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Title: C+ program
Description: C+ programming for computational technique assigent
Description: C+ programming for computational technique assigent
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ASSIGNMENT 8
1
...
Y is X + 10 if X is between
10 and 30
...
Otherwise Y is X − 2
...
h>
#include
10
...
otherwise\n");
printf("4
...
A student is awarded Ex grade if he gets more than 90 marks
...
Similarly range for B, C, D and P are 70-79, 60-69, 50-59, and
35-49 respectively
...
Write a
program, which reads marks of a student and prints his grade
...
h>
#include
80
...
60
...
35
...
exit\n");
printf("your choice?\n");
scanf("%d",&choice);
switch(choice)
{
case 1:
printf("\nEnter marks of the student");
scanf("%d",&x);
printf("Marks=%d",x);
printf("Grade is A\n");
break;
case 2:
printf("\nEnter marks of the student");
scanf("%d",&x);
printf("Marks=%d",x);
printf("Grade is B\n");
break;
case 3:
printf("\nEnter marks of the student");
scanf("%d",&x);
printf("Marks=%d",x);
printf("Grade is C\n");
2
break;
case 4:
printf("\nEnter marks of the student");
scanf("%d",&x);
printf("Marks=%d",x);
printf("Grade is D\n");
break;
case 5:
printf("\nEnter marks of the student");
scanf("%d",&x);
printf("Marks=%d",x);
printf("Grade is P\n");
break;
case 6:
printf("\nEnter marks of the student");
scanf("%d",&x);
printf("Marks=%d",x);
printf("Grade is F\n");
break;
case 7:
exit(0);
default:
printf("Wrong choice!\n");
}
}
return 0;
}
3
...
If income is between 5000 and 6000
then tax is 10% of the amount by which the income exceeds 5000
...
If
income is more than 15000 then the tax is 1900 + 30% of the amount by which the income
exceeds 15000
...
g
...
Write a program, which reads income and calculates the income tax
...
h>
#include
x<5000\n");
printf("2
...
6000
...
exit\n");
printf("your choice?\n");
scanf("%d",&choice);
switch(choice)
{
case 1:
printf("\nEnter Income");
scanf("%d",&x);
printf("No income tax\n");
break;
case 2:
printf("\nEnter Income");
scanf("%d",&x);
y=(x-5000)*10/100;
printf("Tax to be paid is:%d\n",y);
break;
case 3:
printf("\nEnter Income");
scanf("%d",&x);
y=100+(x-6000)*20/100;
printf("Tax to be paid is:%d\n",y);
break;
case 4:
printf("\nEnter Income");
scanf("%d",&x);
y=1900+(x-15000)*30/100;
printf("Tax to be paid is:%d\n",y);
break;
case 5:
exit(0);
default:
printf("Wrong choice!\n");
}
}
return 0;
}
4
...
Y = X + 10 if X is 6
...
Y is 2 ∗ X + 4 if X is 12
...
#include
h>
int main()
{
int choice,x,y;
while(1)
{
printf("\n1
...
x=7\n");
printf("3
...
otherwise\n");
printf("5
...
Write a program, which reads three integers X, Y and Z and prints Y + Z if X is 0
...
If X is 2 then Y ∗ Z is printed
...
#include
h>
int main()
{
int choice,x,y,z,s;
while(1)
{
printf("\n1
...
x=7\n");
printf("3
...
x=3\n");
printf("5
...
Write program, which find the sum of first digits
#include
Write program, which finds sum of those numbers whose first digit is 1
#include
Write program, which finds sum of numbers after deleting first digit
...
h>
main( )
{ int n,i,u,v,sum,digit,x;
scanf("%d",&n);
8
sum=0;
for (i=1;i<=n;i++)
{ scanf("%d",&x);
while (x>0)
{
u=x/10;
v=u*10;
x=x-v;
sum=sum+x;
}
}
printf("Answer=%d",sum);
}
4
...
#include
Write program, which finds sum of numbers after incrementing every even digit
...
h>
main( )
{ int n,i,u,v,w,t,sum,digit,x;
scanf("%d",&n);
sum=0;
for (i=1;i<=n;i++)
{ scanf("%d",&x);
{
u=x/10;
v=x%10;
if(u%2==0)
u=u+1;
if(v%2==0)
v=v+1;
x=(u*10)+v;
{
sum=sum+x;}
}
}
printf("Answer=%d",sum);
}
6
...
h>
10
main( )
{ int n,i,u,v,sum,digit,x;
scanf("%d",&n);
sum=0;
for (i=1;i<=n;i++)
{ scanf("%d",&x);
{
u=x/10;
v=x%10;{
if(u>v)
x=u;
else
x=v;}
sum=sum+x;
}
}
printf("Answer=%d",sum);
}
7
...
#include
Write program, which finds the biggest smallest factor
#include
Write program, which reads a number and finds (sum of digits)smallest factor
...
h>
int main(void)
{
int j,sum=0,n,r=0,q,x;
printf("Enter number");
scanf("%d",&n);
for(j=2;j
if(n%j==0)
{
q=j;
break;
}
}
while(n>0){
r=n%10;
sum+=r;
n/=10;}
x=pow(sum,q);
printf("\nThe smallest factor=%d",q);
printf("\nSum=%d",sum);
printf("\nThe computation of%d power%d =%d",sum,q,x);
return 0;
}
13
10
...
Find sum of greatest common divisors of pair of numbers
...
h>
int main(void)
{
int i,j,sum=0,n,arr[100],gcd=0,k=0;
printf("Enter number");
scanf("%d",&n);
for(i=0;i<2*n;i++)
{
scanf("%d",arr+i);
}
for(i=0;i<2*n;i++)
{
i=i+k;
for(j=1;j<=arr[i]&&arr[i+1];j++)
{
if(arr[i]%j==0&&arr[i+1]%j==0)
{
gcd=j;
break;
}
}
sum+=gcd;
gcd=0;
k=1;}
printf("Sum of all gcds=%d",sum-1);
return 0;
}
14
11
...
h>
int main(void)
{
int a,b,c,s=0,p,m=1,q;
printf("Enter a,b and c");
scanf("%d",&a);
scanf("%d",&b);
scanf("%d",&c);
for(int i=a;i<=b;i++){
p=i;
q=c;
while(q!=0){
m*=p;
p=p+1;
q=q-1;
}
s+=m;
m=1;
}
printf("\nThe output is%d",s);
return 0;
}
15
12
...
h>
#include
Write program, which reads two numbers and computes output in the following manner
...
h>
#include
Write program, which reads n and n numbers
...
#include
Solve the system of equations of example 1 and example 3 using Gauss Jacobi 2x-y+z=
2, -x+3y+3z=3, 2x+y+4z=1
7x+y+2z=10,x+8y+3z=8,2x+3y+9z=6
#include
h>
#include
00001,sum;
int i,j,n,flag;
printf("\nEnter number of variables: ");
scanf("%d",&n);
printf("\nEnter the coefficients row-wise: ");
for(i=0;i
for(j=0;j
scanf("%f",&a[i][j]);
}
}
printf("\nEnter right hand vector: ");
18
for(i=0;i
for(i=0;i
//checking for row dominance
flag=0;
for(i=0;i
sum=0;
for(j=0;j
sum+=fabs(a[i][j]);
if(sum>fabs(a[i][i]))
flag=1;
}
//checking for column dominance
if(flag==1)
flag=0;
for(j=1;j
sum=0;
for(i=1;i
sum+=fabs(a[i][j]);
if(sum>fabs(a[j][j]))
flag=1;
}
if(flag==1)
{
printf("The coefficient matrix is not diagonally dominant \n");
printf("The Gauss-Jacobi method does not converge surely");
exit(0);
}
for(i=0;i
printf("\n");
do
{
for(i=0;i
sum=b[i];
for(j=0;j
sum-=a[i][j]*x[j];
xn[i]=sum/a[i][i];
}
for(i=0;i
...
printf("Solutoin is \n");
for(i=0;i
...
3, 4, 5
...
3
...
h>
void main()
{
float a[10][10],l[10][10],u[10][10],z[10],x[10],b[10];
int i,j,k,n;
printf("\nEnter the size of the coefficient matrix: ");
scanf("%d",&n);
printf("Enter the elements row-wise: ");
for(i=0;i
printf("Enter the right hand vector: ");
for(i=0;i
20
//Computation of L and U matrices
for(i=0;i
for(j=2;j<=n;j++)
u[1][j]=a[1][j]/l[1][1];
for(i=0;i
for(i=2;i<=n;i++)
for(j=2;j<=n;j++)
if(i>=j)
{
l[i][j]=a[i][j];
for(k=1;k<=j-1;k++)
l[i][j]-=l[i][k]*u[k][j];
}
else
{
u[i][j]=a[i][j];
for(k=1;k<=j-1;k++)
u[i][j]=-l[i][k]*u[k][j];
u[i][j]/=l[i][i];
}
printf("\nThe lower triangular matrix L: \n");
for(i=0;i<=n;i++)
{
for(j=0;j
printf("\n");
}
printf("\nThe upper triangular matrix U: \n");
for(i=0;i<=n;i++)
{
for(j=0;j
for(j=i;j
printf("\n");
}
//solve Lz=b by forward substitution
z[1]=b[1]/l[1][1];
for(j=1;i
z[i]=b[i];
for(j=0;j
z[i]/=l[i][i];
}
//solve Ux=z by backward substitution
x[n]=z[n];
for(i=n-1;i>=0;i--)
{
x[i]=z[i];
for(j=i+1;j
}
printf("The solution is: ");
for(i=0;i
21
}
4
...
h>
void main()
{
float a[10][10],l[10][10],u[10][10],z[10],x[10],b[10];
int i,j,k,n;
printf("\nEnter the size of the coefficient matrix: ");
scanf("%d",&n);
printf("Enter the elements row-wise: ");
for(i=0;i
printf("Enter the right hand vector: ");
for(i=0;i
//Computation oof L and U matrices
for(i=0;i
l[i][i]=1;
for(i=2;i<=n;i++)
l[i][1]=a[i][1];
for(j=2;j<=n;j++)
u[1][j]=a[1][j]/l[1][1];
for(i=0;i
if(i>=j)
{
l[i][j]=a[i][j];
for(k=1;k<=j-1;k++)
l[i][j]-=l[i][k]*u[k][j];
}
else
{
u[i][j]=a[i][j];
for(k=1;k<=j-1;k++)
u[i][j]=-l[i][k]*u[k][j];
u[i][j]/=l[i][i];
}
printf("\nThe lower triangular matrix L: \n");
22
}
for(i=0;i<=n;i++)
{
for(j=0;j
printf("\n");
}
printf("\nThe upper triangular matrix U: \n");
for(i=0;i<=n;i++)
{
for(j=0;j
for(j=i;j
printf("\n");
}
//solve Lz=b by forward substitution
z[1]=b[1]/l[1][1];
for(j=1;i
z[i]=b[i];
for(j=0;j
z[i]/=l[i][i];
}
//solve Ux=z by backward substitution
x[n]=z[n];
for(i=n-1;i>=0;i--)
{
x[i]=z[i];
for(j=i+1;j
}
printf("The solution is: ");
for(i=0;i
6,7
...
05 and then modify both programs according to the
modified Euler’s method and R-K’s order 2 method
...
23
#include
h>
double f(double t, double y)
{
return y;
}
int main()
{
double h=0
...
0; y0 = 1
...
0;
x=0;
printf("t Euler Exact\n");
for (i=0; i<= 10; i++)
{
printf("t= %lf %lf %lf\n", t, y, exp(t));
y=y0+(h/2)*(f(x,y0)+f(t,y));
t=t+h;
}
return 0;
}
10
...
h>
#include
05, t, y, k1,k2;
int i;
/* initial value */
t=0
...
0;
for (i=0; i<=10; i++)
{
printf("t= %lf rk= %lf exact=%lf\n", t, y, exp(t));
k1=h*f(t,y);
k2=h*f(t+h, y+k1);
y= y+(k1+k2)/2
Title: C+ program
Description: C+ programming for computational technique assigent
Description: C+ programming for computational technique assigent