Chapter 3: Programming Exercise

Write a function using reference variables as arguments to swap the values of a pair of integers.

#include<iostream>
#include<iomanip>
using namespace std; 
void swap_func(int &a,int &b)
{
 
    cout<<" Before swapping "<<endl
        <<" a = "<<a<<endl<<" b = "<<b<<endl<<endl;
    int temp;
    temp=a;
    a=b;
    b=temp;
    cout<<" After swapping "<<endl
        <<" a = "<<a<<endl<<" b = "<<b<<endl<<endl;
 
}
 
int main()
{
    int x,y;
    cout<<"  Enter two integer value : "<<endl;
    cin>>x>>y;
    swap_func (x,y);
    return 0;
}

Write a function that creates a vector of user given size M using new operator.

#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
    int m;
    int *v;
    cout<<"  Enter vector size : "<<endl;
    cin>>m;
    v=new int [m];
    cout<<" to check your performance  insert "<<m<<" integer value"<<endl;
    for(int i=0;i<m;i++)
    {
        cin>>v[i];
    }
    cout<<" Given integer value are :"<<endl;
    for(i=0;i<m;i++)
    {
 
        if(i==m-1)
        cout<<v[i];
        else
        cout<<v[i]<<",";
 
    }
    cout<<endl;
    return 0;
}

 

Write a program to print the following outputs using for loops

1
22
333
4444
55555
………………

#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
    int n;
    cout<<" Enter your desired number :"<<endl;
    cin>>n;
    cout<<endl<<endl;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=i;j++)
        {
            cout<<i;
        }
        cout<<endl;
    }
    return 0;
}

 

Write a program to evaluate the following investment equation
V = P(1+r)n
and print the tables which would give the value of V for various
of the following values of P, r and n:
P: 1000, 2000, 3000,……………,10,000
r: 0.10, 0.11, 0.12,………………….,0.20
n: 1, 2, 3,…………………………………..,10
(Hint: P is the principal amount and V is the value of money at the end of n years. This equation can be recursively written as
V = P(1 + r)
P = V
In other words, the value of money at the end of the first year becomes the principal amount for the next year and so on)

#include<iostream>
#include<iomanip>
#include<math>
#define  size 8
using namespace std;
int main()
{
    float v,pf;
    int n=size;
    float p[size]={1000,2000,3000,4000,5000,6000,7000,8000};//9000,1000};
    float r[size]={0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18};//,0.19,0.20};
 
    cout<<setw(5)<<"n=1";
        for(int i =2;i<=size;i++)
    cout<<setw(9)<<"n="<<i;
    cout<<"\n";
 
    for(i=0;i<size;i++)
    {
        cout<<setw(-6)<<"p=";
        for(int j=0;j<size;j++)
        {
           if(j==0)
            pf=p[i];
 
            v=pf*(1+r[i]);
 
            cout.precision(2);
                    cout.setf(ios::fixed, ios::floatfield);
            cout<<v<<setw(10);
              pf=v;
         }
         cout<<"\n";
 
     }
    return 0;
}

An election is contested by five candidates. The candidates are numbered 1 to 5 and the voting is done by marking the candidate number on the ballot paper. Write a program to read the ballots and count the vote cast for each candidate using an array variable count. In case, a number read is outside the range 1 to 5, the ballot should be considered as a “spoilt ballot” and the program should also count the numbers of “spoilt ballots”.

#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
    int count[5];
    int test;
    for(int i=0;i<5;i++)
    {
        count[i]=0;
    }
    int spoilt_ballot=0;
    cout<<" You can vot candidate 1 to 5 "<<endl
     <<" press 1 or 2 or 3 or 4 or 5 to vote "<<endl
     <<" candidate 1 or 2 or 3 or 4 or 5 respectively "<<endl
     <<" press any integer value outside the range 1 to 5 for NO VOTE    "<<endl<<" press any negative value to terminate  and see result :"<<endl;
 
    while(1)
        {
            cin>>test;
 
            for(int i=1;i<=5;i++)
            {
                if(test==i)
                {
                    count[i-1]++;
                }
            }
           if(test<0)
                break;
          else if(test>5)
                spoilt_ballot++;
        }
        for(int k=1;k<=5;k++)
        cout<<" candidate "<<k<<setw(12);
        cout<<endl;
        cout<<setw(7);
        for(k=0;k<5;k++)
        cout<<count[k]<<setw(13);
        cout<<endl;
        cout<<" spoilt_ballot "<<spoilt_ballot<<endl;
        return 0;
}

A cricket has the following table of batting figure for a series of test matches:

Write a program to read the figures set out in the above forms, to calculate the batting arranges and to print out the complete table including the averages.

#include<iostream>
#include<iomanip>
char *serial[3]={" FIRST "," SECOND " ," THIRD "};//global declaration
using namespace std;
int main()
{
    int n;
     char name[100][40];
    int *run;
    int *innings;
    int *time_not_out;
    cout<<" How many players' record would you insert ? :";
    cin>>n;
    //name=new char[n];
    run=new int[n];
    innings=new int[n];
    time_not_out=new int[n];
 
    for(int i=0;i<n;i++)
    {
        if(i>2)
        {
            cout<<"\n Input details of "<<i+1<<"th"<<" player's"<<endl;
        }
        else
        {
            cout<<" Input details of "<<serial[i]<<"player's : "<<endl;
        }
 
            cout<<" Enter name : ";
 
            cin>>name[i];
            cout<<" Enter run : ";
            cin>>run[i];
            cout<<" Enter innings : ";
            cin>>innings[i];
            cout<<" Enter times not out : ";
            cin>>time_not_out[i];
    }
 
        float *average;
        average=new float[n];
        for(i=0;i<n;i++)
        {
            float avrg;
            average[i]=float(run[i])/innings[i];
 
        }
        cout<<endl<<endl;
        cout<<setw(12)<<"player's name "<<setw(11)<<"run"<<setw(12)<<"innings"<<setw(16)<<"Average"<<setw(20)<<"times not out"<<endl;
        for(i=0;i<n;i++)
        {
            cout<<setw(14)<<name[i]<<setw(11)<<run[i]<<setw(9)<<innings[i]<<setw(18)<<average[i]<<setw(15)<<time_not_out[i]<<endl;
        }
        cout<<endl;
 
        return 0;
}

Write a program to evaluate the following function to 0.0001% accuracy

(a) sinx = x – x3/3! + x5/5! – x7/7! +…………

(b) SUM = 1+(1/2)2 + (1/3)3 +(1/4)4 + ………

(c) Cosx = 1 –x2/2! + x4/4! – x6/6! + ………

#include<iostream>
#include<math>
#include<iomanip>
#define accuracy 0.0001
#define pi 3.1416
using namespace std;
long int fac(int a)
{
    if(a<=1)
        return 1;
    else
        return a*fac(a-1);
}
int main()
{
    float y,y1,x,fx;
    int n=1;
    int m;
    //const float pi=3.1416;
    cout<<" Enter the value of angle in terms of degree:  ";
        cin>>x;
        float d;
        d=x;
    int sign;
        sign=1;
if(x<0)
{
        x=x*(-1);
        sign=-1;
 }
again:
    if(x>90 && x<=180)
    {
 
        x=180-x;
 
    }
    else if(x>180 && x<=270)
    {
        x=x-180;
        sign=-1;
    }
    else if(x>270 && x<=360)
    {
        x=360-x;
        sign=-1;
    }
 
    else if(x>360)
    {
        int m=int(x);
        float fractional=x-m;
        x=m%360+fractional;
        if(x>90)
            goto again;
        else
        sign=1;
 
    }
    x=(pi/180)*x;
    m=n+1;
    fx=0;
    for(;;)
    {
          long int h=fac(n);
          y=pow(x,n);
          int factor=pow(-1,m);
          y1=y*factor;
          fx+=y1/h;
          n=n+2;
          m++;
          if(y/h<=accuracy)
              break;
    }
 
    cout<<"sin("<<d<<")= "<<fx*sign<<endl;
    return 0;
}

#include<iostream>
#include<math>
#define accuracy 0.0001
using namespace std;
int main()
{
    int n;
    float sum,n1,m;
    n=1;sum=0;
    for(int i=1;;i++)
    {
            n1=float(1)/n;
            m=pow(n1,i);
            sum+=m;
            if(m<=accuracy)
              break;
 
            n++;
            }
    cout<<sum<<"\n";
    return 0;
}
        Sample Output(b)
 
          Solution: (c)
         #include<iostream.h>
#include<math.h>
#define accuracy 0.0001
 
        long int fac(int n)
{
    if(n<=1)
        return 1;
    else
        return n*fac(n-1);
}
 
        int main()
{
 
      float y,y1,x,fx;
      int n=1;
      int m;
      const float pi=3.1416;
      cout<<" Enter the value of angle in terms of degree:  ";
      cin>>x;
      if(x<0)
        x=x*(-1);
      x=(pi/180)*x;
 
        fx=1;
 
         m=2;
        float y2;
        long int h;
        for(;;)
        {
            h=fac(m);
            int factor=pow(-1,n);
            y1=pow(x,m);
            y2=(y1/h)*factor;
            fx+=y2;
            if(y1/h<=accuracy)
                break;
            m=m+2;
            n++;
        }
        cout<<fx<<"\n";
}

Write a program to print a table of values of the function

Y = e-x

For x varying from 0 to 10 in steps of 0.1. The table should appear as follows

TABLE FOR Y =EXP[-X];

X     0.1      0.2       0.3      0.4      0.5      0.6     0.7      0.8      0.900                                    1.0 . . 9.0

 

#include<iostream>
#include<iomanip>
#include<math>
using namespace std;
int main()
{
        float x,y;
        cout<<"                          TABLE FOR  Y=EXP(-X)             :\n\n";
        cout<<"x";
        for(float k=0;k<.7;k=k+0.1)
        cout<<setw(10)<<k;
        cout<<"\n";
        for(k=0;k<10*.7;k=k+0.1)
        cout<<"-";
        cout<<"\n";
               for(float j=0;j<10;j++)
        {
            cout<<j<<setw(4);
            for(float i=0;i<.7;i=i+0.1)
            {
                    x=i+j;
                    y=exp(-x);
                    cout.precision(6);
                                        cout.setf(ios::fixed,ios::floatfield);
                    cout<<setw(10)<<y;
            }
            cout<<"\n";
        }
        return 0;
}

Note: Here we work with 0.4 for a good looking output.

Write a program to calculate the variance and standard deviation of

N numbers

Variance =1/N ∑(xi -x)2

Standard deviation=√1/N ∑(xi -x)2

Where x = 1/N ∑xi

#include<iostream>
#include<math>
using namespace std;
int main()
{
     float *x;
             cout<<" How many number ? :";
     int n;
     cin>>n;
     x=new float[n];
     float sum;
     sum=0;
     for(int i=0;i<n;i++)
     {
            cin>>x[i];
            sum+=x[i];
     }
     float mean;
     mean=sum/n;
     float v,v1;
     v1=0;
     for(i=0;i<n;i++)
     {
            v=x[i]-mean;
            v1+=pow(v,2);
     }
     float variance,std_deviation;
     variance=v1/n;
     std_deviation=sqrt(variance);
     cout<<"\n\n variance = "<<variance<<"\n standard deviation = "<<std_deviation<<"\n";
 
    return 0;
}

An electricity board charges the following rates to domestic users to

discourage large consumption of energy:

For the first 100 units – 60P per unit

For the first 200 units – 80P per unit

For the first 300 units – 90P per unit

All users are charged a minimum of Rs. 50.00. If the total amount is more than Rs. 300.00 then an additional surcharge of 15% is added.

Write a program to read the names of users and number of units consumed and print out the charges with names.

#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
    int unit;
    float charge,additional;
    char name[40];
    while(1)
 
    {
        input:
              cout<<" Enter consumer name & unit consumed :";
                cin>>name>>unit;
                if(unit<=100)
                {
                    charge=50+(60*unit)/100;
                }
                else if(unit<=300 && unit>100)
                {
                    charge=50+(80*unit)/100;
                }
                else if(unit>300)
                {
                    charge=50+(90*unit)/float(100);
                    additional=(charge*15)/100;
                    charge=charge+additional;
                }
                cout<<setw(15)<<"Name"<<setw(20)<<"Charge"<<endl;
                cout<<setw(15)<<name<<setw(20)<<charge<<endl;
                cout<<" Press o for exit / press 1 to input again :";
                int test;
                cin>>test;
                if(test==1)
                goto input;
                else if(test==0)
                break;
    }
    return 0;
}

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