a) AIM: Write a c program to find the sum of individual digits of the given positive integer Start Sum = 0 Read n value While(n>0)

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1 Regulation R10 C PROGRAMMING LAB MANUAL EXERCISE-1: a) AIM: Write a c program to find the sum of individual digits of the given positive integer ALGORITHM: Step 1: Start. Step 2: Take variables namely sum, n. Step 3: Read n value. Step 4: Initialize sum=0. Step 5: While(n>0) then sum=sum+(n%10); n=n/10; Step 6: Write sum of the digits. Step 7: Stop. FLOWCHART: Start Sum = 0 Read n value While(n>0) Sum=sum+(n%10) n = n/10 Write sum Stop Department of Computer Science & Engineering 1

2 PROGRAM: #include<stdio.h> #include<conio.h> void main() int sum=0,n; clrscr(); printf("enter any n value:"); scanf("%d",&n); while(n>0) sum=sum+(n%10); n=n/10; printf("sum of the digits of given number is:%d",sum); getch(); OUTPUT: Enter any n value: 783 Sum of the digits of given number is: 18 Department of Computer Science & Engineering 2

3 b) AIM: A Fibonacci series is defined as follows: the first and second terms in the sequence are 0 and 1. Subsequent terms are found by adding the preceding two terms in the sequence. Write a c program to generate the first n terms of the sequence. ALGORITHM: Step 1: Start. Step 2: Take variables namely n, i,t1, t2,t. Step 3: Initialize variables t1=0 and t2=1. Step 4: Read n value. Step 5: Display t1 and t2. Step 6: for i=1 to i<=n-2 then i++ t=t1+t2. Display t value. t1=t2. t2=t. Step 7: Stop. Department of Computer Science & Engineering 3

4 FLOWCHART: Start Read n value t1 =0 t2=1 Display t1 and t2 i = 1 i <=n-2 t = t1+t2 Display t value. Stop i++ t1 = t2 t2 = t Department of Computer Science & Engineering 4

5 PROGRAM: #include<stdio.h> #include<conio.h> void main() int n,t1=0,t2=1,t,i; clrscr(); printf("enter any n value:"); scanf("%d",&n); printf("fibonancci series is:"); printf("%5d%5d",t1,t2); for(i=1;i<=n-2;i++) t=t1+t2; printf("%5d",t); t1=t2; t2=t; getch(); OUTPUT: Enter any n value: 8 Fibonacci series is: Department of Computer Science & Engineering 5

6 c) AIM: Write a C program to generate all the prime numbers between 1 and n, where n is a value supplied by the user. ALGORITHM: Step 1: Start. Step 2: Take variables namely n,m,i,j,count. Step 3: Read n value. Step 4: for i=1 to i<=n then i++ count=0. Step 5: Stop. for j=1 to j<=i then j++ if(i%j==0) count++ if(count==2) Display i value. Department of Computer Science & Engineering 6

7 FLOWCHART: Start Read n value. i = 1 i <= n count = 0 j = 1 i ++ j <= i i%j= =0 count= =2 Stop count++ Display i j ++ Department of Computer Science & Engineering 7

8 PROGRAM: #include<stdio.h> #include<conio.h> void main() int n,count,i,j; clrscr(); printf("enter any n value:"); scanf("%d",&n); printf("prime numbers from 1 to %d are:",n); for(i=1;i<=n;i++) count=0; for(j=1;j<=i;j++) if(i%j==0) count++; if(count==2) printf("%5d",i); getch(); OUTPUT: Enter any n value: 15 Prime numbers from 1 to 15 are: Department of Computer Science & Engineering 8

9 WEEK-2: a) AIM: Write a C program to calculate the following Sum: Sum=1-x2/2! +x4/4!-x6/6!+x8/8!-x10/10! ALGORITHM: Step 1: Start. Step 2: Take variables namely n,x,i,p; Step 3: Initialize p to 0 and sum to 0 Step 4: Read n and x values. Step 5: for i=1 to i<=n then i++ Step 6: if(i%2!=0) then sum=sum+pow(x,p)/fact(p) goto step9. Step 7: else sum=sum-pow(x,p)/fact(p) goto step9. p=p+2 step 8: Display the sum value. Step 9: fact(int x) if(x<=1) then Return 1 to step5 else Return x*fact(x-1) to step 5. Step 10: Stop. Department of Computer Science & Engineering 9

10 FLOWCHART: Start p=0 x=0 Read n,x values i=1 i<=n i%2!=0 sum=sum+pow(x,p)/fact(p) sum=sum-pow(x,p)/fact(p) Display sum x<1 Stop f= x*fact(x-1) return f p=p+2 i++ Department of Computer Science & Engineering 10

11 PROGRAM: #include<stdio.h> #include<conio.h> #include<math.h> unsigned long fact(int x) if(x<=1) return 1; else return x*fact(x-1); void main() int n,x,i,p=0; float sum=0; clrscr(); printf("enter numer of terms:"); scanf("%d",&n); printf("enter x value:"); scanf("%d",&x); for(i=1;i<=n;i++) if(i%2!=0) sum=sum+pow(x,p)/fact(p); else sum=sum-pow(x,p)/fact(p); p=p+2; printf("sum of given series is:%.2f",sum); getch(); OUTPUT: Enter numer of terms:4 Enter x value:1 Sum of given series is:0.54 Department of Computer Science & Engineering 11

12 b) AIM: Write a c program to find the roots of a quadratic equation. ALGORITHM: Step 1: Start. Step 2: Take variables namely a,b,c,x1,x2. Step 3: Read a,b and c values. Step 4: if b2-4ac<0 then s=abs(b*b-4*a*c) Find real part r=-b/(2*a) value Find imaginary part i=s/(2*a) Display the First root value is (r,i) Display the Second root value is(r,-i) else Find first root value x1=(-b+sqrt(b*b-4*a*c))/(2*a) Find the second root valuex2=(-b-sqrt(b*b-4*a*c))/(2*a) Display x1 and x2. Step 5: Stop. FLOWCHART: Start Read a and b values b2-4ac<0 S=abs(b*b-4*a*c) x1=(float)(-b+sqrt(b*b-(4*a*c)))/(2*a) r=-b/(2*a) X2=(float)(-b-sqrt(b*b-(4*a*c)))/(2*a) i=s/(2*a) Display (r,i) Display x1,x2 Display (r,-i) Stop Department of Computer Science & Engineering 12

13 PROGRAM: #include<stdio.h> #include<conio.h> #include<math.h> void main() int a,b,c; float x1,x2,s,r,i; clrscr(); printf("enter a,b and c values:"); scanf("%d%d%d",&a,&b,&c); if((b*b-4*a*c)<0) s=abs(b*b-4*a*c); r=-b/(2*a); i=s/(2*a); printf("first root value is:(%f,%f)",r,i); printf("\nsecond root value is:(%f,%f)",r,-i); else x1=(-b+sqrt(b*b-4*a*c))/(2*a); x2=(-b-sqrt(b*b-4*a*c))/(2*a); printf("\nfirst root value is:%f",x1); printf("\nsecond root value is:%f",x2); getch(); OUTPUT: Enter a,b and c values: First root value is: Second root value is: Department of Computer Science & Engineering 13

14 WEEK-3: a) AIM The total distance travelled by vehicle in t seconds is given by distance = ut+1/2at2 where u and a are the initial velocity (m/sec.) and acceleration (m/sec2). Write C program to find the distance travelled at regular intervals of time given the values of u and a. The program should provide the flexibility to the user to select his own time intervals and repeat the calculations for different values of u and a. ALGORITHM: Step 1: Start. Step 2: Take variables namely u, a, t and s. Step 3: Read u, a, and t. Step 4: s=u*t+ ½ *a*square(t) Step 5: Write s Step 6: Stop. FLOWCHART: Start Read u, a and t s=u*t+ ½ *a*square(t) display s Stop Department of Computer Science & Engineering 14

15 PROGRAM: #include<stdio.h> #include<conio.h> #include<math.h> void main() float u,a,t,dis; clrscr(); printf("enter initial velocity(u):"); scanf("%f",&u); printf("enter acceleration(a):"); scanf("%f",&a); printf("enter travelled time(t):"); scanf("%f",&t); dis=u*t+((float)1/2*a*pow(t,2)); printf("distance travelled is:%f",dis); getch(); OUTPUT: Enter initial velocity (u): 5 Enter acceleration (a): 2 Enter traveled time: 3 Distance traveled is: Department of Computer Science & Engineering 15

16 b) AIM: Write a C program, which takes two integer operands and one operator form the user, performs the operation and then prints the result. (Consider the operators +,-,*, /, % and use Switch Statement) ALGORITHM: Step 1: Start. Step 2: Take three variables namely a, b and opt. Step 3: Read a, b and opt. Step 4: Switch (opt) Case + then write a+b Case - then write a-b Case * then write a*b Case / then write a/b Case % then write a%b Default then invalid option Step 5: Stop. Department of Computer Science & Engineering 16

17 FLOWCHART: Start Read a, b and opt Case + Case - Case * Case / Case % Write a+b Write a-b Write a*b Write a/b Write a%b Invalid option Stop Department of Computer Science & Engineering 17

18 PROGRAM: #include<stdio.h> #include<conio.h> void main() int a,b; char opt; clrscr(); printf("enter a and b values:"); scanf("%d%d",&a,&b); printf("enter any option:"); flushall(); scanf("%c",&opt); switch(opt) case '+': case '-': case '*': case '/': case '%': default: getch(); printf("addition of given two values is:%d",a+b); break; printf("subtraction of given two values is:%d",a-b); break; printf("multiplication of given two values is:%d",a*b); break; printf("division of given two values is:%d",a/b); break; printf("modulus of given two values is:%d",a%b); break; printf("invalid option"); OUTPUT: Enter a and b values: 10 5 Enter any option: + Addition of given two values is: 15 Department of Computer Science & Engineering 18

19 WEEK-3: i) AIM: Write a c program to find the factorial of given number using recursive function ALGORITHM: Step 1: Start. Step 2: Take variables namely n,f. Step 3: Read n value. Step 4: f=fact(n) Step 5: if(n<=1) then Step 6: return 1 to step4 Step 7: else return x*fact(x-1) to step 4. Step 8: Display f value. Step 9: Stop Start FLOWCHART: Read n f=fact(n) return 1 return x*fact(x-1) Display f n<=1 Stop PROGRAM: #include<stdio.h> #include<conio.h> void main() unsigned long fact(int); int n; unsigned long f; clrscr(); printf("enter any n value:"); scanf("%d",&n); f=fact(n); printf("factorial of given value is:%lu",f); getch(); unsigned long fact(int n) OUTPUT Enter any n value:5 Factorial of given value is:120 if(n<=1) return 1; else return n*fact(n-1); Department of Computer Science & Engineering 19

20 i) AIM: Write a c program to find the factorial of given number using non-recursive function ALGORITHM: Step 1: Start. Step 2: Take variables namely n and f. Step 3: Read n value Step 4: f=factorial(n) Step 5: Take variable fa=1. Step 6: while(n>=1) then fa=fa*n n step 7: Return fa value to step 4 Step 8: Display f value Step 9: Stop FLOWCHART: Start Read n f=factorial(n) fa=1 return fa Display f n>=1 Stop fa=fa*n n-- PROGRAM: #include<stdio.h> #include<conio.h> void main() unsigned long factorial(int); int n; unsigned long f; clrscr(); printf("enter any n value:"); scanf("%d",&n); f=factorial(n); printf("factorial of given value is:%lu",f); getch(); Department of Computer Science & Engineering 20

21 unsigned long factorial(int n) unsigned long fa=1; while(n>=1) fa=fa*n; n--; return fa; OUTPUT Enter any n value:5 Factorial of given value is: 120 ii) AIM: Write a c program to find the GCD (greatest common divisor)of two given integers using non-recursive functions. ALGORITHM: Step 1: Start. Step 2: Take variables namely a,b and g. Step 3: Read two variables namely a,b Step 4: g=gcd(a,b) Step 5: Take vraibles namely m, i and gc Step 6: m=(a>b)?a:b; Step 7: for i=1 to i<=m then i++ if(a%i==0 and b%i==0) then gc=i. step 8: return gc to step 3 step 9: Display g value Step 10: Stop. Department of Computer Science & Engineering 21

22 FLOWCHART: Start Read a and b values g=gcd(a,b) m=(a>b)?a:b for(i=1;i<=m;i++) Display g value if(a%i==0)&& (b%i==0) Stop g = i PROGRAM: #include<stdio.h> #include<conio.h> void main() int gcd(int,int); int a,b,g; clrscr(); printf("enter any two values:"); scanf("%d%d",&a,&b); g=gcd(a,b); printf("g.c.d of given two values is:%d",g); getch(); Department of Computer Science & Engineering 22

23 int gcd(int a,int b) int i,m,gc; m=(a<b)?a:b; for(i=1;i<=m;i++) if(a%i==0&&b%i==0) gc=i; return gc; OUTPUT: Enter any two values: 4 8 G.C.D of given two values is: 4 ii) AIM: Write a c program to find the GCD (greatest common divisor)of two given integers using recursive functions. ALGORITHM: Step 1: Start Step 2: Take variables namely a,b and g. Step 3: Read a and b values Step 4: g=gcd(a,b) Step 5: Take variables namely m,i=1,g Step 6: find minimum among a and b m=(a>b)?a:b step 7: if(a%i==0&&b%i==0) g=i step 8: i++ step 9: if(i<=m) goto step 4 step 10: return g value to step4 step 11: Display g value step 12: Stop. Department of Computer Science & Engineering 23

24 FLOWCHART: Start Read a and b values g=gcd(a,b) m=(a>b)?a:b for(i=1;i<=m;i++) Display g value if(a%i==0)&& (b%i==0) Stop g = i PROGRAM: #include<stdio.h> #include<conio.h> void main() int gcd(int,int); int a,b,g; clrscr(); printf("enter any two values:"); scanf("%d%d",&a,&b); g=gcd(a,b); printf("g.c.d of given two values is:%d",g); getch(); int gcd(int a,int b) static int m,i=1,g; m=(a<b)?a:b; if(a%i==0&&b%i==0) g=i; OUTPUT: i++; Enter any two values: 4 8 if(i<=m) G.C.D of given two values is: 4 gcd(a,b); return g; Department of Computer Science & Engineering 24

25 WEEK 8 a) AIM: Write a program to generate Pascal s triangle. ALGORITHM: Step 1: Start. Step 2: Take variables namely n, i,j,s,ncr. Step 3: Read n value. Step 4: s=n*3. Step 5: for i=0 to i<=n then i++ Display s spaces. for j=0 to j<=i then j++ ncr=fact(i)/fact(j)*fact(i-j). Display ncr. s=s-3. Display \n Step 6: Stop. FLOWCHART: Start Read n value s = n*3 for i=0 to i<=n ;i++ Display s spaces for j=0 to j<=i ;j++ ncr=fact(i)/fact(j)*fact(i-j) Display ncr s=s-3 Display \n Stop Department of Computer Science & Engineering 25

26 PROGRAM: #include<stdio.h> #include<conio.h> unsigned long fact(int x) if(x<=1) return 1; else return x*fact(x-1); void main() int n,i,j,s,ncr; clrscr(); printf("enter any n value:"); scanf("%d",&n); s=n*3; for(i=0;i<=n;i++) printf("%*c",s,32); for(j=0;j<=i;j++) ncr=fact(i)/(fact(j)*fact(i-j)); printf("%5d",ncr); s=s-3; printf("\n"); getch(); OUTPUT: Enter any n value: Department of Computer Science & Engineering 26

27 b) AIM: Write a c program to construct a pyramid of numbers ALGORITHM: Step 1: Start. Step 2: Take variables namely i, j, m, n,s. Step 3: Read n value. Step 4: s=n*4. Step 5: for i=1 to i<=n then i Display s spaces. 5.2 m=i. 5.3 for j=1 to j<2*i then j if(j<i) then Display m++ value else then Display m-- value 5.4 Display \n. 5.5 s=s-4. Step 6: Stop. Department of Computer Science & Engineering 27

28 FLOWCHART: Start Read n value s=n*4 for i=1 to i<=n i++ Display s spaces m=i Display m-- for j=1 to j<=i i++ if(j<i) Display m++ Display \n s=s-4 Stop Department of Computer Science & Engineering 28

29 PROGRAM: #include<stdio.h> #include<conio.h> void main() int i,m,j,n,s; clrscr(); printf("enter any n value:"); scanf("%d",&n); s= n*4; for(i=1;i<=n;i++) printf("%*c",s,32); m=i; for(j=1;j<2* i;j++) if(j<i) else printf("\n\n"); s=s-4; getch(); OUTPUT: Enter any n value: printf("%4d",m++); printf( %4d,m--); Department of Computer Science & Engineering 29

30 WEEK 9 AIM: Write a C function to read in two numbers, x and n, and then compute the sum of this geometric progression: 1+x+x2+x3+.+xn ALGORITHM: Step 1: Start Step 2: Take variables namely n,i,x,sum=0 Step 3: Read n value Step 4: if(n<0) Display number of terms must be a positive value Goto Step Step 5: Read x value Step 6: for(i=0;i<=n;i++) Sum=sum+power(x,i) Step 7: Display the sum value Step 8: Stop FLOWCHART: Start sum=0 Read n value if(n<0) n must be positive Read x value for(i=0;i<=n;i++) Display sum sum=sum+pow(x,i) Stop Department of Computer Science & Engineering 30

31 PROGRAM: #include<stdio.h> #include<conio.h> #include<math.h> void main() int n,i; float x,sum=0; clrscr(); lb: printf("\nenter number of terms value:"); scanf("%d",&n); if(n<0) printf("number of terms value must be positive"); goto lb; printf("enter any x value:"); scanf("%f",&x); for(i=0;i<=n;i++) sum=sum+pow(x,i); printf("sum of the given series is:%.2f",sum); getch(); OUTPUT: Enter number of terms value: 5 Enter any x value: 1 Sum of the given series is: 6.00 Department of Computer Science & Engineering 31