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Thursday, 2 August 2012

C program for electricity bill


#include<iostream>
using namespace std;
#define cost1 0.60
#define cost2 0.80
#define cost3 0.90
#define min 50
#include<conio.h>
void main()
{
char name[10][20];
float amount[10];
int units[10],n;

cout<<"Enter Number of Customers: ";
cin>>n;

cout<<"Enter Names and Number of Units Consumed\n";
for(int i=0;i<n;i++)
{
cout<<"-------------------------\n";
cout<<"Enter Customer Name: ";
cin>>name[i];
cout<<"Enter Number of units= ";
cin>>units[i];
cout<<"-------------------------\n";

if(units[i]<=100)
{
amount[i]=(units[i]*cost1)+min;
}
else if(units[i]>100&&units[i]<=300)
{
amount[i]=(((units[i]-100)*(cost2))+(100*cost1))+min;

}
else if(units[i]>300)
{
amount[i]=(((units[i]-300)*(cost3))+(200*cost2)+(100*cost1))+min;
amount[i]=amount[i]+(amount[i]*(15.0/100.0));
}
}

cout<<"**************BILL**************\n";
cout<<"Name\t\t"<<"Units\t\t"<<"Amount\n";
for(int i=0;i<n;i++)
{
cout<<name[i]<<"\t\t"<<units[i]<<"\t\t"<<"Rs."<<amount[i];
cout<<"\n";
}
getch();
}









C program to implement Newton Raphson method


1)F(x)=x^2-3x+2

#include<stdio.h>
#include<math.h>
#include<conio.h>
#define e 0.001
#define F(x) (2*x)-3
float frac(float a)
{
       float f1;
       f1=a*a-3*a+2;
       return f1;
}
int main()
{
       float x1,x2,f1=0,f2,er,d;
       printf("F(x) = x^2-3x+2\n\n");
       printf("Enter the value of x1: ");
       scanf("%f",&x1);
       printf("\nx1 = %f",x1);
       printf("\n________________________________________________________________________________\n");
        
       printf("     x1      |       x2      |      f1       |       f'1      |  |(x2-x1)/x2|  |  \n");
       printf("--------------------------------------------------------------------------------\n");
       do
       {
              f1=frac(x1);
              d=F(x1);
              x2=x1-(f1/d);
              er=fabs((x2-x1)/x2);
           printf("  %f   |    %f   |     %f  |     %f  |      %f  |   \n",x1,x2,f1,d,er);
              x1=x2;
       }
       while(er>e);
       printf("--------------------------------------------------------------------------------\n\n");
       printf("\n  Root of the equation is: %f",x2);
       getch();
}


Sunday, 29 July 2012


1) F(x) = X^3 – X - 1

#include<stdio.h>
#include<conio.h>
#include<math.h>
#define e 0.001
float frac(float a)
{
       float f1;
       f1=a*a*a-a-1;
       return f1;
}
void main()
{
       float x1,x2,x0,f0,f1,f2;
       do
       {
       printf("Enter values of x1 and x2\n");
       scanf("%f%f",&x1,&x2);
       printf("\nx1 = %f\n",x1);
       printf("x2 = %f\n",x2);
       f1=frac(x1);
       f2=frac(x2);
       }
       while((f1*f2)>0);
       printf("_______________________________________________________________________________________\n");
        
        printf("     x1      |      x2     |      x0     |      f1       |      f2      |      f0     |\n");
        printf("---------------------------------------------------------------------------------------\n");
        
       do
       {
              f1=frac(x1);
              f2=frac(x2);
              x0=(x1+x2)/2;
              f0=frac(x0);
              if((f1*f0)<0)
              {
                     printf("  %f   |  %f   |  %f   |  %f    |  %f    |   %f  | \n",x1,x2,x0,f1,f2,f0);
                     x2=x0;
                     f2=f0;
              }
              else
              {
                     printf("  %f   |  %f   |  %f   |  %f    |  %f    |  %f  | \n",x1,x2,x0,f1,f2,f0);
                     x1=x0;
                     f1=f0;
                    
              }
       }
       while(fabs(x2-x1)>e);
       printf("---------------------------------------------------------------------------------------\n\n");
       printf("\n  Root of the equation is: %f",x0);
       getch();
}