Code 1:
1. Warp
to check a number is Armstrong
2. C program to check whether a number is Armstrong or not
3. Simple
c program for Armstrong number
4. Armstrong
number in c with output
#include<stdio.h>
int main(){
int num,r,sum=0,temp;
printf("Enter a
number: ");
scanf("%d",&num);
temp=num;
while(num!=0){
r=num%10;
num=num/10;
sum=sum+(r*r*r);
}
if(sum==temp)
printf("%d is an
Armstrong number",temp);
else
printf("%d is
not an Armstrong number",temp);
return 0;
}
Sample output:
Enter a number: 153
153 is an Armstrong number
The time complexity of a program that determines
Armstrong number is: O (Number of digits)
Code 2:
1. Write
a c program for Armstrong number
2. C
program for Armstrong number
generation
3. How
to find Armstrong number in c
4. Code
for Armstrong number in c
#include<stdio.h>
int main(){
int num,r,sum,temp;
int min,max;
printf("Enter
the minimum range: ");
scanf("%d",&min);
printf("Enter
the maximum range: ");
scanf("%d",&max);
printf("Armstrong
numbers in given range are: ");
for(num=min;num<=max;num++){
temp=num;
sum
= 0;
while(temp!=0){
r=temp%10;
temp=temp/10;
sum=sum+(r*r*r);
}
if(sum==num)
printf("%d
",num);
}
return 0;
}
Sample output:
Enter the minimum range: 1
Enter the maximum range: 200
Armstrong numbers in given range
are: 1 153
Code 3:
1. Armstrong
number in c using for loop
#include<stdio.h>
int main(){
int num,r,sum=0,temp;
printf("Enter a
number: ");
scanf("%d",&num);
for(temp=num;num!=0;num=num/10){
r=num%10;
sum=sum+(r*r*r);
}
if(sum==temp)
printf("%d is an
Armstrong number",temp);
else
printf("%d is
not an Armstrong number",temp);
return 0;
}
Sample output:
Enter a number: 370
370 is an Armstrong number
Logic of Armstrong number in c
Code 4:
1. C
program to print Armstrong numbers from 1 to 500
2. C
program for finding Armstrong
numbers
#include<stdio.h>
int main(){
int num,r,sum,temp;
for(num=1;num<=500;num++){
temp=num;
sum
= 0;
while(temp!=0){
r=temp%10;
temp=temp/10;
sum=sum+(r*r*r);
}
if(sum==num)
printf("%d
",num);
}
return 0;
}
Output:
1 153 370 371 407
Definition of Armstrong number or what is an
Armstrong number:
Definition according to c programming point of view:
Those numbers which sum of the cube of its digits is equal to that number
are known as Armstrong numbers. For example 153 since 1^3 + 5^3 + 3^3 = 1+ 125
+ 9 =153
Other Armstrong
numbers: 370,371,407 etc.
In general definition:
Those numbers which sum of its digits to power of number of its digits is
equal to that number are known as Armstrong numbers.
Example 1: 153
Total digits in 153
is 3
And 1^3 + 5^3 + 3^3 =
1 + 125 + 27 = 153
Example 2: 1634
Total digits in 1634 is 4
And 1^4 + 6^4 + 3^4 +4^4 = 1 + 1296 + 81 + 64 =1634
Examples of Armstrong numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634,
8208, 9474, 54748, 92727, 93084, 548834, 1741725
18. Write a c program which passes structure to function.
50 comments:
this will get you only 3 digit armstrong nos..
lets work out this with example
take num = 153
while 153!=0 satisfies
r = 153%10 = 3
num = 153/10 = 15
sum=0+(3*3*3) = 27
now num = 15
15!=0
r=15%10=5
num=15/10=1
sum = 27+(5*5*5)=152
now num=1
1!=0
r=1%10=1
num=1/10=0
sum = 152+1=153
so now the condition sum==temp or num satisfies
I have written this code for any no. of digits. It works. I have used above code. Need Help. Help me reduce no of lines as well as no of variables. you can reply on arpann72@gmail.com
#include
void main()
{
int num,exp=1,rem,sum=0;
int i,temp,temp2,count=0,j;
scanf("%d",&num);
temp=num;
while(num/exp>0)
{
count=count+1;
exp*=10;
}
j=count;
while(j)
{
temp2=1;
rem=num%10;
num=num/10;
for(i=0;i<count;i++)
{
temp2*=rem;
}
sum+=temp2;
exp*=10;
j--;
}
if(sum==temp)
printf("\nThe number %d is an armstrong number",temp);
else
printf("\nThe number %d is not an armstrong number",temp);
}
Armstrong checking for any no
#include
int main(){
int num,r,s,sum=0,temp,c=0;
printf("\nEnter a number:-");
scanf("%d",&num);
temp=num;
while(num!=0){
r=num%10;
num=num/10;
c++;
}
printf("\nno of digits:\n%d",c);
num=temp;
printf("\nno after assignment :\n%d",num);
while(num!=0){
s=num%10;
num=num/10;
sum=sum+power(s,c);
}
printf("\nsum is\n%d",sum);
if(sum==temp)
printf("\nThe number %d is an armstrong number",temp);
else
printf("\nThe number %d is not an armstrong number",temp);
getch();
return 0;
}
int power(int a,int b){
int i=1,mul=1;
while(i<=b){
mul*=a;
i++;
}
return mul;
}
/*the below program holds good for any no. of digits*/
#include
#include
int main()
{
int m,t,i=1,s=0,n,r,j=1;
printf("enter the no: ");
scanf("%d", &n);
m=t=n;
while(j!=0)
{
r=n%10;
if(n<10)
{
break;
}
n=n/10;
j++;
}
printf("no of digits=%d\n",j);
while(i<=j)
{
r=m%10;
m=m/10;
s=s+pow(r,j);
i++;
}
printf("sum=%d\n",s);
if(s==t)
{
printf("valid");
}
else
{
printf("invalid");
}
getch();
return 0;
}
in the above program directives are stdio.h and math.h
how it works for 407......
merging 2 1-d arrays into 3rd array
how to find armstrong numbers between 1 and 1000 using only while?
ethilum bedam njana
nice explanation yaar...thanx alot
Guys can u help me out knowing why "temp" is used in this program???
you can visit www.codecast.org for more codes,can register and upload there
There is a rail road to different towns from one town. It is considered that the rail road is a one way because of budgetary issues. If there is city A and it has cities B and C cities connected by rail roads, it is not mandatory that A has direct rail road to both B and C. The route to C from A can be A to B and then B to C. The above scenario is depicted as a graph. The nodes are the towns and edges are the distances between them. The input given will be a graph and also the route to towns. The output must be the total rail road distance between the towns and if no route exists, it must say 'No Route exists'.
PLEASE SEND ME A C PROGRAMME FOR THIS QUESTION. PLEASE ITS URGENT.
There is a mistake in the explanation of the Armstrong section: its 3^3=27 not 9 (the first part of the explanation)
A correction on this part. Its seems that if the Armstrong num. is anything other than three, then this wont work (except for 1). I think the number of r's has to be adjusted to how many digits the natural number inputted has. the way to do this is to create a loop to see how many times is the number entered divisible by 10 at the beginning of the program, and then multiply by that many r's everytime. Please notify me if I am wrong. Thanks
good , but small mist.
go to Definition. 1^3 + 5^3 + 3^3 = 1+ 125 + 27 =153
thats it.
4*4*4=64
7*7*7=343
343+64=407
int i=1,temp,s;
while(i<=1000)
{
temp=i;
s=0
i++;
while(temp!=0)
{
r=temp%10;
temp=temp/10;
s=s+temp;
}
if(s==i)
printf("%d",i);
}
return 0;
}
because by using temp or any another variable we can get the single value.
for example:
153
step 1 step 2 step 3
153%10=3 15%10=5 1%10=1
153/10=15 15/10=1 1/10=0
0+(3*3*3)=27 27+(5*5*5)=152 152+(1*1*1)=153
yahh!! 1-9 all r armstrong but it won't show it
i have searched in net that there are only 4 three digited amstrong nos
153,370,371,407
but if 153 is amstrong then 351 and 531 are also amstrong numbers
if u find answer for this plz mail@mani sai
@mani4sai@gmail.com
how can 531 be an amstrong no,
3^3+5^3+1^3=27+125+1=153 not 351 so 351 cant be an amstrong no.An amstrong no is sum of cubes of the number is the no itself.
same goes with 531.
thanks
is it for amstrong numbers??? or for perfect numbers...??
above all code implemented by admin is wrong
#include
#include
int main(void)
{
int sum,term,policytype;
float bonus;
printf("enter the sum \n");
scanf("%d",&sum);
printf("enter 1 for Endownment \n 2 for Money Bank \n 3 for Term Assurance \n");
scanf("%d",&policytype);
switch(policytype)
{
case 1: printf("enter the term: \n");
scanf("%d",&term);
if(term>=25)
{
bonus=((float)sum/1000)*60;
printf("the bonus rate per thousand is %f \n",bonus);
}
else if((term>=20)&&(term<25))
{
bonus=((float)sum/1000)*55;
printf("the bonus rate per thousand is %f \n",bonus);
}
else if((term>=10)&&(term<20))
{
bonus=((float)sum/1000)*50;
printf("the bonus rate per thousand is %f \n",bonus);
}
else if(term<10)
{
bonus=((float)sum/1000)*45;
printf("the bonus rate per thousand is %f \n",bonus);
}
break;
case 2:printf("enter the term: \n");
scanf("%d",&term);
if(term>=25)
{
bonus=((float)sum/1000)*55;
printf("the bonus rate per thousand is %f \n",bonus);
}
else if((term>=20)&&(term<25))
{
bonus=((float)sum/1000)*50;
printf("the bonus rate per thousand is %f \n",bonus);
}
else if((term>=10)&&(term<20))
{
bonus=((float)sum/1000)*45;
printf("the bonus rate per thousand is %f \n",bonus);
}
break;
case 3:printf("the bonus is not applicable for this policy type \n");
break;
default: printf("please enter valid policy type \n");
break;
}
getch();
return 0;
}
/* */
#include //directive for input output
#include //directive for displaying output
#include
int compound_interest(float p,float t, float r,float n);
void main()
{
float comp_intrest,principal_amt,rate,time,number_interest;
float result;
printf("enter the amount, time, rate and number\n");
scanf("%f%f%f%f",&principal_amt,&time,&rate,&number_interest);
result = compound_interest(principal_amt,time,rate,number_interest);
printf("the compound intreset is %f\n",result);
getch();
}
int compound_interest(float p,float t, float r,float n)
{
float amount, ci,temp=1,nt;
int i;
nt=n*t;
for(i=0;i<nt;i++)
{
temp=temp*(1+r/n);
}
amount=p*temp;
ci=amount-p;
return ci;
}
nice explnation
dont know why m nt gtng ans..plz any1 can notify me wher m going wrng
result iz blank..?????
#include
#include
int main()
{
int n,sum=0,a,i;
printf("armstrng no are \n\n");
for(i=1;i<=1000;i++)
{
n=i;
sum=0;
while(n>0)
{
a=n%10;
n=n/10;
sum=sum+(a*a*a);
}
if(n==sum)
{
printf("%d",n);
}
}
getch();
return 0;
}
change condition of if, if(i==sum) not n==sum, bcoz at this point n 'll be zero...
Maybe not the best but works for 3 digit armstrong number :D
#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
main()
{
int a,b,c,whole,choice,n=0,count=0,onedigit=0,twodigit=0,threedigit=0;
printf("Enter the number you want to test :");
scanf("%d",&whole);
n=whole;
while(n!=0)
{
n/=10;
++count;
}
printf("The number of digits entered are %d\n",count);
if(count==1)
{
onedigit=(whole*whole*whole);
if(onedigit==whole)
printf("This is an armstrong number");
else
printf("Not an armstrong number");
}
else if(count==2)
{
a=whole%10;
b=whole/10;
twodigit=(a*a*a)+(b*b*b);
if(twodigit==whole)
printf("This is an armstrong number");
else
printf("Not an armstrong number");
}
else if (count==3)
{
a=whole%10;
b=(whole/10)%10;
c=whole/100;
threedigit=(a*a*a)+(b*b*b)+(c*c*c);
if(threedigit==whole)
printf("This is an armstrong number");
else
printf("Not an armstrong number");
}
else
printf("Not a valid input");
getche();
}
mis print of 3^3 = 9
the above mentioned codes wont work for 4 digit Amstrong numbers so those are wrong codes, please review my code and it will work for all Amstrong numbers:
#include
#include
#include
#include
void main()
{
char num[8];int len,temp,rem,sum=0,temp1;
printf("enter number:");
gets(num);
len=strlen(num);
temp=atoi(num);
temp1=temp;
while(temp>0)
{
rem=temp%10;
sum=sum+pow(rem,len);
temp=temp/10;
}
if(sum==temp1)
{printf("%d is an Amstrong number",temp1);}
else
{printf("%d is not an Amstrong number",temp1);}
}
send email to me at chvijay.1990@gmail.com regarding the review of this code....!
thank u sooooo much
armstrong number program
#include
#include
void main()
{
int sum=0, num,temp,n=0;
printf("Enter Number");
scanf("%d",&num);
temp=num;
while(temp>0)
{
n++;
temp/=10;
}
temp=num;
while(temp>0)
{
sum+=pow(temp%10,n);
temp/=10;
}
if(sum==num)
{
printf("%d is an armstrong Number",num);
}
else
{
printf("%d is Not an Armstrong Number",num);
}
}
for Armstrong number program in java visit:
http://programmerminds.blogspot.in/2013/11/armstrong-number-program-in-java.html
Thank you very much...
//This will work for any number.
#include
int main ()
{
int i=0,y,t,sum=0,m,x,num;
scanf("%d",&num);
m=num;
while(m!=0)
{
m=m/10;
i++;
}
while(num!=0)
{
x=num%10;
y=1;
t=i;
while(t!=0)
{
y=y*x;
t--;
}
sum =sum + y;
num = num/10;
}
printf("%d",sum);
}
Why is temp used?
You r worng because Armstrong number are those numbers which power is 3 and get same value like 1^3 =1, 153= 1×1×1+5×5×5+3×3×3
Given a string of word with the combination of words number and symbol.write a program to read the number in the string and add it print like below.Exampleinput : ci2t2r1is@y$%soutput citrisys5@$%
write a program for given logic belowstep 1: find the unitdigit of the number Example 34246 is 6(unit digit)Step 2: (3*2 + 4*2 + 2*2 + 4*2)/4 = 6step 3: check unit dight and above condition equals if TRUE print SUCCESS elseFAIL
check and remove duplicate in Array(very Easy ANS put the value in Hashset collection)Given array ={1,2,3,4,1,2,3,5,1,2}ouput={1,2,3,4,5}
given a condition 2^n+1 taake some value of n like n=100check number beween the condion is prime or notfor example2^0+1=2 is prime2^1+1=3 is prime2^2+1=5 is prime2^3+1=9 is not prime
if anybody knows answer my questions please send me....
how to find given number is armstrong num or not using strings]
An armstrong number is an n-digit number that is equal to the sum of the nth powers of its digits. Write a program that accepts an integer and determines if it is an armstrong number or not.
Example:
153 is a 3- digit number so each digit should be raised to 3 and add the results.
1 3 ++ 5 3 + 33 = 1 + 125 + 27 = 153
153 = 153
Therefore, 153 is an armstrong number.
25 is a 2-digit number so each digit should be raised to 2 and add the results.
22 + 52 = 4 + 25 = 29
25 is not equal to 29
Therefore, 25 is not an armstrong number.
Requirement: do not use any math function (math.h)E.g. pow( ), sqrt() etc.
Create a loop that:
1.)counts the number of digits in an n-digit number (for determining exponent)
2.) loop that extracts individual digit
3.) loop that computes the power of x raise to n.
please reply to me in an e-mail: hannahdelacruz009@gmail.com. Thank you
sir your programming using for loop of Armstrong no is not running in blue j
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