[PDF] SSCM 1313 C++ COMPUTER PROGRAMMING Chapter 5





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SSCM 1313

C++ COMPUTER PROGRAMMING

Chapter 5:

Structure and Array Applications

Authors:

Farhana Johar

Professor Dr. Shaharuddin Salleh

Structure A structure groups variables into a hierarchical tree based on their common ancestral origin. Under this arrangement, a structure starts with a parent which includes several children.

x Constructed using tydef struct x A structure must have a name x Declared in the pre-processing area

Example

typedef struct char name[10]; members int age; double cpa; } CAMPUS; name of structure object of the structure

CAMPUS student;

strcpy(student.name,"Michael"); assignment of values student.age=35; to members student.cpa=3.15;

Points in Cartesian Coordinates

typedef struct { double x,y; } POINT; POINT pt[10];

4].5[4

1].5[1

5 5 ypty xptx yipty xiptx i i

Code5A.cpp: Structure representing a point.

#include #define n 4 using namespace std; void main() { int i,j; typedef struct double x,y; } POINT; POINT pt[n+1]; double Eu[n+1][n+1]; for (i=1;i<=n;i++) { pt[i].x=(double)2*i; } cout << "Euclidean distance between points i and j" << endl; cout << "(" << pt[i].x << "," << pt[i].y << ") and ("<Nested Structure typedef struct { double x,y; } POINT; typedef struct { POINT b,e; } LINE;

Description:

A line has two end points.

Each point in a line is represented as (x,y) coordinates.

LINE Ln; Ln.b.x=3; Ln.b.y=Ǧ1; e

Ln.e.x=7; Ln.e.y=0; (e.x,e.y)

b (b.x,b.y) typedef struct { double x,y; } POINT; typedef struct { POINT b,e; } LINE; typedef struct { LINE p,q,r; } TRIANGLE;

A triangle has three points.

Each line in a triangle has two end points.

Each point in a line is represented as (x,y) coordinates. t.p.e.x=7; t.p.e.y=0; t.q.b.x=7; t.q.b.y=0; (5,-8) t.r.b t.q.e t.r t.r.e t t.q (3,-1) t.p.b t.p t.q.b (7,0) t.p.e

Figure 5.1. TRIANGLE and its nested structures.

Code5B.cpp: Structure representing a triangle.

#include #include #include #define n 3 using namespace std; void main() { int i; srand(time(0)); typedef struct { double x,y; } POINT; typedef struct { POINT b,e; double length; } LINE; typedef struct { LINE p,q,r; double area; } TRIANGLE;

TRIANGLE t[n+1];

for (i=1;i<=n;i++) { t[i].p.b.x=(double)(rand()%50); t[i].p.b.y=(double)(rand()%50); t[i].p.e.x=(double)(rand()%50); t[i].p.e.y=(double)(rand()%50); t[i].q.e.x=(double)(rand()%50); t[i].q.e.y=(double)(rand()%50); t[i].q.b.x=t[i].p.e.x; t[i].q.b.y=t[i].p.e.y; t[i].r.b.x=t[i].q.e.x; t[i].r.b.y=t[i].q.e.y;

t[i].p.b.x*t[i].q.e.y+ t[i].p.e.x*t[i].q.e.y); cout << "Triangle #" << i << endl; cout << "(" << t[i].p.b.x << "," << t[i].p.b.y << ")" << setw(15);

cout << "(" << t[i].q.b.x << "," << t[i].q.b.y << ")" << setw(15); cout << "(" << t[i].r.b.x << "," << t[i].r.b.y << ")" << endl; cout << "Area is " << t[i].area << endl << endl; } cin.get(); CS1 (Case Study 1): Finding the root of a function Given f (x) 0 , find the value of x , assuming the value exists. f (a0) c 1 c 0 b 0 x a 0 c 2 f (b0) a3 b3 i = 3 a 2 b 2 i = 2 a 1 b 1 i = 1 a0 b0 i = 0

Algorithm 5.1. Bisection Method.

Given f (x) 0 , ˢ and the initial end-points [a0 , b0 ] where f (a0 ) f (b0 ) 0 ;

Given max =maximum number of iterations;

For i 0 to max

a b

Compute ci

i 2 i ;

If f ai f ci 0

Update bi 1 ci and ai 1 ai ;

If f ai f ci 0

Update ai 1 ci and bi 1 bi ;

c i ෥ c i ෥1 İ If

Solution= ci ;

Endfor

Stop the iterations;

Code5C.cpp: Finding the root of a function.

#include #define N 20 #define EPSILON 0.005 using namespace std; void main() { double *a, *b, *c; a=new double [N+1]; b=new double [N+1]; c=new double [N+1]; a[0]=1; b[0]=2; "<0) a[i+1]=c[i]; b[i+1]=b[i]; // update a else b[i+1]=c[i]; a[i+1]=a[i]; // update b if (i>0) cout << i << setw(15)<< a[i]<for (i=1;i<=N;i++) cout << v[i] << " "; cout< (i=1;i<=N;i++) cout << w[i] << " "; cout << endl; cin.get(); }

CS3: Detecting the Upper Triangular Matrix

0*** 00** 000* *000 **00 ***0 LU The following code segment determines whether a given matrix is upper triangular method for detection:

If flag is 0 then the matrix is not upper triangular. If flag is 1 then the matrix is upper triangular.

This code below showing how the input data is read. ifstream InFile;

InFile.open("Code5E.in");

for (i=1;i<=N;i++) { for (j=1;j<=N;j++) { InFile >> a[i][j]); cout << endl;

Code5E.cpp: Detecting upper triangular matrix

#include #include #define N 4 using namespace std; void main() { bool flag=1; double a[N+1][N+1]; int i,j; // read matrix A values and display them to confirm ifstream InFile; InFile.open("Code5E.in"); for (i=1;i<=N;i++) for (j=1;j<=N;j++) {

InFile >> a[i][j];

cout << a[i][j] << " "; } cout << endl; }

InFile.close();

for (i=2;i<=N;i++) { for (int j=1;jcout<< "the matrix is upper triangular"< cin.get(); }

Code5E.in

8 0 0 0

-2 1 0 0

2 -1 7 0

1 8 1 -2

Detecting a diagonal matrix

Detecting a tridiagonal matrix

CS4: Matrix Reduction into an Upper Triangular Form Method: Row Operations through Gaussian Elimination Method Example 5.2. Find the upper triangular matrix U of the following matrix: 2181
1712
8312
2128
A

8.000 2.000 -1.000 2.000

0.000 1.500 -3.250 -7.500

0.000 -1.500 7.250 -1.500

0.000 -8.250 1.125 -2.250

Operations with respect to the second row.

8.000 2.000 -1.000 2.000

0.000 1.500 -3.250 -7.500

0.000 0.000 4.000 -9.000

0.000 0.000 -16.750 -43.500

m a21 / a11 , m a31 / a22 ,quotesdbs_dbs8.pdfusesText_14
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