Practical no 1 Aim: Write a program in C/C++ to find the root of a quardatic equaltion and perform the following on it: Boundary Value Analysis (BVA). Consider the program for the determination of nature of roots of a quadratic equation. Its input is a triple of postive intergers a,b,c and values may be from interval [0,100]. The pr ogram output may have one of the following words: Not a QE, Real Roots, Imaginary Roots or Equal Roots.
#include
#include void main() { int a,b,c,d; clrscr(); cout<<"The Quadratic equation is of the type a(x^2)+bx+c=0"<>a; cout<<"Enter the value of b: "<>b; cout<<"Enter the value of c: "<>c; d=(b*b)-4*a*c; if((a<0)||(b<0)||(c<0)||(a>100)||(b>100)||(c>100)) cout<<"Invalid input"<
The quadratic equation is of the type a(x^2) +bx +c = 0 Enter the value of a: 50 Enter the value of b: 50 Enter the vaule of c: 0 Real roots
The quadratic equation is of the type a(x^2) +bx +c = 0 Enter the value of a: 50 Enter the value of b: 50 Enter the vaule of c: 50 Imaginary roots
The quadratic equation is of the type a(x^2) +bx +c = 0 Enter the value of a: 10 Enter the value of b: 40 Enter the vaule of c: 40 Roots are equal
The quadratic equation is of the type a(x^2) +bx +c = 0 Enter the value of a: 0 Enter the value of b: 50 Enter the vaule of c: 50 Not a quadratic equation Boundary Value analysis: The basic idea of boundary value analysis is to use input variable at their manimum, just above manimum, normal value, just below maximum and maximum. In this program, we consider the value as 0 (minimum), 1(just above minimum), 50 (normal), 99 (just below maximum) and 100 (maximum). Test case ID 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.
a 50 50 50 50 50 50 50 50 50 0 1 99 100
b 50 50 50 50 50 0 1 99 100 50 50 50 50
c 0 1 50 99 100 50 50 50 50 50 50 50 50
Expected output Real roots Real roots Imaginary roots Imaginary roots Imaginary roots Imaginary roots Imaginary roots Imaginary roots Equal roots Not a QE Real roots Imaginary roots Imaginary roots
The quadratic equation is of the type a(x^2) +bx +c = 0 Enter the value of a: 50 Enter the value of b: 50 Enter the vaule of c: 50 Imaginary roots
The quadratic equation is of the type a(x^2) +bx +c = 0 Enter the value of a: 10 Enter the value of b: 40 Enter the vaule of c: 40 Roots are equal
The quadratic equation is of the type a(x^2) +bx +c = 0 Enter the value of a: 0 Enter the value of b: 50 Enter the vaule of c: 50 Not a quadratic equation Boundary Value analysis: The basic idea of boundary value analysis is to use input variable at their manimum, just above manimum, normal value, just below maximum and maximum. In this program, we consider the value as 0 (minimum), 1(just above minimum), 50 (normal), 99 (just below maximum) and 100 (maximum). Test case ID 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.
a 50 50 50 50 50 50 50 50 50 0 1 99 100
b 50 50 50 50 50 0 1 99 100 50 50 50 50
c 0 1 50 99 100 50 50 50 50 50 50 50 50
Expected output Real roots Real roots Imaginary roots Imaginary roots Imaginary roots Imaginary roots Imaginary roots Imaginary roots Equal roots Not a QE Real roots Imaginary roots Imaginary roots
Practical no 2 Aim: write a program in C\C++ to find the area of a Triangle, Square, Rectangle and a Circle. Perform Equivalence class testing on it.
#include #include #include void main() { clrscr(); int ch; char c; float b,h,a; 1: cout<<"Enter your choice"; cout<<"\n 1. Triangle"; cout<<"\n 2. Square"; cout<<"\n 3. Rectangle"; cout<<"\n 4. Circle"; cout<<"\n 5. Exit\n"; cin>>ch; switch(ch) { case 1: b: cout<<"\Enter the base of the triangle (1-200)"; cin>>b; if((b<=0)||(b>200)) { cout<<"\n Invalid entry for base \n"; goto b; } h: cout<<"\n Enter the height of the triangle (1-200)"; cin>>h; if((h<=0)||(h>200)) { cout<<"\n Invalid height\n Enter the valid height (1-200)"; goto h; } a= 0.5*b*h; cout<<"\n The area is : "<>c; if((c=='y')||(c=='y')) goto 1; break; case 2 : s: cout<<"\n Enter the side of the squre (1-200)"; cin>>b; if((b<=0)||(b>200)) { cout<<"\n Invalid entry for base \n"; goto s; }
a=b*b; cout<<"\n The area is :"<>c; if((c=='y')||(c=='y')) goto 1; break; case 3: d: cout<<"\n Enter the Base of the rectangle (1-200)"; cin>>b; if((b<=0)||(b>200)) { cout<<"\n Invalid entry for base \n"; goto d; } p: cout<<"\n Enter the height of the rectangle (1-200)"; cin>>h; if((h<=0)||(h>200)) { cout<<"\n Invalid Height \n Enter the height (1-200)"; goto p; } a=b*h; cout<<"\n The area is :"<>c; if((c=='y')||(c=='y')) goto 1; break; case 4 : t: cout<<"\n Enter the radius of the circle (1-200)"; cin>>b; if((b<=0)||(b>200)) { cout<<"\n Invalid entry for radius \n"; goto t; } a=3.14*b*b; cout<<"\n The area is :"<>c; if((c=='y')||(c=='y')) goto 1; break; case 5 : exit(0); break; default : cout<<"\n Wrong Choice:"; goto 1; } getch(); }
Test cases: In Equivalence class testing, we find two types of equivalence classes; input domain and output domain; Input domain is formed from one valid sequence and two invalid sequences. The output domain is obtained from different types of output of a problem. For Triagle: Input domain:::
I1 = {h : h<=0} I2 = {h : H>200} I3 = {h : 1<=h<=200} I4 = {b : b<=0} I5 = {b : b>200} I6 = {b : 1<=b<=200} Test cases: Test case ID 1. 2. 3. 4. 5. 6.
h 0 100 201 100 100 100
Output domain:::
O1 = { : triangle in h > 0,b>0} O2 = { : Not a triangle if h<=0, b<=0} Output screen shots for the triangle is:
Enter your choice: 1. Triangle 2. Square 3. Rectangle 4. Circle 5. Exit 1 Enter the base of the triangle (1 – 200) 0 Invalid entry for base Enter the base of the triangle (1 – 200) 100 Enter the height of the triangle (1 – 200) 201 Invalid height Enter the Height(1-200) Enter the height of the triangle(1 – 200) 100 The area is 5000 Want to enter more? (y/n)y
b 100 100 100 0 100 201
Expected output Invalid input 5000 Invalid input Invalid input 5000 Invalid input
For square: Input domain:::
I1={s : s<=0} I2={s : s>200} I3={s : 1<=s<=200} Test cases: Test case ID 1. 2. 3.
s 0 100 201
Output domain:::
O1 = { : square if s>0} O2 = { : Not a square if s<=0} Output screen shots for the square is:
Enter your choice: 1. Triangle 2. Square 3. Rectangle 4. Circle 5. Exit 2 Enter the side of the square (1 – 200) 0 Invalid entry for side Enter the side of the square (1 – 200) 201 Invalid entry for side Enter the side of the square(1 – 200) 100 The area is 10000 Want to enter more? (y/n)y For Rectangle: Input domain:::
I1 = { l : l <=0} I2 = { l : l>200} I3 = { I : 1<=l <=200} I4 = { b : b<=0} I5 = { b : b>200} I6 = { b : 1<=b<=200}
Expected output Invalid input 10000 Invalid input
Test cases:
Test case ID 1. 2. 3. 4. 5. 6.
l 0 100 201 100 100 100
b 100 100 100 0 100 201
Expected output Invalid input 10000 Invalid input Invalid input 10000 Invalid input
Output domain:::
O1 = { : rectangle if l > 0,b>0} O2 = { : Not a triangle if l<=0, b<=0} Output screen shots for the Rectangle is:
Enter your choice: 1. Triangle 2. Square 3. Rectangle 4. Circle 5. Exit 3 Enter the length of the rectangle (1 – 200) 0 Invalid entry for length Enter the length of the rectangle (1 – 200) 201 Invalid entry for length Enter the length of the rectangle(1 – 200) 100 Enter the breadth of the rectangle (1 – 200) 100 The area is 10000 Want to enter more? (y/n)y
For Circle: Input domain:::
I1 = {r: r<=0} I2 = {r : r>200} I3 = { r: 1<=r<=200} Test cases: Test Cases 1. 2. 3.
r 0 100 201
Expected output Invalid input 31400 Invalid input
Output domain:
O1 : {: Circle if 1<=r<=200} O2 : {: not a circle if r<=0} Output screen shots for the Circle is:
Enter your choice: 1. Triangle 2. Square 3. Rectangle 4. Circle 5. Exit 4 Enter the radius of the circle (1 – 200) 0 Invalid entry for radius Enter the radius of the circle (1 – 200) 201 Invalid entry for radius Enter the radius of the circle(1 – 200) 100 The area is 31400 Want to enter more? (y/n)y
Practical No 3 Aim: write a program in C/C++ to calculate the value of a b. Perform Decision Table based testing on it.
#include #include #include void main() { clrscr(); int a,b; float c; char ch; 1: cout<<"To Calculate 'a to the power b' \n"; cout<<"Enter the value of 'a' \n"; cin>>a; cout<<"Enter the value of 'b' \n"; cin>>b; c=pow(a,b); cout<>ch; if((ch=='y')||(ch=='y')) goto 1; getch(); } Test cases: Decision Table Based testing is useful for describing situations in which a number of combination of actions are taken for different conditions. There are four parts of a decision table; condition stub, action stub, condition entries and action entries. Test case ID 1. 2. 3. 4. 5. 6.
Decision table is: Conditions are::: C1 : a = 0, b = 0 C2 : a = -ve, b = +ve C3 : a = +ve, b = -ve C4 : a = -ve, b = -ve C5 : a = +ve, b = +ve C6 : a = 0,b = integer
A 2 -1 -2 0 0 -1
B 3 3 -4 1 0 -0.6
Expected output +ve result -ve result +ve result Result is 0 Domain Error Result is 1
C7 : b = 0, a = integer C8 : a = -ve, b= -ve odd Actions:::
A1 : Domain error A2 : Negative output A3 : output =1 A4 : positive output A5 : output = 0 Condition C1 C2 C3 C4 C5 C6 C7 C8 Action A1 A2 A3 A4 A5
R1 T --
R2 T -
R3 -T
R4 T -
R5 T -
R6 T -
R7 T -
R8 T -
X X
X X
X
The output for the above program is:
To calculate’a to the power of b’ Enter the value of ‘a’ 0 Enter the value of ‘b’ 0 Pow: Domain error 1 Want to enter again? (y/n) y To calculate’a to the power of b’ Enter the value of ‘a’ 2 Enter the value of ‘b’ 3 8 Want to enter again? (y/n)n
X
X X
Practical No 4 Aim: write a program in C/C++ to compute previous data. We are given the present date as input. Also perfome the Boundary value analyses on it.
#include #include int main() { int day, month, year, flag=0; printf("Enter the day value:"); scanf("%d",&day); printf("Enter the month value:"); scanf("%d",&month); printf("Enter the year value:"); scanf("%d",&year); if(year>=1900 && year<=2025) { if(month==1||month==3||month==5||month==7||month==8||month==10||month==12) { if(day>=1 && day<=31) { flag=1; } else { flag=0; } } else if(month==2) { int rval=0; if(year%4==0) { rval=1; if((year%100)==0 && (year%400)!=0) { rval=0; } } if(rval==&&(day>=1 &&day <=29)) { flag=1; } else if (day>=1 &&day <=28) { flag=1; } else { flag=0;
} } if(flag) { if(day==1) { if(month==1) { year--; day=31; month=12; } else if(month==3) { int rval=0; if(year%4==0) { rval=1; if((year%100)==0 &&(year%400)!=0) { rval=0; } } if(rval==1) { day=29; month--; } else { day=28; month--; } } else if(month==2||month==4||month==6||month==9||month==11) { day=31; month--; } else { day=30; month--; } } else { day--; } printf(" The next data is : %d-%d-%d",day,month,year); }
else { printf(" The entered data is invalid: %d-%d-%d",day,month,year); } getch(); return 1; }} Test cases: Test Case ID 1. 2. 3. 4. 5. 6.
Month June June May March August Februry
The output for the above program is:
Enter the day value; 3 Enter the month value ; 3 Enter the year value ; 2000 The previous day is 2 – 3 – 2000
Day 1 31 1 31 29 29
Year 1964 1984 1945 2007 2007 1962
Expected output 31 May, 1964 Impossible 30 April, 1945 30 March, 2007 28 August, 2007 impossible
Practical no 5 Aim: Write a program in C/C++ to read three sides of a triangle and to check whether the triangle is Isoceles, equilateral or scalane. Perform path testing on it and deduce: 1. 2. 3. 4.
Flow graph DD path graph Independent path Cyclomatic complexity
#include #include int main() { int a,b,c,validinput=0; printf("Enter the side ‘a’ value"); scanf("%d",&a); printf("Enter the side ‘b’ value"); scanf("%d",&b); printf("Enter the side ‘c’ value"); scanf("%d",&c); if((a>0)&&(a<100)&&(b>0)&&(b<100)&&(c>0)&&(c<100)) { if(((a+b)>c)&&((c+a)>b)&&((b+c)>a)) { validinput=1; } } else { validinput=-1; } if(validinput==1) { if((a==b)&&(b==c)) { printf("The triangle is equilateral"); } else if((a==b)||(b==c)||(c==a)) { printf("The triangle is isosceles"); } else { printf("The triangle is scalene"); } } else if(validinput==0) { printf("The value do not constitute the triangle"); }
else { printf("The input belongs to invalid range"); } getch(); return 1; }
1. Flow Graph: It shows that how the control and data is flow from one node to another node. The
flow graph for the following program is given below:
2. DD path graph:
Flow Graph Nodes 1 to 9 10 11 12,13 14 15,16,17 18 19 20,21 22 23,24 25,26,27 28 29 30,31
DD Path Graph Corr. Nodes A B C D E F G H I J K L M N O
Remarks Sequential Decision Decision Sequential Two edges are joined here Sequentian nodes Decision nodes plus joining of two edges Decision nodes Sequential nodes Decision nodes Sequential nodes Sequential nodes Three edges combined here Decision nodes Sequential nodes
32,33,34 35 36,37
P Q R
Sequential nodes Three edges comibined here Sequential nodes with exit node
3. Independent path : It is a path in the flow graph that has at least one edge that has not been
traversed before in other path. 1. 2. 3. 4. 5. 6. 7.
ABFGNPQR ABFGNOQR ABCEGNPQR ABCDGINOQR ABFGHIMQR ABFGHJKMQR ABFGHJLMQR
4. Cyclomatic complexity: To calculate the cyclomatic complexity we calculate the following
formula: V(G) = e – n + 2P
Here : e = 23 n = 18 and p=1 so the total no of independent paths are 23 – 18 + 2 = 7 The output for the above program is :
Enter the side ‘a’ value: 12 Enter the side ‘b’ value: 12 Enter the side ‘c’ value: 12 The triangle is Equilateral Enter the side ‘a’ value: 45 Enter the side ‘b’ value: 45 Enter the side ‘c’ value: 23 The triangle is isosceles
Practical no 6 Aim: write a program in C\C++ to compute total salary of an employee given his/her basic salary. The slab is given below: HRA = 30 % of basic DA=80% of basic MA = Rs. 100/TA = Rs. 800/I.tax = Rs. 700/PF = Rs. 780/- Also find out the path graph and cyclomatic complexity.
#include #include #include void main() { clrscr(); cout<<"Enter the basic salary of the employee"; float bs; cin>>bs; int hra,da,ma=100,itax=700, pf=780,ta=800; hra=0.3*bs; da=0.8*bs; cout<<"\nHouse allowance: rs."<
Enter the basic salary of the employee10000 House allowance: rs.2999 Darkness allowance: rs.8000 medical allowance: rs.100 travel allowance: rs.800 Income tax rs.700 providend fund: rs.780 The net selary of the employee is : 20319
The Cyclomatic complexity for the above path graph is: To calculate the cyclomatic complexity we use the formula: V(G) = e - n + 2P Here: e = 15 n = 16 p=1 hence V(G) = 1
Practical 7: Aim: Draw the DD Path graph for the practical no 6 ( write a program in C\C++ to compute total salary of an employee given his/her basic salary. The slab is given below: HRA = 30 % of basic DA=80% of basic MA = Rs. 100/TA = Rs. 800/I.tax = Rs. 700/PF = Rs. 780/- ) The Flow graph for the given program is :
Here we can see that in the flow graph all the nodes are arrenged in the sequential order. So we have only one DD node Flow Graph Nodes 1 to 9
DD Path Graph Corr. Nodes A
Remarks Sequential
Program no 8 Aim: Write a program to read the marks of 10 students in five subjects. Find the average and allot the grades. Draw its graph matrix and find its V(G).
#include #include #include #include struct student { char sname[10]; float marks[5]; float total,avg; char grade; }s[10]; void main() { clrscr(); int i; for(i=1;i<10;i++) { cout<<"\n Enter the name of the student"<>s[i].marks[j]; } for(j=0;j<5;j++) { s[i].total=s[i].total+s[i].marks[j]; } s[i].avg=s[i].total/5; cout<<"\n The average of student"<90.0) s[i].grade='O'; else if((s[i].avg<90.0)&&(s[i].avg>=85.0)) s[i].grade='A'; else if((s[i].avg<85.0)&&(s[i].avg>=80.0)) s[i].grade='B'; else if((s[i].avg<80.0)&&(s[i].avg>=70.0)) s[i].grade='C'; else s[i].grade='D'; cout<<"\n The grade of student" <
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
Graph matrix:
A B C D E F G H I J K L
A --
B 1 ---
C 1 -
D 1 --
E -1 -
F 1 -
G -
H --
I 1 -
J 1 -
K -1 --
L 1 1 1 1 1 -
The cyclomatic complexity is V(G) = 5. That means there are five indepent path in the flow graph. The output of the above program is :
Enter the name of student 0: A Enter the marks in subject1 99 Enter the marks in subject2 98 Enter the marks in subject3 97 Enter the marks in subject4 96 Enter the marks in subject5 95 The average of student 0 is : 97 The grade of student 00 enter the name of student1 The average of student 0 is : 97 The grade of student 00 enter the name of student1 Enter the marks in subject0 Enter the marks in subject1 Enter the marks in subject2 Enter the marks in subject3 Enter the marks in subject4
88 85 86 87 84
The average of student 1 is : 86 The grade of student1A enter the name of the student2
B
1 1 0 1 0 0 1 0 0 4+1=5
Program no 9 Aim: Write a program to read the marks of 10 students in five subjects. Find the average and allot the grades. Apply data flow testing on the above program.
#include #include #include #include struct student { char sname[10]; float marks[5]; float total,avg; char grade; }s[10]; void main() { clrscr(); int i; for(i=1;i<10;i++) { cout<<"\n Enter the name of the student"<>s[i].marks[j]; } for(j=0;j<5;j++) { s[i].total=s[i].total+s[i].marks[j]; } s[i].avg=s[i].total/5; cout<<"\n The average of student"<90.0) s[i].grade='O'; else if((s[i].avg<90.0)&&(s[i].avg>=85.0)) s[i].grade='A'; else if((s[i].avg<85.0)&&(s[i].avg>=80.0)) s[i].grade='B'; else if((s[i].avg<80.0)&&(s[i].avg>=70.0)) s[i].grade='C'; else s[i].grade='D'; cout<<"\n The grade of student" <
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
Variable Sname Marks Total Avg Grade I J
Defined At Nodes 3 4 5 6 7 12 17
Used At Nodes 16 20,24 24,26 26,27,28,30,32,34 29,31,33,35,37,38 13,16,20,24,26,27,28,29,30,31,32,33,34,35,37,38 17,20,22,24
The du-paths are identified and are named by their beginning and end nodes using the variable as follows: Variable Sname Marks
Total Avg
Grade
I
J
Path(beginning, End) Nodes 3,16 4,20 4,24 5,24 5,26 6,26 6,27 6,28 6,30 6,32 6,34 7,29 7,31 7,33 7,35 7,37 7,38 12,13 12,16 12,20 12,24 12,26 12,28 12,29 12.30 12,31 12,32 12,33 12,34 12,35 12,37 12,38 17,17 17,20 17,22 17,24
Definition clear ? yes Yes yes Yes yes yes yes yes yes yes yes Yes Yes Yes Yes Yes yes Same node: yes Yes Yes yes
The output of the above program is :
Enter the name of student 0: A Enter the marks in subject1 99 Enter the marks in subject2 98 Enter the marks in subject3 97 Enter the marks in subject4 96 Enter the marks in subject5 95 The average of student 0 is : 97 The grade of student 00 enter the name of student1 The average of student 0 is : 97 The grade of student 00 enter the name of student1 Enter the marks in subject0 Enter the marks in subject1 Enter the marks in subject2 Enter the marks in subject3 Enter the marks in subject4
88 85 86 87 84
The average of student 1 is : 86 The grade of student1A enter the name of the student2
B
Practical no 10 Aim: consider the triangle probem. Its input is a triplet of 3 positive integers (say a, b, c) from the interval (1 to 100). The output may be one of the following words – scalene, isosceles, equilateral, not a triangle. Also draw the DD Graph, Decision table, and cause effect graph for the same.
#include #include 1 int main() 2 { 3 int a,b,c,valid_I=0; 4 printf("\nEnter a:"); 5 scanf("%d",&a); 6 printf("\nEnter b:"); 7 scanf("%d",&b); 8 printf("\nEnter c:"); 9 scanf("%d",&c); 10 if((a>0) && (a<=100)&& (b>0) && (b<=100)&& (c>0) && (c<=100)) { 11 if((a+b)>c) && ((c+a)>b) &&((b+c)>a)){ 12 valid_I=1; 13 } 14 } 15 else { 16 valid_I = -1; 17 } 18 if(valid_I==1) { 19 if((a==b)&&(b==c)) 20 printf("Traingle is equilateral"); 21 } 22 else if(a==b)||(b==c)||(c==a)) { 23 printf("The triangle is isoceles"); 24 } 25 else { 26 printf("The triangle is scalene"); 27 } 28 } 29 else if(valid_I==0) { 30 printf("\nNot a traingle"); 31 } 32 else { 33 printf("\nInvalid triangle"); 34 } 35 getch(); 36 return 0; 37 }
Flow graph for the triangle problem
To find the du-path and dc-path we perform the following steps: Step 1: We prepare a teble of define and use nodes for all variables used in this program:
Variables a b c Valid_I
Defined at node 5 7 9 3, 12, 16
Used at node 10, 11 ,19, 22 10, 11, 19, 22 10, 11, 19, 22 18, 29
Step 2: To calculate the du-path we prepare another table:
Variables a a
Path (Beg, End) 5, 10 5, 11
Defination Clear Yes Yes
a a b b b b c c c c Valid_I Valid_I Valid_I Valid_I Valid_I Valid_I
5, 19 5, 22 7, 10 7, 11 7, 19 7, 22 9, 10 9, 11 9, 19 9, 22 3, 18 3, 29 12, 18 12, 29 16, 18 16, 29
Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes No No No No Yes Yes
Out of the 18 du – path, 4 paths namely (3,18), (3, 29), (12,18), (12,29) are not dc-path. The cause effect graph for the triangle problem is : Step 1 : Firstly, we must identify the causes and its effects. The causes are C1 = Side X is less than sum of Y and Z. C2 = Side Y is less than sum of X and Z C3 = Side Z is less than sum of X and Y. C4= Side X is equal to side Y, C5= Side X is equal to side Z, C6= Side Y is equal to side X,
The effects are E1= Not a triangle. E2= Scalene Triangle E3= Isosceles triangle E4= Equilateral Triangle E5= Impossible
Step 2 : The cause – effect graph for the following is
The Decision table for the Above graph is:
C1 C2 C3 C4 C5 C6 E1 E2 E3 E4 E5
R1 F -----X
R2 T F ----X
R3 T T F ---X
R4 T T T T T T
R5 T T T T T F
R6 T T T T F T
R7 T T T T F F
R8 T T T F T T
X X X
X
X
R9 T T T F T F
R10 T T T F F T
X
X
R11 T T T F F F X
Practical no 11. Aim: (Quadratic Equation Problem). This program reads a, b and c as the three coefficients of a quadratic equation ax2 + bx + c = 0. It determines the nature of the roots of this equation. Draw its flow graph
#include #include 1 int main() 2 { 3 int a,b,c,boolean=0; 4 double D; 5 printf("\nEnter a coefficient:"); 6 scanf("%d",&a); 7 printf("\nEnter b coefficient:"); 8 scanf("%d",&b); 9 printf("\nEnter c coefficient"); 10 scanf("%d",&c); 11 if((a>=0) && (a<=100)&& (b>=0) && (b<=100)&& (c>=0) && (c<=100)) { 12 boolean=1; 13 if(a==0) 14 boolean = -1 15 } 16 } 17 if(boolean == 1){ 18 d=b*b - 4 * a * c; 19 if(d==0){ 20 printf("Roots are equal"); 21 } 22 else if(d>0) { 23 D=sqrt(d); 24 printf("Roots are real" 25 } 26 else{ 27 D=sqrt(-d)/(2*a); 28 printf("Roots are imaginary"); 29 } 30 } 31 printf("Not a quadratic equation"); 32 } 33 else { 34 printf("Invalid input range ..."); 35 } 36 getch(); 37 return 0; 38 }
