C PROGRAMMING EXERCISE 5
Exercise 2: Loop Control Structure
Youtube Video:
CODE IS AS FOLLOWS:
//Hello and welcome everyone
//Lets start with Exercise 5 but first let me tell u that ex 5 is based on LOOP CONTROL STRUCTURES
//Lets get started
//completinf the code
#include
#include
void main()
{
int opt,sum=0,m,n,c,temp,i,j,b=1,p,x,fd,ld,num_even=0,num_odd=0,num_0=0;
clrscr();
while(1)
{
printf("\nEnter your choice\n");
printf("1.\t2.\t3.\n4.\t5.\t6.\n7.\t8.\t9.\n10.\t11.\t12.\nPress 12 for exit");
scanf("%d",&opt);
switch (opt)
{
case 1: printf("Q1:-WRITE A PROGRAM TO CALCULATE SUM OF DIGITS OF A GIVEN INPUT NUMBER");
printf("\nSolution");
printf("Enter the digit");
scanf("%d",&n);
while(n>0)
{
m=n%10;
sum=sum+m;
n=n/10;
}
printf("%d is the sum of digit",sum);
break;
case 2: printf("Q2:-ACCEPT TWO NUMBERS AS RANGE AND DISPLAY SUM OF ALL NUMBERS BETWEEN THAT RANGE");
printf("\nSolution");
printf("Enter tow numbers as range");
scanf("%d%d",&m,&n);
for(i=m;i<=n;i++)
{
sum=sum+i;
}
printf("%d is sum of all numbers between the ramge",sum);
break;
case 3: printf("Q3:-WRITE A PROGRAM TO CHECK WHETHER A INPUT IS ARMSTRONG OR NOT");
printf("\nSolution");
printf("Enter the number");
scanf("%d",&n);
temp=n;
while(n>0)
{
m=n%10;
c=m*m*m;
sum=sum+c;
n=n/10;
}
//lets check if the number is a armstrong or not
if(sum==temp)
{
printf("Number is armstrong");
}
else
{
printf("Not an armstrong");
}
break;
case 4: printf("Q4:-WRITE A PROGRAM TO ACCEPT BINARY NUMBER AND CONVERT IT IN DECIMAL");
printf("\nSolution");
printf("Enter binary num");
scanf("%d",&n);
while(n>0)
{
m=n%10;
sum=sum+n*b;
n=n/10;
b=b*2;
}
printf("%d is the decimal number",sum);
break;
case 5: printf("Q5:-WRITE A PROGRAM TO CHECK WHETHER THE INPUT NUMBER IS PERFECT NUMBER OR NOT");
printf("\nSolution");
printf("Enter the number");
scanf("%d",&n);
temp==n;
for(i=1;i0)
{
m=n%10;
sum=sum*10+m;
n=n/10;
}
if(sum==temp)
{
printf("Pallindrome");
}
else
{
printf("Non palindrome");
}
break;
case 8: printf("Q8:-WRITE A C PROGRAM TO DISPLAY MULTIPLICATION OF TWO INPUT NUMBERS WITHOUT USING * OPERATOR");
printf("\nSolution");
printf("Enter two numbers");
scanf("%d%d",&n,&m);
for(i=1;i<=m;i++)
{
sum=sum+n;
}
printf("%d*%d=%d",m,n,sum);
break;
case 9: printf("Q9:-WRITE A PROGRAM TO CALCULATE SUM OF FIRST AND LAST DIGIT OF A NUMBER");
printf("\nSolution");
printf("Enter a number");
scanf("%d",&n);
ld=n%10; //n%10 gives us the last digit
//to find first digit,we will divide n by 10 until n>=10
while(n>=10)
{
n=n/10;
}
fd=n;
//now we have first digit so lets add them
sum=fd+ld;
//displaying the numbers
printf("%d is the sum of first and last number",sum);
break;
case 10: printf("Q10:-WRITE A PROGRAM TO ACCEPT A NUMBER AND COUNT NUMBER OF EVEN,ODD,ZERO DIGITS WHITHIN THAT NUMBER");
printf("\nSolution");
printf("Enter the number");
scanf("%d",&n);
//we will use while loop
//num_odd,num_even,num_0 are new variables
while(n>0)
{
m=n%10;
//now we will use nested if else statement
if(m==0)
{
num_0++;
}
else if(m%2==0)
{
num_even++;
}
else
{
num_odd++;
}
n=n/10;
}
printf("even number digits except 0=%d\nodd number digits=%d\nnumber of zero digit=%d",num_even,num_odd,num_0);
break;
case 11: printf("Q11:-WRITE A C PROGRAM WHICH ACCEPTS A NUMBER N AND DISPLAYS EACH DIGIT IN WORDS");
printf("\nSolution");
printf("Enter number");
scanf("%d",&n);
while(n>0)
{
m=m*10+n%10;
n=n/10;
}
while(m>0)
{
p=m%10;
//rem=rem/10;
switch(p)
{
case 1: printf(" one");
break;
case 2: printf(" two");
break;
case 3: printf(" three");
break;
case 4: printf(" four");
break;
case 5: printf(" five");
break;
case 6: printf(" six");
break;
case 7: printf(" seven");
break;
case 8: printf(" eight");
break;
case 9: printf(" nine");
break;
}
}
break;
case 12:exit(1);
}
//completed coding....time to jump to check errors...yes jumping to another app :)
//okay done...stay tuned my future programmer;)
}
getch();
}
0 Comments