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Theme: General Programs Program No: 1 Objective: Write a program to calculate the area of triangle using formula at=√s (s- a) (s-b) (s-c) Logic Given a triangle with side lenths a, b and c. Then is area can be calculated as==√s (a+b+c)/2where s=(a+b+c)/ Example: If the input is: a = 3. b = 5. c = 7. Triangle area = 6. Algorithm:
- Start
- Enter the value of a, b and c
- Calculate the value of s as s=(a+b+c)/
- Calculate the value of Area as area=sqrt(s(s-1) (s-b) (s-c)
- Print the area
- Stop Program Code: #include<stdio.h> #include<conio.h> #include<math.h> void main() { float a, b, c, s, area; clrscr(); printf(“Enter the value of a) scanf(%f,&a) printf(“Enter the value of b”); scanf(“ %f”,&b); printf(“Enter the value of c”); scanf(“%f ”, &d); s = (a+b+c)/2;
area= sqrt(s(s-a)(s-b)*(s-c); print(“ Area of the triangle %f”, area); getch(); Output Window :
getch(); } Output Window:
THEME: Conditional Branching
Program No: Objective: Write a program to determine the roots of quadratic equation. Logic: The general form of a quadratic equation is (ax^2 + bx + c = 0). The formula to find the roots of a quadratic equation is given as follows x = [-b +/- sqrt (-b^2 - 4ac)]/2a The discriminant of the quadratic equation is d= (b^2 - 4ac). Depending upon the nature of the discriminant, the roots can be found in different ways.
- If the discriminant is positive, then there are two distinct real roots.
- If the discriminant is zero, then the two roots are equal.
- If the discriminant is negative, then there are two distinct complex roots. For example, consider the following equation 2x^2 8x + 3 = 0. a = 2, b = - 8, c = 3 Discriminant value, d =b^2 - 4ac = 8^2 - 4(-8) * 3 = 40 The discriminant value is positive. Hence, the roots are real and distinct. r1 = (-b +d)/ 2a= (8 +40) /22 = 2. r2 = (b +d)/ 2a =(-8 +40) /2*2 = - 0. r1 = 2.3875andr2 = - 0.3875 are the two roots.
real_part = - b/2a; imag_part = sqrt(-discriminant)/2a; printf(First root of the equation is %lf+%lf”, real_part, imag_part); printf(Second root of the equation is %lf-%lf”, real_part, imag_part); getch(); } } Output Window :
Program No: Objective: Write a program to find the largest of three numbers using nested if else Logic: Given three numbers A, B and C The task is to find the largest number among the three. Input: A = 2, B = 8, C = 1 Output: Largest number = 8 Input: A = 231, B = 4751, C = 75821 Output: Largest number = 75821 Algorithm:
- Start
- Read the three numbers to be compared, as A, B and C.
- Check if A is greater than B. 3.1 If true, then check if A is greater than C. 3.1.1 If true, print 'A' as the greatest number. 3.1.2 If false, print 'C' as the greatest number. 3.2 If false, then check if B is greater than C. 3.1.1 If true, print 'B' as the greatest number. 3.1.2 If false, print 'C' as the greatest number.
- Stop Program Code: #include<stdio.h> #include<conio.h> void main() { float a, b, c: clrscr(); printf(“Enter the value of a) scanf(%f,&a) printf(“Enter the value of b”); scanf(“ %f”,&b); printf(“Enter the value of c”); scanf(“%f ”, &d); if(a>b && a>c) { printf( a is largest of three number %/f”, a); }
Program No: 3 Objective: Write a program to receive marks of physics, chemistry & maths from user & check its eligibility for course if a) Marks of physics > 40 b) Marks of chemistry > 50 c) Marks of math’s > 60 d) Total of physics & math’s marks > 150 or Total of three subjects marks > 200 Logic: Take the marks of Physics , Chemistry and Math from user Case1: if Marks of Physics > Case2: if Marks of Chemistry > Case3: if Marks of Math > Case 4: if Addition of Physics and maths marks > Case5: if Addition of Physics , chemistry and Maths> If all five cases are satisfied then candidate is eligible. Algorithm:
- Start
- Input marks of Physics , Chemistry and Math from user
- If Marks of Physics is greater than 40 and
- If Marks of Chemistry is greater than 50 and
- If Marks of Maths is greater than 60 and
- If Total marks of Physics and Maths is greater than 150
- If total marks of Physics , Chemistry and Maths is greater than 200
- If all step 9, 10, 11 12 13 is satisfied than candidate is eligible for course admission
- Otherwise not eligible for admission in course
- end Program Code : #include<stdio.h> #include<conio.h> void main() {
int physics, chemistry, maths, total1,total2; clrscr(); printf(“ Enter the marks of Physics”); scanf(“%d”, & physics); printf(“ Enter the marks of Chemistry”); scanf(“%d”, & chemistry); printf(“ Enter the marks of Maths”); scanf(“%d”, & maths); toatl1=physics+maths; total2=physics+chemistry+maths; if(physics>40) if(chemistry>50) if(maths>60) if(total1>150|| total2>200) printf(“ Applicants is eligible for admission”); else printf(“ Applicants is not eligible for admission”); else printf(“ Applicants is not eligible for admission”); else printf(“ Applicants is not eligible for admission”); else printf(“ Applicants is not eligible for admission”); } getch() }
Program - 4 Objective: Write a program to find the value of y for a particular value of n. The a, x, b, n is input by user (switch) if n=1 y=ax%b if n=2 y=ax2+b if n=3 y=a-bx if n=4 y=a+x/b Logic: Use the switch feature to choose one condition from multiple condition. Case 1: n=1, calculate using formula y=ax%b Case2: n=2 calculate using y=ax2+b Case3: n=3 calculate using y=a-bx Case 3 n=4 calculate using y=a+x/b Example: Input: If a=5, b=1, x=1, n= Output=4.0 beacause as n=3 case 3 will be executed as y=a-bx=5-1*1= Algorithm:
- Start
- Enter the value of a, b,x and n ,
- Apply the switch case to select only one match case.
- If case label is n=1 then compute using formula y=ax%b
- If case label is n=2 then compute using formula y=ax2+b
- If case label is n=1 then compute using formula y=a-bx
- If case label is n=1 then compute using formula y=a+x/b
- Stop Program Code: #include<stdio.h> #include<conio.h> void main() { int a, b, x , n; float y; clrscr(); printf(" Eneter the value of a\n"); scanf("%d",&a); printf(" Eneter the value of b\n");
scanf("%d",&b); printf(" Eneter the value of x\n"); scanf("%d",&x); printf(" Eneter the value ofn\n"); scanf("%d",&n); switch(n) { case 1: y=(ax%b); printf(" The value of y is=%f" ); break; case 2: y=(ax2+b2); printf( " The value of y is=%f" ,y); break; case 3: y=(a-b*x); printf( " The value of y is=%f" ,y); break; case 4: y=(a+x/b); printf( " The value of y is=%f" ,y); break; } getch(); } Output Window :
#include<conio.h> void main() { int a ,b ,c , i, terms; clrscr(); printf(“ Enter the number the terms”); scanf(“%d”, &terms); a = 0; b= 1; c=0; printf(“Fibonacci series is :\n”); for(i=0;i<+terms;i++) { printf(“\t%d”,c); a=b; b=c; c=a+b; getch(); } Output Window :
Program No: Objective: Write a program to find whether the number is Armstrong number. Logic: If Sum of cubes of every digit of number is equal to itself number is known as Armstrong Number like 0, 1, 153, 370, 371 and 407 153= 1^3 +5^3 +3^3 Determine Remainder of Given number as Remainder=Number%10.Then calculate the last digit of given number as Number=Number/10. Algorithm:
- Start
- Input any integer number
- Set Original number = Number
- Calculate the Remainder= Original number %
- Compute the Sum=Sum+RemainderRemainderRemainder
- Original number=Original number /10;
- Repeat steps 4 ,5 and 6 while Number>
- If Sum= Number Print Number is Armstrong Otherwise print number is not Armstrong.
- Stop Program Code: #include<stdio.h> #include<conio.h> void main() { int Number, Original_number, Remainder, Sum=0; clrscr(); printf("enter the number of three digit integer\n"); scanf("%d",& Number); Original_number=Number; while(Original_number!=0) { Remainder=Original_number%10;
Program No: Objective: Write a program to generate sum of series 1!+2!+3!+ n!. Logic: Apply loop to Compute the summation of every Integer value.Initialize the value of sum=0 and loop counter by 1.At each iteration factorial of each term and then compute Sum. Example: Input: Terms= So, Sum=1! +2! +3! +4! +5!=1+2+6+24+120= Output: 153 Algorithm:
- Start
- Enter the value of terms ie n
- Calculate the value of P=P*I
- Calculate the value Sum=Sum+P
- Repeat steps 3 ,4 for i=0 to term
- Print the value of Sum
- Stop Program Code: #include<stdio.h> #include<conio.h> void main() { long i , term, Sum=0, P=1; printf( “Enter the terms”) scanf(“%d”, &term); for(i=0;i<term;i++) { P= P*i; Sum=Sum+P; }
printf(“ The sum of the series is: %d”,Sum); getch(); } Output Window :
