programming language_01(CS)

4. Write a program that will print the following for a 4 by 4 matrix A

input:

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

output:

1
5 6
9 10 11
13 14 15 16

solution:

#include< stdio.h >
#define row 4
#define column 4
int main(){
int A[row][column]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
int i,j,k;
for( i = 0 ; i < row ; i++ ){
        for(  j = 0 ; j <=i; j++ )
            {
                printf("%d  ",A[i][j]);
            }

        printf("\n\n");

    }
    return 0;
}

4. Write a program that will print the following for a 4 by 4 matrix A

input:

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

output:

 1  2  3  4
    6  7   8
       11  12
             16

solution

#include< stdio.h >
#define ROW 6
#define COLUMN 6
int main( void ){
    int A[ROW][COLUMN] = {1,2,3,4,5,6,
                          7,8,9,10,11,12,
                          13,14,15,16,17,18,
                          19,20,21,22,23,24,
                          25,26,27,28,29,30,
                          31,32,33,34,35,36};
    //here we are printing the lower left
int i,j,k;
    for(  i = 0 ; i < ROW ; i++ ){
        for(  j = 0 ; j < COLUMN ; j++ )
            if( i ==j ) {
                for(  k = j ; k < COLUMN ; k++ )
                    printf("\t%d",A[i][k]);
            }else{
                printf("\t" );
            }
        printf("\n\n");
    }
    return 0;
}

 3. Implement the following using function.
a. Input: Two numbers x and n, where x (must be decimal) is the value of to be used in
the series and n (must be integer) is the no of times the series will run.
b. Desired Output: calculate the series for Sin(x) up to n.
 solution:

#include< stdio.h >
int myfunc(int x){
    printf("\n\nthe value of sin is:\n\n");
printf("sin%d=%d-%d^3/3!+%d^5/5!-%d^7/7!+....\n\n",x,x,x,x,x);
  printf("the value of cos is:\n\n");
printf("cos%d=%d-%d^2/2!+%d^4/4!-%d^6/6!+....\n\n",x,x,x,x,x);
return 0;
}
int main(){
int n;
printf("put the value of x");
scanf("%d",&n);
myfunc(n);
return 0;
}
 

programming language _01(CS)

One Dimensional Array:

1. Sort an integer array of 10 elements in descending order (large to small) using bubble sort,

selection sort and insertion sort.

a. Variables: integer array A[10].

b. Initialize the array with values: 56,34,5,-68,-1,0,6,4,2,19.

c. You must print the array before sorting and also after sorting to help understand

that the sorting really happened.

solution:

Bubble sorting:

#include< stdio.h >
int main( void ){
    int temp;
    int A[ 10 ] = { 56,34,5,-68,-1,0,6,4,2,19 };
    printf("Before Bubble Sort:\n\n");
    int i,m,n,j;
    for(  i = 0 ; i < 10 ; i++ ) printf( "%d " , A[ i ] );
    printf("\n\n");
    for(  m = 0 ; m < 10 - 1 ; m++ ){
        for(  n = 0 ; n < 10 - 1 - m ; n++ ){
            if( A[ n ] < A[ n + 1 ] ){
                temp = A[ n ];
                A[ n ] = A[ n + 1 ];
                A[ n + 1 ] = temp;
            }
        }
    }
    printf("After Bubble Sort:\n\n");
    for(  j = 0 ; j < 10 ; j++ ) printf( "%d " , A[ j ] );
    return 0;
}

insertion sorting:

#include< stdio.h >
int main( void ){
    int temp;
    int A[ 10 ] = { 56,34,5,-68,-1,0,6,4,2,19 };
    printf("Before Insertion Sort:\n\n");
    int i,m,n,j;
    for(  i = 0 ; i < 10 ; i++ ) printf( "%d " , A[ i ] );
    printf("\n\n");
    for( m = 1 ; m <= 10 - 1 ; m++ ){
        temp = A[ m ];
        for( n = m - 1 ; n >= 0 && A[ n ] < temp ; n-- ){
            A[ n + 1 ] = A[ n ];
            A[ n ] = temp;
        }
    }
    printf("After Insertion Sort:\n\n");
    for(  j = 0 ; j < 10 ; j++ ) printf( "%d " , A[ j ] );
    return 0;
}
 

selection sorting:

#include< stdio.h >
int main( void ){
    int temp;
    int A[ 10 ] = { 56,34,5,-68,-1,0,6,4,2,19 };
    printf("Before Selection Sort:\n\n");
    int i,m,n,j;
    for( i = 0 ; i < 10 ; i++ ) printf( "%d " , A[ i ] );
    printf("\n\n");
    for(  m = 0 ; m < 10 - 1 ; m++ ){
        for(  n = m + 1 ; n < 10 ; n++ ){
            if( A[ m ] < A[ n ] ){
                temp = A[ m ];
                A[ m ] = A[ n ];
                A[ n ] = temp;
            }
        }
    }
    printf("After Selection Sort:\n\n");
    for(  j = 0 ; j < 10 ; j++ ) printf( "%d " , A[ j ] );
    return 0;
}

2. Find the median number of an integer array A and B.
a. Algorithm: If the size of array is even (N) then median = average of Array[N/2] +
A[N/2-1]. If the size of array is odd then median = Array[N/2].
b. Variables: integer array A[10], B[11].
c. Initialize the array with values: A = 1,2,3,4,5,6,7,8,9,10 and B =
11,12,13,14,15,16,17,18,19,20,21.
d. You must print the median for both arrays.

solution:

#include< stdio.h >
int main( void ){
    int temp;
    int A[ 10] = { 1,2,3,4,5,6,7,8,9,10 };
    int B[11]={11,12,13,14,15,16,17,18,19,20,21};
    printf("Before Selection Sort of A[11]:\n\n");
    int i,m,n,j;
    for( i = 0 ; i < 10 ; i++ ) printf( "%d " , A[ i ] );
    printf("\n\n");
    printf("Before Selection Sort of B[11]:\n\n");
    for( i = 0 ; i < 10 ; i++ ) printf( "%d " , B[ i ] );
    printf("\n\n");
    for(  m = 0 ; m < 10 - 1 ; m++ ){
        for(  n = m + 1 ; n < 10 ; n++ ){
            if( A[ m ] > A[ n ] ){
                temp = A[ m ];
                A[ m ] = A[ n ];
                A[ n ] = temp;
            }
        }
    }
    for(  m = 0 ; m < 10 - 1 ; m++ ){
        for(  n = m + 1 ; n < 10 ; n++ ){
            if( B[ m ] > B[ n ] ){
                temp = B[ m ];
                B[ m ] = B[ n ];
                B[ n ] = temp;
            }
        }
    }
    printf("After Selection Sort of A[10]:\n\n");
    for(  j = 0 ; j < 10 ; j++ ) printf( "%d " , A[ j ] );
    printf("\n\n");
    printf("After Selection Sort of B[11]:\n\n");
    for(  j = 0 ; j < 10 ; j++ ) printf( "%d " , B[ j ] );
    printf("\n\n");
    int N=10,L=11;
    float median;
    median=(A[N/2]+A[(N/2)-1]);
    median=median/2;
    printf("median of A[10]=%f",median);
    printf("\n\n");
    median=B[L/2];
    printf("median of B[11]=%f",median);
    return 0;
}

3. Write a program that will transpose (converting the row values to column values and viceversa)
4 by 4 matrix.
a. Variable: integer array A[4][4].
b. Desired Output: transpose of A[4][4]

solution:

#include< stdio.h >
#define ROW 3
#define COLUMN 3
int main(){
    int A[ROW][COLUMN]={1,2,3,4,5,6,7,8,9};
    int i,k;
    int B[ROW][COLUMN];


    for ( i=0 ; i < ROW ;  i++){
    for ( k=0 ; k < COLUMN ; k++){
    printf("%d\t",A[i][k]);
    }
    printf("\n");
    }


printf("\n\n");
for ( k=0 ; k < ROW ; k++){
    for ( i=0 ; i < COLUMN ; i++){
        B[k][i]=A[i][k];
    printf("%d\t",B[k][i]);
    }
    printf("\n");
    }
    return 0;
}

2.Write a program that will add two 4 by 4 matrix A and B respectively and store in another 4
by 4 matrix C.
a. Variables: integer array A[4][4], B[4][4] and C[4][4].
b. Desired Output: C[4][4] = A[4][4] + B[4][4]

solution:

#include< stdio.h >
#define ROW 4
#define COLUMN 4
int main(){
    int A[ROW][COLUMN]={1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
    int B[ROW][COLUMN]={1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
    int C[ROW][COLUMN];
    int i , sum , k;
    for ( i=0 ; i < ROW ; i++){
    for ( k=0 ; k < COLUMN ; k++){
    C[i][k]=A[i][k]+B[i][k];
    }
    }
    for ( i=0 ; i < ROW ; i++){
    for ( k=0 ; k < COLUMN ; k++){
    printf("C[%d][%d]:%d\t",i,k,C[i][k]);
    }
    printf("\n");
    }
    return 0;
}

1. Implement the following series using function.
a. Input: an integer number.
b. Desired Output: The series will be printed as

input:4

output:1+1+3+1+3+5+1+3+5+7

c. Note: You need to print the series not the summation result.

solution:

#include< stdio.h >
int series(int n){
    int i , k , m , s;
    for ( k=1 ; k <= n ; k++){
    for (i=1 ; i <= k ; i++){
    m=((2*i)-1);
    printf(" %d " , m);
    printf(" + ");
    }
    }
    return 0;
}
int main(){
int x;
scanf("%d",&x);
series(x);
return 0;
}

2. Implement the following using function.
a. Input: an integer number.
b. Desired Output: The series will be printed as

input:5

output:

1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

c. Note: You must show the output as given in the sample.

solution:

#include< stdio.h >
int myfunc(int x){
    int  i ,j;
for ( i=1 ; i  <= x ; i++){
int   z=1;
for (j=1 ; j <= i ;  j++ ){
printf(" %d " , z);
z++;
}
printf(" \n ");
}
return 0;
}
int main (){
    int  n;
    printf(" no of input:\n\n ");
    scanf(" %d ", &n);
    myfunc(n);
return 0;
}