"4.Exit"<<endl;cout<<"Enter your choice:";cin>>choice;switch(choice){case 1:t1.inc();goto l1;break;case 2:t1.display();goto l1;break;case 3:t1.displaynop();goto l1;break;

case 4:exit(0);break;}return 0;}OUTPUT:

Q4. Write a program to accept information about a person and then display it.

#include<iostream.h>#include<conio.h>#include<stdio.h>#include<string.h>class person{char name[30];long phone;public:person() {cout<<"\nConstructor initialised\n"; }void getname() {cout<<"\nEnter the name\n";gets(name);cout<<"\nEnter the phone number\n";cin>>phone; }void display()

{cout<<"\nName\t";puts(name);cout<<"\nPhone number\t"<<phone;}~person() {cout<<"\n******************\n"; }};

class spouse:public person{char spousename[40];public:void getsp();void display();};

void spouse::getsp(){ person::getname();cout<<"\nEnter the spouse name\n";gets(spousename);}void spouse::display(){ person::display();cout<<"Spouse name\t";puts(spousename);}void main(){clrscr();spouse obj;obj.getsp();obj.display();getch();}

OUTPUT:

Q5. Write a program to record a score of a cricket match.One array stores information of batting team such as atman's name runs scored, etc. The other array stores information about bowling team. The program reads in above information and depending on user's choice, it displays either batting team's information or bowling team's information.

#include<iostream.h>#include<conio.h>#include<stdio.h>struct bat{char name[20],modeout[70], indica;int runs, score, totruns, totove, xtras;};struct bowl{char name[20];int ttvrs,rnsgvn,wktstkn;};void main(){clrscr();int plno;int plytyp;bat pl1[3];bowl pl2[3];cout<<"Enter the batsmen details:"<<endl;

for (int i=0;i<3;i++){cout<<"Enter name of player "<<i+1<<endl;gets (pl1[i].name);cout<<"enter the runs scored by player "<<i+1<<endl;cin>>pl1[i].runs;cout<<"Enter the overs played by the player"<<i+1<<endl;cin>>pl1[i].totove;cout<<"Enter the status of the player if out (N)or not(Y)"<<endl;cin>>pl1[i].indica;}cout<<"Enter the bowlers details "<<endl;for (i=0;i<3;i++){cout<<"Enter the name of the bowler "<<i+1<<endl;gets(pl2[i].name);cout<<"Enter the runs given by the bowler "<<i+1<<endl;cin>>pl2[i].rnsgvn;cout<<"Enter the wickets taken by the bowler "<<i+1<<endl;cin>>pl2[i].wktstkn;cout<<"Enter the total overs played by the bowler "<<i+1<<endl;cin>>pl2[i].ttvrs;}cout<<"Thank you all details recd"<<endl;xyz:cout<<"Select between batsmen(1) or bowlers(2) to see their details"<<endl;abc:cin>>plytyp;switch (plytyp){case 1:cout<<"Enter the batsman number to see his details "<<endl<<endl<<endl;cin>>plno;plno--;cout<<"Batsman number :"<<plno+1<<endl;cout<<"Batsman name :";puts(pl1[plno].name);cout<<"Runs scored by the batsman :"<<pl1[plno].runs<<endl;cout<<"Total overs played by the batsman :"<<pl1[plno].totove<<endl;cout<<"Player status out "<<pl1[plno].indica<<endl;break;case 2:

cout<<"Enter the bowlers number to see his details "<<endl<<endl<<endl;cin>>plno;plno--;cout<<"Bowlers name :";puts(pl2[plno].name);cout<<"Runs given by the player is :"<<pl2[plno].rnsgvn<<endl;cout<<"Total overs played by the player :"<<pl2[plno].ttvrs<<endl;cout<<"Total wickets taken by the user :"<<pl2[plno].wktstkn<<endl;break;default:cout<<"Idiot enter a decent value"<<endl;goto abc;}cout<<endl<<endl<<endl<<"Do you wish to continue? Y-1 N-2"<<endl;cin>>plno;if (plno==1)goto xyz;elsecout<<"Thank you Press any key to exit";getch();}OUTPUT:

Q6. Write a program to calculate the no. of types of guests i.e.gentlemen or ladies or children using a class.

#include<iostream.h>#include<conio.h>#include<stdio.h>class counter{int count;public:counter(){count=0;}void incount(){count++;}int givecount(){return (count);}}c1,c2,c3;void main(){clrscr();char guest[10];int i,a,b,c;for(i=0;i<10;i++){cout<<"enter gntlmn(g),ladies(l),chldrn(c) "<<endl;cin>>guest[i];if(guest[i]=='g'||guest[i]=='G'){c1.incount();a=c1.givecount();}else if(guest[i]=='l'||guest[i]=='L'){c2.incount();b=c2.givecount();}else{c3.incount();c=c3.givecount();}

}cout<<"GENTLEMEN :"<<a<<endl;cout<<"LADIES :"<<b<<endl;cout<<"CHILDREN :"<<c<<endl;getch();}OUTPUT:

Q7. Write a program to compute tax of a person using a class taxpayer. Class taxpayer should the person's details of account including pan no.

#include<iostream.h>#include<conio.h>#include<stdio.h>class taxpayer{int pan;char name[20];long float tableinc;double tax;public:void inputdata();void display();double computetax();};void taxpayer::inputdata(){cout<<"enterpersonal acct num:";cin>>pan;cout<<"\nenter the name of the person:";

gets(name);cout<<"\nenter total annual taxable income:";cin>>tableinc;}double taxpayer::computetax(){if(tableinc<=60000)tax=0;else if((tableinc>60000)||(tableinc<=150000))tax= tableinc *0.05;else if(tableinc>150000||tableinc<=500000)tax=tableinc*0.1;elsetax=tableinc*0.15;return (tax);}void taxpayer::display(){cout<<"\npan num:"<<pan<<"\tname:";puts(name);cout<<"\ttotal annual income:"<<tableinc<<"\tTax payable:"<<tax;}void main(){clrscr();taxpayer a;a.inputdata();clrscr();a.computetax();a.display();getch();}OUTPUT:

Q8. Write a c++ program to calculate the no. of objects created of a particular class type.

#include<iostream.h>#include<conio.h>class just{int x;public:void enter(){ cout<<"Enter a value of x:";cin>>x;}void out(){cout<<"The value is:"<<x<<endl;}};int main(){just j;char ch='y';int i=0;while(ch=='y'){j.enter();i++;cout<<"Do you eant to continue:";cin>>ch;}cout<<"The number of objects created are:"<<i;return 0;}OUTPUT:

Q9. Write a c++ program using a class student to calculate the percentage of marks obtained by him.

#include<iostream.h>#include<conio.h>class student{int roll_no;char name[20];char class_st[8];int marks[5];float percentage;float calculate();public:void readmarks();void displaymarks();};float student::calculate(){percentage=0;for(int i=0;i<5;i++)percentage+=marks[i];percentage=(percentage/5);return percentage;}void student::readmarks(){cout<<"Enter the roll no.:";cin>>roll_no;cout<<"Enter the name:";cin>>name;cout<<"Enter the class studing in:";cin>>class_st;cout<<"Enter the marks:"<<endl;for(int j=0;j<5;j++){cout<<"\tEnter mark "<<j+1<<":";cin>>marks[j];}}void student::displaymarks(){cout<<"Roll no:"<<roll_no<<endl;cout<<"Name:"<<name<<endl;cout<<"Class:"<<class_st<<endl;cout<<"Percentage:"<<calculate()<<endl;}int main()

{student s1;s1.readmarks();s1.displaymarks();return 0;}OUTPUT:

Q10.Write a c++ program using class serial to store and display serial's title, duration, number of episodes, serial code.

#include<iostream.h>#include<conio.h>#include<stdio.h>class serial{int code;char title[20];float duration;int noe;public:serial(){duration=30;noe=103;}void newserial();void otherentries(int dur,int no);void displaydata();};

void serial::newserial(){cout<<"\n\nEnter the serial code:";cin>>code;cout<<"Enter the title:";gets(title);}void serial::otherentries(int dur,int no){duration=dur;noe=no;}void serial::displaydata(){cout<<"\tSerial code is:"<<code<<endl;cout<<"\tTitle is:"<<title<<endl;cout<<"\tDurations is:"<<duration<<endl;cout<<"\tNo. of episodes are:"<<noe<<endl<<endl;}int main(){clrscr();char ch='y';int i=0,dur,no;serial s1[10];while(ch=='y'){s1[i].newserial();cout<<"Enter the duration and the no. of episodes:";cin>>dur>>no;s1[i].otherentries(dur,no);i++;cout<<"\n\nDo you want to continue:";cin>>ch;}cout<<"\n\nThe details you have entered are:"<<endl<<endl;for(int j=0;j<i;j++){cout<<"Data of serial "<<j+1<<" is:"<<endl;s1[j].displaydata();}return 0;}

OUTPUT:

Data Structu

res

Based Progra

ms

Q1. Write a program in c++ to find the sum of diagonals, rows and columns of a given matrix.#include<iostream.h>#include<conio.h>#include<process.h>

int sum1(int a[50][50],int r){

int s=0;for(int i=0;i<r;i++){ s=s+a[i][i];}return(s);}int sum2(int a[50][50],int r){int s=0;int j;j=r-1;for(int i=0;i<r;i++)s=s+a[i][j--];return(s);}void row(int a[50][50],int r){int s=0;for(int i=0;i<r;i++){s=0;for(int j=0;j<r;j++){s=s+a[i][j];}cout<<"Sum of Row "<<i+1<<" = "<<s<<endl;}}void col(int a[50][50],int r){ int s=0;for(int i=0;i<r;i++){

s=0;for(int j=0;j<r;j++){s=s+a[j][i];}cout<<"Sum of Column "<<i+1<<" = "<<s<<endl;}}void main(){clrscr();int i,s,j,r,c,ch,a[50][50];x:cout<<"Entr Array Limit--(Enter only Row as R=C)"<<endl;cin>>r;

cout<<"Enter Array"<<endl;for(i=0;i<r;i++) {

for(j=0;j<r;j++) {cin>>a[i][j]; } }y:cout<<endl<<endl<<" Enter Choice :"<<endl<<"Sum of---- 1:main diagonal 2:Secondary diagonal 3.Rows 4.Columns 5.Re-enter 6.Exit "<<endl;cin>>ch;switch(ch){case 1:s=sum1(a,r);cout<<"Sum = "<<s<<endl;goto y;case 2 :s=sum2(a,r);cout<<"Sum = "<<s<<endl;goto y;case 3:row(a,r);goto y;case 4:col(a,r);goto y;case 5:goto x;case 6:exit(0);default :cout<<"Wrong Choice"<<endl;break;

}getch();}

OUTPUT:

Q2. Write a program in c++ to implement circular queue.#include<iostream.h>#include<conio.h>#include<process.h>

class queue {int data[10];int front,rear;public:queue() {front=-1;rear=-1; }void add();void remove();void display(); };

void queue::add() {if((rear+1==front)||(rear==9&&front==0)) {cout<<"Overflow "; }else {if((rear==-1) &&(front==-1))

{rear=0;front=0;}

else if(rear==9){rear=0;

}else

{rear++;}

cout<<"Enter the element ";cin>>data[rear]; } }

void queue::remove() {if(front==-1&&rear==-1) {cout<<"Underflow "; }else {if(front==9) {front=0; }else if(front==rear) {front=-1;

rear=-1; }else {front++; } } }

void queue::display() {int i=0,n=10;if(rear==-1) {cout<<"No elements.."<<endl; }else{ if(rear>=front) {for(i=0;i<front;i++)

{cout<<"_ ";}

for(i=front;i<=rear;i++){cout<<data[i]<<" ";}

for(i=rear+1;i<n;i++){cout<<"_ ";}

}else {for(i=0;i<=rear;i++)

{cout<<"_ ";

}for(i=rear+1;i<front;i++)

{cout<<data[i]<<" ";}

for(i=front;i<n;i++){cout<<"_ ";}

} } }

void main() {clrscr();int ch;queue queue; X:cout<<"\nEnter your choice\n1.Insert\n2.Delete\n3.Display\n4.Exit\n";cin>>ch;switch(ch) {case 1:queue.add();

goto X;case 2:queue.remove();

goto X;case 3:queue.display();

goto X;case 4:exit(0); }getch(); }OUTPUT:

Q3. Write a program in c++ to sort the given array using bubble sort.#include<iostream.h>#include<conio.h>#include<process.h>void main(){clrscr();int ch;int i,j,x,k,z,l,m,n,o,p,a[50],small;q :cout<<"Enter the choice 1:Selection 2:Bubble 3:Exchange Selection 4:Insertion 5:Exit"<<endl;cin>>ch;switch(ch){case 1:cout<<"Enter the limit "<<endl;cin>>n;cout<<endl;cout<<"Enter the elements"<<endl;

for(i=0;i<n;i++) {cin>>a[i];cout<<endl; }

for(j=0;j<n;j++) {small=a[j]; p=j;for(i=j;i<n;i++) {if(small>a[i]) {small=a[i]; p=i; } }for(k=p;k>j;k--) {a[k]=a[k-1]; }a[j]=small; }cout<<"Result"<<endl;

for(z=0;z<n;z++) {cout<<a[z];cout<<endl;}goto q;case 2:cout<<"Enter the limit"<<endl;cin>>n;cout<<"Enter the elements"<<endl;for(i=0;i<n;i++)cin>>a[i];for(j=0;j<n;j++) {for(i=0;i<n-1;i++) {if (a[i]>a[i+1]) { x=a[i+1];a[i+1]=a[i];a[i]=x; } } }cout<<"Result"<<endl;for (i=0;i<n;i++) {cout<<a[i];cout<<endl; }break;case 3 :cout<<"Enter the limit "<<endl;cin>>n;cout<<endl;cout<<"Enter the elements"<<endl;

for(i=0;i<n;i++) {cin>>a[i];cout<<endl; }

for(j=0;j<n;j++) {small=a[j]; p=j;for(i=j;i<n;i++)

{if(small>a[i]) {small=a[i]; p=i; } }a[p]=a[j];a[j]=small; }cout<<"Result"<<endl;for(z=0;z<n;z++) {cout<<a[z];cout<<endl; }goto q;case 4 :int m=-32767;cout<<"enter the no. of elements"<<endl;cin>>n;cout<<"enter the array"<<endl;for(i=1;i<=n;i++) {cin>>a[i];}a[0]=m;for(i=1;i<=n;i++) {for(j=0;j<i;j++) {if(a[i]<a[j]) {p=a[i];for(k=i-1;k>=j;k--) {a[k+1]=a[k];}a[j]=p;}}}for(i=1;i<=n;i++) {cout<<a[i];}goto q;case 5:exit(0);default:cout<<"Wrong choice";goto q;

}getch();}

OUTPUT:

Q4. Write a program in c++ to implement stack using linked list.#include<iostream.h>#include<conio.h>#include<process.h>#include<stdio.h>

struct node {char name[20];int age;node *link; }*ptr=NULL,*save=NULL;class stack {node *top;public:stack() {top=NULL; }void stackpush();void stackpop();void display();}st;

void stack::stackpush() {ptr=new node;if(ptr==NULL) {cout<<"Overflow "; }

else {cout<<"Enter the name ";gets(ptr->name);cout<<"Enter the age ";cin>>ptr->age;ptr->link=NULL;if(top==NULL)

{top=ptr;}

else{ptr->link=top;top=ptr;}

} }

void stack::stackpop() {if(top==NULL) {cout<<"Underflow "; }else {save=top;top=top->link;cout<<"Name ";puts(save->name);cout<<"Age "<<save->age;delete save; } }

void stack::display() {if(top==NULL) {cout<<"No elements.."<<endl; }else {ptr=top;while(ptr!=NULL) {cout<<"\nName ";

puts(ptr->name);cout<<"Age "<<ptr->age;ptr=ptr->link;

} } }void main() {clrscr();int ch;

X:cout<<"\nEnter your choice\n1.Insert\n2.Delete\n3.Display\n4.Exit\n";cin>>ch;switch(ch) {case 1:st.stackpush();

goto X;case 2:st.stackpop();

goto X;case 3:st.display();

goto X;default:cout<<"Wrong choice ";

goto X;case 4:exit(0); }getch(); }OUTPUT:

Structured

Query

Language

Queries

Q1. Consider the following tables Product and Client. Write SQL commands for the statement (i) to (iv) and give outputs for SQL queries (v) to (viii).

Table: PRODUCT

P_ID Product Name

Manufacturer

Price

TP01 TalcomPowder LAK 40

FW05 Face Wash ABC 45BS01 Bath Soap ABC 55SH06 Shampoo XYZ 120FW12 Face Wash XYZ 95

Table: CLIENT

C_ID Client Name City P_ID

01 TalcomPowder Delhi FW05

06 Face Wash Mumbai BS0112 Bath Soap Delhi SH0615 Shampoo Delhi FW12

16 Face Wash Banglore TP01

(i) To display the details of those Clients whose city is Delhi.

Ans: Select all from Client where City=”Delhi”;

(ii) To display the details of Products whose Price is in the range of 50 to 100(Both values included).

Ans: Select all from product where Price between 50 and 100;

(iii) To display the ClientName, City from table Client, and ProductName and Price from table Product, with their corresponding matching P_ID.

Ans: Select ClientName,City,ProductName, Price from Product,Client where Product.P_ID=Client.P_ID;

(iv) To increase the Price of all Products by 10

Ans: Update Product Set Price=Price +10;

(v) SELECT DISTINCT Address FROM Client.

Ans: 

CityDelhiMumbaiBangalore

(vi) SELECT Manufacturer, MAX(Price), Min(Price), Count(*) FROM Product GROUP BY Manufacturer;

Ans:

Manufacturer

Max(Price)

Min(Price)

Count(*)

LAK 40 40 1ABC 55 45 2XYZ 120 95 2

(vii) SELECT ClientName, ManufacturerName FROM Product, Client WHERE Client.Prod_Id=Product.P_Id;

Ans:

ClientName ManufacturerName

CosmeticShop

ABC

TotalHealth ABCLiveLife XYZPrettyWoman XYZDreams LAK

(viii) SELECT ProductName, Price * 4 FROM Product.

Ans:

ProductName

Price*4

TalcomPoweder

160

FaceWash 180 BathSoap 220 Shampoo 480 Face Wash 380

Q2. Consider the following tables Item and Customer. Write SQL commands for the statement (i) to (iv) and give outputs for SQL queries (v) to (viii). Table: ITEM

C_ID ItemName Manufacturer Price

PC01 Personal Computer ABC 35000

LC05 Laptop ABC 55000

PC03 Personal Computer XYZ 32000

PC06 Personal Computer COMP 37000

LC03 Laptop PQR 57000

Table: CUSTOMER

C_ID CustomerName City P_ID

01 N.Roy Delhi LC0306 H.Singh Mumbai PC0312 R.Pandey Delhi PC0615 C.Sharma Delhi LC03

16 K.Agarwalh Banglore PC01

(i) To display the details of those Customers whose city is Delhi.

Ans: Select all from Customer Where City=”Delhi”;

(ii) To display the details of Item whose Price is in the range of 35000 to 55000 (Both values included).

Ans: Select all from Item Where Price>=35000 and Price <=55000;

(iii) To display the CustomerName, City from table Customer, and ItemName and Price from table Item, with their corresponding matching I_ID.

Ans: Select CustomerName,City,ItemName, Price from Item,Customer where Item.I_ID=Customer.I_ID;

(iv) To increase the Price of all Items by 1000 in the table Item.

Ans: Update Item set Price=Price+1000;

(v) SELECT DISTINCT City FROM Customer.

Ans:   

City Delhi Mumbai Bangalore

(vi) SELECT ItemName, MAX(Price), Count(*) FROM Item GROUP BY ItemName;

Ans:

ItemName Max(Price)    

Count(*)

PersonalComputer

37000  3 

Laptop  57000 2

(vii) SELECT CustomerName, Manufacturer FROM Item, Customer WHERE Item.Item_Id=Customer.Item_Id;

Ans:

CustomerName

Manufacturer Name

N.Roy PQRH.Singh XYZR.Pandey COMPC.Sharma PQRK.Agarwal ABC

(viii) SELECT ItemName, Price * 100 FROM Item WHERE Manufacturer = ‘ABC’;

Ans:

ItemName Price*100

PersonalComputer

3500000

Laptop 5500000

Q3. Consider the following tables Consignor and Consignee. Write SQL command for the statements(i)to(iv) And give outputs for the SQL quries (v) to ( viii).

TABLE : CONSIGNOR

CnorID CnorName CnorAddress City

ND01 R singhal 24,ABC Enclave New Delhi

ND02 AmitKumar 123,Palm Avenue New Delhi

MU15 R Kohil 5/A,South,Street Mumbai

MU50 S Kaur 7-K,Westend Mumbai

TABLE : CONSIGNEE

CneeID CnorID CneeName CneeAddress CneeCity

MU05 ND01 RahulKishore 5,Park Avenue Mumbai

ND08 ND02 P Dhingr a 16/j,Moore Enclave New Delhi

KO19 MU15 A P Roy 2A,Central/ avenue Kolkata

MU32 ND0 2 S mittal P 245, AB Colony MumbaiND48 MU5 0 B P jain 13,Block d,a,viha New Delhi

(i) To display the names of all consignors from Mumbai.

Ans: Select CnorName from Consignor where city=”Mumbai”;

(ii) To display the cneeID, cnorName, cnorAddress, CneeName, CneeAddress for every Consignee.

Ans: Select CneeId, CnorName, CnorAddress, CneeName, CneeAddress from Consignor,Consignee where Consignor.CnorId=Consignee.CnorId;

(iii) To display the consignee details in ascending order of CneeName.

Ans: Select * from Consignee Orderby CneeName Asc;

(iv) To display number of consignors from each city.

Ans: Select city, count(*) from Consignors group by city;

(v) SELECT DISTINCT City FROM CONSIGNEE;

Ans:CneeCityMumbaiNew DelhiKolkata

(vi) SELECT A.CnorName, B.CneeName FROM Consignor A, Consignee B WHERE A.CnorID=B.CnorID AND B.CneeCity=’Mumbai’;

Ans:

CnorName CneeName

(vii) SELECT CneeName,CneeAddress FROM Consignee WHERE CneeCity Not IN (‘Mumbai’, ‘Kolkata’);

Ans:

CneeName CneeAddressP Dhingr a 16/j,Moore Enclave

B P jain 13,Block d,a,viha

(viii) SELECT CneeID, CneeName FROM Consignee WHERE CnorID = ‘MU15’ OR CnorID = ‘ND01’;

senderCityNew DelhiMumbai

Ans: 

CneeID CneeName

MU05 Rahul KO19 A P Kishore

Roy

Q4.Consider the following tables. Write SQL command for the statements (i)to(iv)and give outputs for the SQL quries (v) to (viii).

TABLE : SENDER

SenderID SenderName

Sender Address

Sender City

ND01 R jain 2,ABC Appts New Delhi

MU02 H sinha 12, Newton Mumbai

MU1 5 S haj 27/ A,Park Street

New Delhi

ND5 0 T Prasad 122-K,SDA Mumbai

TABLE :RECIPIENT

RecID SenderID

ReCName RecAddress ReCCity

KO05 ND01 RBajpayee 5,Central Avenue Kolkata

ND08 MU0 2 S Mahajan 116, A Vihar NewDelh

i

MU19 ND01 H sing 2A,Andheri East Mumbai

MU32 MU1 5 PK Swamy B5, CS erminus Mumbai

ND48 ND50 S Tripathi 13, B1 D,Mayur Vihar

NewDelhi

(i) To display the names of all senders from Mumbai.

Ans: Select * from Sender where SenderCity =’Mumbai’;

(ii) To display the recID, senderName, senderAddress, RecName, RecAddress for every recipt.

Ans: Select recID, SenderName, SenderAddress, RecName, RecAddress from Sender, Recipient where Sender.Senderid=Recipient.RenderId;

(iii) To display the sender details in ascending order of SenderName.

Ans: Select * from Sender order by SenderName;

(iv) To display number of Recipients from each city.

Ans: Select RecCity,Count(*) from Recipient group by RecCity;

 (v) SELECT DISTINCT SenderCity FROM Sender;

Ans:Sender CityNew DelhiMumbai

(vi) SELECT A.SenderName A, B.RecName FROM Sender A, Recipient B WHERE A.SenderID=B. SenderID AND B.RecCity=’Mumbai’;

Ans:     

SenderName

RecName

R.Jain H.Singh    S.Jha P.K.Swamy

(vii) SELECT RecName,RecAddress FROMRecipient WHERE RecCity Not IN (‘Mumbai’,Kolkata’);

Ans:   

RecName RecAddressSMahajan 116,A ViharS Tripati 13, B1 D, Mayur Vihar

(viii) SELECT RecID, RecName FROM Recipient WHERE SenderID = ‘MU02’ OR SenderID = ‘ND50’;

Ans:

RecID RecNameND08 S Mahajan ND48 S Tripathi

Q5. Study the following tables FLIGHTS and FARES and write SQL commands for the questions (i) to (iv).

TABLE: FLIGHTS

FL_NO STARTING ENDING NO_ FLGHTS

NO_ STOPS

IC301 MUMBAI DELHI 8 0

IC799 BANGALORE DELHI 2 1

MC101 INDORE MUMBAI 3 0IC302 DELHI MUMBAI 8 0AM812 KANPUR BANGLORE 3 1IC899 MUMBAI KOCHI 1 4

AM501 DELHI TRIVENDRUM 1 5

MU499 MUMBAI MADRAS 3 3IC701 DELHI AHMEDABAD 4 0

TABLE:FLIGHTS

FL_NO AIRLINES FARE TAX%IC701 INDIAN 6500 10

AIRLINESMU499 SAHARA 9400   5AM501 JET AIRWAYS 13450   8

IC899 INDIAN AIRLINES 8300   4

IC302 INDIAN AIRLINES 4300 10

IC799 INDIAN AIRLINES 1050 10

MC101 DECCAN AIRLINES 3500   4

(i) Display FL_NO and NO_FLIGHTS from “KANPUR” TO “BANGALORE” from the table FLIGHTS.

Ans: Select FL_NO, NO_FLIGHTS from FLIGHTS where Starting=”KANPUR” AND ENDING=”BANGALORE”;

(ii) Arrange the contents of the table FLIGHTS in the ascending order of FL_NO.

Ans: SELECT * From FLIGHTS ORDER BY FL_NO ASC;

(iii) Display the FL_NO and fare to be paid for the flights from DELHI to MUMBAI using the tables FLIGHTS and FARES, where the fare to paid = FARE+FARE+TAX%/100.

Ans: Select FL_NO, FARE+FARE+(TAX%/100) from FLIGHTS, FARES where Starting=”DELHI” AND Ending=”MUMBAI”;

(iv) Display the minimum fare “Indian Airlines” is offering from the tables FARES.

Ans: Select min(FARE) from FARES Where AIRLINES=”Indian Airlines”;

Q6. Study the following tables DOCTOR and SALARY and write SQL commands for the questions (i) to (iv) and give outputs for SQL queries (v) to (vi) :

TABLE: DOCTOR

ID NAME DEPT SEX EXPERIENCE

101 Johan ENT   M   12

104 Smith ORTHOPEDIC   M     5

107 George CARDIOLOGY   M   10

114 Lara SKIN   F     3

109 K George MEDICINE   F     9

105 Johnson

ORTHOPEDIC   M   10

117 Lucy ENT   F     3111 Bill MEDICINE   F   12

130 MurphyORTHOPEDIC   M   15

TABLE: SALARY

ID  BASIC

ALLOWANCE

CONSULTAION

101 12000   1000   300

104 23000   2300   500

107 32000   4000   500

114 12000   5200   100

109 42000   1700   200

105 18900   1690   300

130 21700   2600   300

(i) Display NAME of all doctors who are in “MEDICINE” having more than 10 years experience from the Table DOCTOR.

Ans: Select Name from Doctor where Dept=”Medicine” and Experience>10;

(ii) Display the average salary of all doctors working in “ENT”department using the tables. DOCTORS and SALARY Salary =BASIC+ALLOWANCE.

Ans: Select avg(basic+allowance) from Doctor,Salary where Dept=”Ent” and Doctor.Id=Salary.Id;

(iii) Display the minimum ALLOWANCE of female doctors.

Ans: Select min(Allowance) from Doctro,Salary where Sex=”F” and Doctor.Id=Salary.Id;

(iv) Display the highest consultation fee among all male doctors.

Ans: Select max(Consulation) from Doctor,Salary where Sex=”M” and Doctor.Id=Salary.Id;

(v) SELECT count (*) from DOCTOR where SEX = “F”

Ans: 4

(vi) SELECT NAME, DEPT , BASIC from DOCTOR, SALRY Where DEPT = “ENT” AND DOCTOR.ID = SALARY.ID

Ans:   

  

Name

Dept

Basic

Jonah Ent 12000

Q7. Consider the following tables EMPLOYEES and EMPSALARY. write SQL commands for the Statements (i) to (iv) and give outputs for SQL quires (v) to (viii).

 

EMPID FIRSTNAME

LASTNAME ADDRESS CITY

010  GEORGE Smith 83 First Street Howard

105 MARY Jones 842VineAve Losantiville152 SAM Tones 33 Elm st Paris215  SARAH Ackerman 440 U.S.110 Upton

244  MANILA Sengupta 24 FriendsStreet New Delhi

300  ROBERT Samuel 9 Fifth Cross Washington

335  HENRY Williams 12 Moore Street Boston

400 RACHEL Lee 121 Harrison New York441  PETER Thompson 11 Red road Paris

EMPSALRAY

EMPID

SALARY

BENEFITS

DESIGNATION

010 75000 15000 Manager105 65000 15000 Manager152 80000 25000 Director215 75000 12500 Manager244 50000 12000 Clerk300 45000 10000 Clerk335 40000 10000 Clerk400 32000   7500 Salesman441 28000   7500 Salesman

(i) To display Firstname, Lastname, Address and City of all employees living in Paris from the table EMPLOYEES.

Ans: Select Firstname,Lastname,Address,City from Employees where City=”Paris”;

(ii) To display the content of EMPLOYEES table in descending order of FIRSTNAME.

Ans: Select * from Employees Order By Firstname Desc;

(iii) To display the Firstname, Lastname, and Total Salary of all managers from the tables, where Total Salary is calculated as Salary+Benifts.

Ans: Select Firstname,Lastname,Salary+Benefits from Employees, Empsalary where Designation=”Manager” and Employees.EmpId=EmpSalary.EmpId;

(iv) To display the Maximum salary among Managers and Clerks from the table EMPSALARY.

Ans: Select Designation,max(Salary) from EmpSalary where Designation=”Manager” or Designation=”Clerk”;

 (v) SELECT FIRSTNAME,SALARY FROM EMPLOYEES,EMPSALARY WHERE DESTINATION =’Salesman’AND EMPOLYEES.EMPID=EMPSALARY.EMPID ;

Ans:   

Firstname

Salary

Rachel 32000Peter 28000

(vi) SELECT COUNT (DISTINT DESIGNATION ) FROM EMPSALARY

Ans: 4

(vii) SELECT DESIGNATION , SUM(SALARY) FROM EMPSALARY GROUP BY DESIGNATION HAVING COUNT(*)>2;

Ans:    

Designation

Sum(Salary)

 Manager 215000Clerk 135000

(viii) SELECT SUM (BENEFITS) FROM EMPSALARY WHERE DESIGNATION=’Clerk’;

Ans: 32000

Q8. Consider the following tables WORKERS and DESIG. Write SQL commands for the statements (i) to (iv) and give outputs for SQL queries (v) to (viii).

WORKERS

W_ID  FIRSTNAME

LASTNAME ADDRESS CITY

102  Sam Tones 33 Elm St. Paris105  Sarah Ackerman 44 U.S.110 NewYork

144  Manila Sengup ta 24 Friends Street New Delhi

210  George Smith 83 First Street Howard

255  Mary Jones 842 Vine Ave. Losantiville300  Robert Samuel 9 Fifth Cross Washington

335  Henry Williams 12Moore Street Boston

403 Ronny Lee 121 Harrison St. New York

451  Pat Thomps on 11 Red Road Paris

DESIG

W_ID  SALARY BENEFITS DESIGINA TION

102 75000 15000 Manager105 85000 25000 Director144 70000 15000 Manager210 75000 12500 Manager255 50000 12000 Clerk300 45000 10000 Clerk335 40000 10000 Clerk400 32000 7500 Salesman451 28000 7500 Salesman

(i) To display W_ID Firstname, address andCity of all employees living in New York fromthe Table WORKERs

Ans: select W_ID ,firstname,address,city from workers where city=”New York”;

(ii) To display the content of workers table in ascending order of LASTNAME.

Ans:Select * from Worker Order By lastname Asc;

(iii) To display the FIRSTNAME, LASTNAME and Total Salary of all Clerks from the tables WORKERS And DESIG, where Total salary is calculated as Salary + benifts.

Ans: Select firstname, lastname, salary+benefits where worker.w_id=desg.w_id and Designation=”Clerk”;

(iv) To display the minimum salary among managers and Clerks from the tables DESIG.

Ans: Selet DESIGNATION,min(SALARY) From DESIG Group By DESIGNATION Having DESIGNATION In(“Clerk”,”Manager”);

(v) SELECT FIRSTNAME, SALARY FROM WORKERS, DESIG WHERE DESIGINATION = “MANAGER” AND WORKERS.W_ID = DESIGN.W_ID

Ans:    

FIRSTNAME

SALARY

 Sam 75000Manila 70000

George 75000

(vi)SELECT COUNT(DISTINCT DESIGNATION) FROM DESIGN ;

Ans: 4

(vii) SELECT DESIGNATION, SUM(SALARY) FROM DESIG GROUP BY DESIGNATION HAVING COUNT (*) < 3;

Ans:

   Designation

Sum(Salary)

  Director 85000     Salesman 60000

(viii) SELECT SUM(BENIFTS) FROM DESIG WHERE DESIGINATION =”salesman”;

Ans: 15000

 

Q9. Give the following table for database a LIBRARY.TABLE : BOOKS

BOOK_ID

BOOK_NAME

AUTHONAME

PUBLISHER

PRICE TYPE QUANTIT

Y

F0001  The Tears William Hopkins First Publ 750 Fictio

n    10

F0002  Thund erbolts Anna Roberts First Publ. 700 Fictio

n      5

T0001 My first C+ +

Brains & Brooke EPB 250 Text    10

T0002  C++ Brain works

A.W.Ros saine TDH 325 Text      5

C001  Fast Cook Lata Kapoore EPB 350 Cookery      8

TABLE:ISSUED

Write SQL queries from b to g.

(i)To show Book name, Author name and Price of books of EPB publisher.

Ans: select Book_name,Author_name, price from books where Publisher =”EPB”;

(ii) To list the names of the books of FICTIONS type.

Ans: Select Book_name from books where type=”FICTION”;

(iii) To display the names and prices of the books in descending order of their price.

Ans: Select Book_name, price from books order by price desc;

BOOK_ID

QUANTITY_ISSUED

F0001       3T0001        1C0001       5

(iv) To increase the price of all books of First Pub.by 50.

Ans: update books set price= price+50 where publishers = “First Publ”;

(v) To Display the Book_ID, Book_Name and Quantity Issued for all books Which have been issued.

Ans:Select Book_ID, Book_Name, Quantity_Issued from Books,Issued where Books.BookId=Issued.BookId;

(vi) To insert a new row in the table Issued having the following data: “F0002”,4

Ans: insert into Issued values(“F0002”,4);

(vii) Give the output of the following queries on the above tables:

(1) Select Count(Distinct Publishers) From Books;

Ans: 3

(2) Select Sum(Price) From Books Where Quantity>5;

Ans: 1350.

(3) Select Book_Name,Author_Name From Books Where Price<500;

Ans: Book_Name Author_Name My First C++ Brian & Brooks C++ Brainworks A.W.Rossaine Fast Cook Lata Kapoor.

(4) Select Count(*) From Books;

Ans: 5

Q10. Write SQL commands for (b) to (g) and write the outputs for (h) on the basis of tables TNTERIORS and NEWONES.

TABLE: INTERIORS

NO  ITEM NAME TYPE DATEOFSTOC

KPRICE

DISCOUNT

1 Red rose DoubleBed   23/02/02 32000   152  Soft touch Baby cot   20/01/02 9000   10

3  Jerry’shome Baby cot   19/02/02 8500   10

4  Rough wood Office Table   01/01/02 20000   20

5  Comfort zone Double Bed   12/01/02 15000   20

6  Jerry look Baby cot   24/02/02 7000   197  Lion king Office Table   20/02/02 16000   208  Royal tiger Sofa   22/02/02 30000   25

9  Park sitting Sofa   13/12/01 9000   15

10  Dine DinningTab   19/02/02 11000   15

paradise le

TABLE:NEWONES

NOITEMNAME TYPE DATEOFSTO

CKPRICE

DISCOUNT

11  White wood 

Doublebed   23/03/03  20000    20

12  James007 Sofa   20/02/03 15000    1513  Tom look  Baby cot   21/02/03 7000    10

(i) To show all information about the sofas from the INTERIORS table.

Ans: Select * from INTERIORS where type= “sofa”;

(ii) To list ITEMNAME and TYPE of those items, in which DATEOFSTOCK is before 22/01/02 from the INTERIORS table in descending order of ITEMNAME.

Ans: Select Itemname,Type From Interiors Where Dateofstock<{22/01/02} order by Itemname;

(iii) To display ITEMNAME and DATEOFSTOCK of those items in which the Discount percentage is more than 15 from INTERIORS.

Ans: Select Itemname,Dateofstock from Interiors Where Discount>15;

(iv) To count the number of items whose type is “Double bed”;

Ans: Select Count(*) from Interiors Where Type=”Double Bed”;

(v) To insert new row in the NEWONES table with the following data:14, “True Indian “, “Office Table “, {28/03/03},15000,20.

Ans: Insert into Newones values(14,”True Indian”,”Office Table”,”{28/03/03},15000,20);

(vi) Give the outputs for the following SQL statements.

(1) Select COUNT (distinct TYPE) from INTERIORS;

Ans: 5

(2) Select AVG(DISCOUNT)from INTERIORS where TYPE =”Baby cot”;

Ans: 13

(3) Select SUM(price)from INTERIORS where DATEOFSTOCK<{12/02/02};

Ans: 53000