I was always looking for explanations behind such programs, a source from where I can learn the concepts, and I finally found this channel, I really needed this as I use to feel alienated in computer science, but this already made me excited about everything related to computer science again, can't thank enough!! ♥️🙏🏾
i know Im randomly asking but does someone know a trick to get back into an Instagram account?? I somehow lost my account password. I appreciate any tips you can give me!
alternative int main() { int num,count=0,sum=0,rem,result=0; printf("enter a number "); scanf("%d",&num); int kela=num; int lela=num; while(num!=0){ num=num/10; count++; } while(lela!=0){ rem=lela%10; sum=sum+pow(rem,count); lela/=10; } if(kela==sum) { printf("its a armstrong number i.e %d",kela); } else{ printf("its not a armstrong number coz %d",sum); } return 0; }
i love you explanation each step i learn programming for while many tutorial go fast without clear explanation like this make me exhaust to figure it out what going on in every line of code, now i'm back to go to basic and found this channel love it
that was really a presentation which cleared my concept on what Armstrong number is and what algorithm should be used to find a number whether is it armstrong number or not
I would code it like this int main() { int n, number, o, number1=0, temp, i; printf("Enter your number:"); scanf("%d", &number); number1=number; temp=number; int result=0; i=0; while(number!=0) { number=number/10; i++; } while(number1!=0) { n=number1%10; result=result+pow(n,i); number1=number1/10; } if(temp==result) { printf("Your number is armstrong"); } else { printf("Your number is not armstrong bro"); }
For single digit values it is wrong program. Because it is showing 2,3,4... As Armstrong numbers but they are not. The definition is sum of the cubes of the each digit of the number is equal to the number itself
@Rajesh Sanagala We have initialize mul=1 as 1 is the only number after multiply with we get the same number and moreover we have to use the mul variable each time the loop run and If we don't initialize it with 1 then we will be multiplying the new digit with the previous result Like Nu=153 mul=mul*rem // (1*3=3**3=9) For next digit we need to have the mul variable as 1 only mul=mul*rem //(1*5=5) Otherwise we will get 9*5 but we have to add the digit power 3..
what is the need of count the digits of number? we can do it without counting the digits. # include int main() { int num,n,r,sum=0; printf("Enter Your Number: "); scanf("%d",&num); n=num; while(n>0) { r=n%10; sum=sum+r*r*r; n=n/10; } if(sum==num) printf("Yes. NO is Armstrong "); else printf("NO. No is not Armstrong "); }
I tried before seeing the answer int n; int main() { printf("Armstrong number? "); scanf("%d", &n); int counter = 0; int holder = n; while(n != 0) { n = n / 10; counter++; } n = holder; int i = counter; int sum = 0; int rem; int mul = 1; while(holder != 0) { rem = holder % 10; //most left number holder = holder / 10; while(counter > 0) { mul = mul*rem; counter--; } sum = sum + mul; counter = i; mul = 1; } if(sum == n) { printf("Yes!", sum); } else printf("No!", sum); }
#include #include int main() { int num,temp,digits,count=0; printf("enter a number: "); scanf("%d",&num); temp=digits=num; //counting the number of digits while(digits!=0) { digits=digits/10; count++; } int r; int s=0; //multiplying each digits "count" times and adding them while(num!=0) { r=num%10; r=pow(r,count);//multiplying using pow function s=s+r;//adding here num=num/10; } if(temp==s) { printf("armstrong number"); } else { printf("not an armstrong number"); } return 0; }
I think actual code is this. Here cnt = count will be inside of second while loop. #include int main() { //Find order of the number int number, count=0, q, cnt, rem, mul=1, result=0; printf("Please enter a number: "); scanf("%d", &number); printf("Checking Number: %d ", number); q = number; while(q!=0) { q = q/10; count++; } printf("Order of that number: %d ", count); q = number; while(q!=0) { cnt = count; rem = q%10; while(cnt !=0) { mul = mul*rem; cnt--; } result = result + mul; q = q/10; mul = 1; } printf("After multiplication with Order of that number: %d ", result); //Is Armstrong Number or Not if(result == number) printf("%d is an Armstrong Number ", number); else printf("%d is not an Armstrong Number ", number); }
#include int main() { int num=0; int x; printf("Enter the number: "); scanf("%d",&x); int num0=x; int a=x; int i=0; while(a>0) { a=a/10; i++; } while(x!=0) { num=num+pow(x%10,i); x=x/10; } if(num0==num) printf("The entered number is a armstrong"); else printf("The entered number is NOT a armstrong"); return 0; }
😀 before i saw your result i tried to find the solution by myself and here are it's . long but worked 😂😂 #include #include int power (int *y,int *n); int main () { unsigned int b,y,n,input,result =0; printf("Enter a number:"); scanf("%i",&b); input = b; n = log10(b)+1; for(int i = 1; i
this my code >> # include int len(int n) { int size; size = 0; while (n > 0) { size++; n /= 10; } return (size); } void is_armstrong(int n) { int size ; int nb = n; int reminder = 0; int check; int total = 0; while (nb > 0) { reminder = nb % 10; check = 1; size = len(n); while (size > 0) { check *= reminder; size--; } total += check; nb /= 10; } if (total == n) printf("is armstrong"); else printf("its not armstrong"); } int main() { is_armstrong(3); }
int main() { int number,rev,arm,result=0; int q; int digit=0; printf("Enter a number "); scanf("%d",&number); rev=number; arm=number; while(rev!=0) { digit++; rev=rev/10; } while(arm!=0) { int mult=1; q=arm%10; for(int i=1;i
It's so funny that this is the second exercise in Exercism for C right after Hello World lol. As a newbie, I thought there would be more a gradual escalation
count=n; //count n again, for next digit q/=10; //to omit the used digit :) omitting starts from ones place :) //mul=1; like it if u like :) so that others can see :)
//code for Armstrong Number right or wrong?sir #include #include int main() { int n , q , rem; int result = 0; printf("enter value of n:"); scanf("%d" , &n); q = n; while(q != 0){ rem = q % 10; result += pow(rem , 3); q = q / 10; } printf("%d " , result); if(result==n) printf("its a armstrong number!"); else printf("its not a armstrong number!"); return 0; }
Sir Why this code shows every number , an Armstrong number...?? there is some mistake in this code. (Sir can you fix it). #include #include int main() { int number,count=0,result=0,mul=1,cnt,rem; printf("Enter a number "); scanf("% d ",&number); int q = number; while(q !=0) { q = q/10; count++; } cnt = count; q = number; while(q!=0) { rem=q%10; while(cnt!=0) { mul= mul*rem; cnt--; } result = result + mul; cnt=count; q = q/10; mul = 1; } if(result == number) { printf("%d is an Armstrong no", number); } else { printf("%d is not an Armstrong no", number); } return 0; }
Which cnt for first one we have given the value of count to cnt variable so that changing anything in cnt doesn't affect count For second one Count is = 3 we initialised cnt to 3 by equating it to count as while loop mein cnt ki value humne 0 kardi thi last process mein
Guys i tried it myself also this lecture but that was not running . I had done a blunder by putting ; after while😂 but i am happy . now my program is running successfully.
!! Well if you include math.h (By which you can include pow) in the header files it can be easier !! :) #include #include int main(){ int num,temp,rem,result=0,count=0; printf("enter the number that you want to check "); scanf("%d",&num); temp=num; while(temp!=0){ temp=temp/10; count++; } temp=num; while(num>0){ rem=num%10; result=result+pow(rem,count); num=num/10; } if(temp==result) { printf("It is an armstrong number "); } else{ printf("It is not an armstrong number "); } return 0; }#include #include int main(){ int num,temp,rem,result=0,count=0; printf("enter the number that you want to check "); scanf("%d",&num); temp=num; while(temp!=0){ temp=temp/10; count++; } temp=num; while(num>0){ rem=num%10; result=result+pow(rem,count); num=num/10; } if(temp==result) { printf("It is an armstrong number "); } else{ printf("It is not an armstrong number "); } return 0; }
struggled to understand ur explaination using example, but successfully understood. thanks alot for ur contribution in sharing huge amouts of knowledge.
Can u tell me which part of code is wrong? I am getting the wrong output. #include #include #include int main() {int n,x=0,count=0,temp,temp1,rem,sum=0; printf("enter no."); scanf("%d",&n); temp=n; while(temp!=0) { temp=temp/10; count++; } temp1=n; x=count; while(x!=0) { rem=temp1%10; sum=sum+pow(rem,count); x--; } if(sum==n) printf("yes"); else printf("NO"); return 0; }
Because we've use ‘count' to count the digit of the number at first step and in second step we're using cnt instead of count as count variable was used earlier
06:54 Can someone tell me how would that while(cnt!=0) is gonna work bec the mul variable inside it is initialised below that loop, so isn't it that basically the mul inside the while(cnt!=0) is still initialised?
as we got the value of cnt from previous function and before entering the while loop the condition will be checked and if the condition will be true so code will be executed. and mul is initialised in main function recheck the full snippet and ya we again initialised mul as 1 so that we can again multiply the left digits with their respective order . thanku and watch this video again.
Suppose n is the number you want to count the digits then let q= n then follow the code given, we don't want the number n which is given as input to change so we are using another variable q. Is my explanation clear ?
(is my code correct for armstrong number?), #include #define p(a) printf("%d", a) #define scan(x) scanf("%d", &x) int main() { int a, b,c, save; scan(a); save = a; while(a>0){ b = a%10; c = c + (b*b*b); a = a/10; } //p(c); if(c == save){ p(1); } else{ p(2); } return 0; } expand_more
I was always looking for explanations behind such programs, a source from where I can learn the concepts, and I finally found this channel, I really needed this as I use to feel alienated in computer science, but this already made me excited about everything related to computer science again, can't thank enough!! ♥️🙏🏾
i know Im randomly asking but does someone know a trick to get back into an Instagram account??
I somehow lost my account password. I appreciate any tips you can give me!
alternative
int main() {
int num,count=0,sum=0,rem,result=0;
printf("enter a number
");
scanf("%d",&num);
int kela=num;
int lela=num;
while(num!=0){
num=num/10;
count++;
}
while(lela!=0){
rem=lela%10;
sum=sum+pow(rem,count);
lela/=10;
}
if(kela==sum) {
printf("its a armstrong number i.e %d",kela);
}
else{
printf("its not a armstrong number coz %d",sum);
}
return 0;
}
Are u btech cse student?
its amazing and very interesting to learn c programming from neso academy
Wow.... Thank you so so much.. Concept is clearly explained.. Voice is very soothing.. Lots of love ❤❤❤❤❤❤.. Thank you so much 😇😇
i love you explanation each step i learn programming for while many tutorial go fast without clear explanation like this make me exhaust to figure it out what going on in every line of code, now i'm back to go to basic and found this channel love it
Using
digits = (int) log(10) + 1 //digits is count in this case
that was really a presentation which cleared my concept on what Armstrong number is and what algorithm should be used to find a number whether is it armstrong number or not
Sir.. In 2nd step.. We hv to initialize q value again. q=number, because.. In 1st step.. q becoming 0 ..
It can be done in simpler way
t=n;
sum=0;
while(n>0)
{
r=n%10;
sum=sum+(r*r*r);
n=n/10;
}
if (t==n)
{
printf("Armstrong Number
");
}
this code only works with a 3 digits number
what if we take number in string and then use "strlen" to find number of input ?
Good explanation ❤
Thank u sir for great explanation 🎉🎉❤
sir shouldn't we initialize mul to 1 initially
If it is four digits number means how to write the program .... I struggle to write that program.... Please tell me sir ....
Lovee u man....Really u are a legend...If i meet u in person then ill hug u and give u some gifts.. :-)
Thank you so much sir 💥💥
I would code it like this
int main()
{
int n, number, o, number1=0, temp, i;
printf("Enter your number:");
scanf("%d", &number);
number1=number;
temp=number;
int result=0;
i=0;
while(number!=0)
{
number=number/10;
i++;
}
while(number1!=0)
{
n=number1%10;
result=result+pow(n,i);
number1=number1/10;
}
if(temp==result)
{
printf("Your number is armstrong");
}
else
{
printf("Your number is not armstrong bro");
}
sir can you tell me plzz that how we are getting numbers in mul variable
I have a doubt. Is 2-9 an Armstrong number? because this code says they are?
thank you so much
7:33 i caught you stanger >:)
For single digit values it is wrong program. Because it is showing 2,3,4... As Armstrong numbers but they are not.
The definition is sum of the cubes of the each digit of the number is equal to the number itself
thankyou sir 🙏
Can anyone plz explain me that why he initialized mul=1 in the last of second step
@Rajesh Sanagala
We have initialize mul=1 as 1 is the only number after multiply with we get the same number and moreover we have to use the mul variable each time the loop run and If we don't initialize it with 1 then we will be multiplying the new digit with the previous result
Like
Nu=153
mul=mul*rem // (1*3=3**3=9)
For next digit we need to have the mul variable as 1 only
mul=mul*rem //(1*5=5)
Otherwise we will get 9*5 but we have to add the digit power 3..
N my clg they don't teach concepts never they just copy aloud programs ND hence we have to cram them up
Tnks sir g
what is the need of count the digits of number? we can do it without counting the digits.
# include
int main()
{
int num,n,r,sum=0;
printf("Enter Your Number: ");
scanf("%d",&num);
n=num;
while(n>0)
{
r=n%10;
sum=sum+r*r*r;
n=n/10;
}
if(sum==num)
printf("Yes. NO is Armstrong
");
else
printf("NO. No is not Armstrong
");
}
Why i am getting 1278 as answer for any input 🤬🤬L
I tried before seeing the answer
int n;
int main() {
printf("Armstrong number? ");
scanf("%d", &n);
int counter = 0;
int holder = n;
while(n != 0) {
n = n / 10;
counter++;
}
n = holder;
int i = counter;
int sum = 0;
int rem;
int mul = 1;
while(holder != 0) {
rem = holder % 10; //most left number
holder = holder / 10;
while(counter > 0) {
mul = mul*rem;
counter--;
}
sum = sum + mul;
counter = i;
mul = 1;
}
if(sum == n) {
printf("Yes!", sum);
}
else
printf("No!", sum);
}
#include
#include
int main()
{
int num,temp,digits,count=0;
printf("enter a number: ");
scanf("%d",&num);
temp=digits=num;
//counting the number of digits
while(digits!=0)
{
digits=digits/10;
count++;
}
int r;
int s=0;
//multiplying each digits "count" times and adding them
while(num!=0)
{
r=num%10;
r=pow(r,count);//multiplying using pow function
s=s+r;//adding here
num=num/10;
}
if(temp==s)
{
printf("armstrong number");
}
else
{
printf("not an armstrong number");
}
return 0;
}
Why cnt=3?
Because if its 3. it reduces to 0 and it can not be looped again for the remaining values we need to multiply
so after it becomes 0 you basically need to set it to 3 again, so it continues that process of Looping. I hope I could help
Is 2,3,4 armstrong number???
Numbers from 0-9 are Armstrong numbers
❤️❤️
3%10=0.3
remainder is 0
@@saikiran5345 a small number % a bigger one is the smal one ==> so 3%10 is 3 bro (modulo property)
As clearly the left-hand side number is left than right hand side number....the result would be left hand side number ...without doubt
no its remainder so it would be 3
I think actual code is this. Here cnt = count will be inside of second while loop. #include
int main()
{
//Find order of the number
int number, count=0, q, cnt, rem, mul=1, result=0;
printf("Please enter a number: ");
scanf("%d", &number);
printf("Checking Number: %d
", number);
q = number;
while(q!=0)
{
q = q/10;
count++;
}
printf("Order of that number: %d
", count);
q = number;
while(q!=0)
{
cnt = count;
rem = q%10;
while(cnt !=0)
{
mul = mul*rem;
cnt--;
}
result = result + mul;
q = q/10;
mul = 1;
}
printf("After multiplication with Order of that number: %d
", result);
//Is Armstrong Number or Not
if(result == number)
printf("%d is an Armstrong Number
", number);
else
printf("%d is not an Armstrong Number
", number);
}
#include
int main()
{ int num=0;
int x;
printf("Enter the number: ");
scanf("%d",&x);
int num0=x;
int a=x;
int i=0;
while(a>0)
{
a=a/10;
i++;
}
while(x!=0)
{
num=num+pow(x%10,i);
x=x/10;
}
if(num0==num)
printf("The entered number is a armstrong");
else
printf("The entered number is NOT a armstrong");
return 0;
}
😀 before i saw your result i tried to find the solution by myself and here are it's . long but worked 😂😂
#include
#include
int power (int *y,int *n);
int main ()
{
unsigned int b,y,n,input,result =0;
printf("Enter a number:");
scanf("%i",&b);
input = b;
n = log10(b)+1;
for(int i = 1; i
this my code >> # include
int len(int n)
{
int size;
size = 0;
while (n > 0)
{
size++;
n /= 10;
}
return (size);
}
void is_armstrong(int n)
{
int size ;
int nb = n;
int reminder = 0;
int check;
int total = 0;
while (nb > 0)
{
reminder = nb % 10;
check = 1;
size = len(n);
while (size > 0)
{
check *= reminder;
size--;
}
total += check;
nb /= 10;
}
if (total == n)
printf("is armstrong");
else
printf("its not armstrong");
}
int main()
{
is_armstrong(3);
}
i cheated and i used the math library so i can use pow instead of while
Thanks a lot ❤️❤️❤️because of you I am in love with c❤️
Yar how can you love it...I'm unable to understand c language..can you help me
@@PoojaMishra-xc6lp start from the first lecture of c programming of neso,eventually u will understand everything
So true in mine case
@@ramarajujaya2480 yes it is true in my case also ❤️❤️
Neso is the best ❤️❤️
sir how to store this complex logic inside my head for ever? i understood it but im sure i will forget it after a day or two
you can determinate the length of number with length (type int ) length=log10(number)+1
int main()
{
int number,rev,arm,result=0;
int q;
int digit=0;
printf("Enter a number
");
scanf("%d",&number);
rev=number;
arm=number;
while(rev!=0)
{
digit++;
rev=rev/10;
}
while(arm!=0)
{
int mult=1;
q=arm%10;
for(int i=1;i
How simple u made that program !!!
Aapki wajah se samjha muze ye...🤵♀️
thank you so much for this video
helped me clearing the concept of Armstrong no...
thanks
Inside the nested WHILE loop, can we use a FOR loop instead of the WHILE loop????
Like for example:
for(cnt=count;cnt>0;cnt--)
{
mul=mul*rem;
}
sir should've shown the whole code too before/after the snippets 🥲🥲
Flew over my head
It's so funny that this is the second exercise in Exercism for C right after Hello World lol. As a newbie, I thought there would be more a gradual escalation
can't we directly use pow function instead of multiplying?
You can use that but that will not actually give u the feeling of the logic of making a number raise to some power
I'm soooooooo happy rn I've understood very well
Tsym
Baawli gaand😂
Thank you so much sir...now next one should be about Strong number...
Easier Way
int main()
{
int InitialNum, FinalNum = 0, tempcontianer = 0, count = 0;
cin >> InitialNum;
count = log10(InitialNum) + 1;
tempcontianer = InitialNum;
while (tempcontianer != 0)
{
FinalNum += pow(tempcontianer%10, count);
tempcontianer /= 10;
}
cout
what happen if user give four digit no.?
while loop will run four time mean in place of cube we get four time multiple?
Baad mein kyu cnt or mul ko intialize kiye h koi baatayega hmko
your explanation is crystal clear. But, what is the use of again assigning the cnt=count &mul=1 in the while loop??
see the value changed inside while is permanent...not temprary
@@its_SR07 if its permanent then again mul value is changed into 343 right then is that 3 want to multiple with 343?
count=n; //count n again, for next digit
q/=10; //to omit the used digit :) omitting starts from ones place :)
//mul=1;
like it if u like :) so that others can see :)
//code for Armstrong Number right or wrong?sir
#include
#include
int main()
{
int n , q , rem;
int result = 0;
printf("enter value of n:");
scanf("%d" , &n);
q = n;
while(q != 0){
rem = q % 10;
result += pow(rem , 3);
q = q / 10;
}
printf("%d
" , result);
if(result==n)
printf("its a armstrong number!");
else
printf("its not a armstrong number!");
return 0;
}
header lines are obvious in every programs. i guess thats why he didnt mention
Sir Why this code shows every number , an Armstrong number...?? there is some mistake in this code. (Sir can you fix it).
#include
#include
int main()
{
int number,count=0,result=0,mul=1,cnt,rem;
printf("Enter a number
");
scanf("% d ",&number);
int q = number;
while(q !=0)
{
q = q/10;
count++;
}
cnt = count;
q = number;
while(q!=0)
{
rem=q%10;
while(cnt!=0)
{
mul= mul*rem;
cnt--;
}
result = result + mul;
cnt=count;
q = q/10;
mul = 1;
}
if(result == number)
{
printf("%d is an Armstrong no", number);
}
else
{
printf("%d is not an Armstrong no", number);
}
return 0;
}
@Darshan patil did you find out ?
In scanf remove that gap between % and d then it's working. I checked.
@@paraprannoy3732 yeah thank you
You are awesome man! 😵
Sir upload videos on arrays , strings, pointers
👌👌👌👌👌👌👌👌👌👌
Can also ** operator for exponentiation of each digit to its order and then adding simultaneously
I didn't understood why did you wrote this :
cnt = count
and q = number
Which cnt for first one we have given the value of count to cnt variable so that changing anything in cnt doesn't affect count
For second one Count is = 3 we initialised cnt to 3 by equating it to count as while loop mein cnt ki value humne 0 kardi thi last process mein
Same with q while counting the no. Of digits for our given no. Humne q ki value 0 kardi thi we initialised it back to our original no.
It's better to use pow( ) under math.h💁
But, you are not allowed in all cases
Guys i tried it myself also this lecture but that was not running . I had done a blunder by putting ; after while😂 but i am happy . now my program is running successfully.
What if you have a number with more digists then 3? can u use pow somehow in the code?
yep
!! Well if you include math.h (By which you can include pow) in the header files it can be easier !! :)
#include
#include
int main(){
int num,temp,rem,result=0,count=0;
printf("enter the number that you want to check
");
scanf("%d",&num);
temp=num;
while(temp!=0){
temp=temp/10;
count++;
}
temp=num;
while(num>0){
rem=num%10;
result=result+pow(rem,count);
num=num/10;
}
if(temp==result)
{
printf("It is an armstrong number
");
}
else{
printf("It is not an armstrong number
");
}
return 0;
}#include
#include
int main(){
int num,temp,rem,result=0,count=0;
printf("enter the number that you want to check
");
scanf("%d",&num);
temp=num;
while(temp!=0){
temp=temp/10;
count++;
}
temp=num;
while(num>0){
rem=num%10;
result=result+pow(rem,count);
num=num/10;
}
if(temp==result)
{
printf("It is an armstrong number
");
}
else{
printf("It is not an armstrong number
");
}
return 0;
}
Mine output is not coming
It is showing else statement
Sir plz make video on automorphic number
And I can easily understand your explanation
I don’t understand the use of cnt=count here
struggled to understand ur explaination using example, but successfully understood. thanks alot for ur contribution in sharing huge amouts of knowledge.
thanks sirbut can u make videsos onn microprocessors arm7
Canu pls share Ur code
Awesome explanation sir
After line 30 , cnt should again initialize to count
Sir also make videos on java for beginners
Nice video sir
Thanks alot
Thank u
thank you sir, you explained very well
Can u tell me which part of code is wrong? I am getting the wrong output.
#include
#include
#include
int main()
{int n,x=0,count=0,temp,temp1,rem,sum=0;
printf("enter no.");
scanf("%d",&n);
temp=n;
while(temp!=0)
{
temp=temp/10;
count++;
}
temp1=n;
x=count;
while(x!=0)
{
rem=temp1%10;
sum=sum+pow(rem,count);
x--;
}
if(sum==n)
printf("yes");
else
printf("NO");
return 0;
}
instead of x-- use temp1=temp1/10; and in the condition in while use : temp1!=0.
@@aakashmudigonda3375 thank you
Very nice video sir but sir jb 37/10=30
Count =7 hoga
30/10=3
Count=0
3/10=0
Count=3
We are storing the value of 37/10 in 'q' not in count and in next step we are simply incrementing the count variable to find the order of number.
37/10 = 3
Why we are using cnt =count ?
We can declare int count =0
Because we've use ‘count' to count the digit of the number at first step and in second step we're using cnt instead of count as count variable was used earlier
3/10 =0?
Yes if the variable in which you store it is integer
06:54
Can someone tell me how would that while(cnt!=0) is gonna work bec the mul variable inside it is initialised below that loop, so isn't it that basically the mul inside the while(cnt!=0) is still initialised?
as we got the value of cnt from previous function and before entering the while loop the condition will be checked and if the condition will be true so code will be executed. and mul is initialised in main function recheck the full snippet and ya we again initialised mul as 1 so that we can again multiply the left digits with their respective order . thanku and watch this video again.
@@harmankaur3938 Thanks for help and info mate!
Appreciated 👏
Can anyone tell me , why we have to initialize q separately ?
Suppose n is the number you want to count the digits then let q= n then follow the code given, we don't want the number n which is given as input to change so we are using another variable q. Is my explanation clear ?
Sir i am new at programming pls suggest me do we have to memorize the codes or from time we'll have it practiced
pratice makes a man prefect you havent head that. dont use your memory here. experience is the mother of all wisdom
this is my code
#include
#include
int main()
{
int num,result=0,rem,p,order=0;
printf("Enter Num:
");
scanf("%d",&num);
p=num;
int q=num;
while(q!=0)
{
order++;
q=q/10;
}
while(p>0)
{
rem=p%10;
result =result+pow(rem,order);
p=p/10;
}
if(num==result)
{
printf("ArmStrong");
}else{
printf("LegStrong");
}
return 0;
}
(is my code correct for armstrong number?), #include
#define p(a) printf("%d", a)
#define scan(x) scanf("%d", &x)
int main()
{
int a, b,c, save;
scan(a);
save = a;
while(a>0){
b = a%10;
c = c + (b*b*b);
a = a/10;
}
//p(c);
if(c == save){
p(1);
}
else{
p(2);
}
return 0;
}
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