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Write the program form Armstrong no in c++?
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Answer / vinay k bharadwaj
dear frd your ans is wrong.
#include<stdio.h>
#include<conio.h>
void main()
{
int sum=0,n,m,x;
clrscr();
scanf("%d",&n);
m=n;
while(n>0)
{
x=n%10;
n=n/10;
sum=sum+(x*x*x);
}
if(sum==m)
printf("number is armstrong");
else
printf("number is not armstrong");
getch();
}
| Is This Answer Correct ? | 87 Yes | 47 No |
Answer / saravanan
#include<stdio.h>
#include<conio.h>
void main()
{
int sum=0,n,m;
clrscr();
scanf("%d",&n);
m=n;
while(n>0)
{
m=n%10;
sum=sum+(m*m*m);
n=n/10;
}
if(m==sum)
{
printf("the given number is amstrong");
else
printf("the given number is not amstrong");
}
getch();
}
| Is This Answer Correct ? | 97 Yes | 70 No |
Answer / roshan patel
#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int x,y,d;
cout<<"\nEnter the no : ";
cin>>x;
y=x;
int sum=0;
while(y>0)
{
d=y%10;
y/=10;
sum=sum+d*d*d;
}
if(x==sum)
cout<<"\n "<<x<<" is Armstrong ";
else
cout<<"\n"<<x<<" is not Armstrong";
getch();
}
| Is This Answer Correct ? | 17 Yes | 5 No |
Answer / rahul puri
#include<stdio.h>
#include<iostream.h>
#include<conio.h>
void main()
{
int sum=0,n,m;
clrscr();
cin>>n;
m=n;
while(n>0)
{
m=n%10;
sum=sum+(m*m*m);
n=n/10;
}
if(m==sum)
{
cout<<"the given number is amstrong";
}
else
{
cout<<"the given number is not amstrong";
}
getch();
}
| Is This Answer Correct ? | 48 Yes | 41 No |
Answer / astitva srivastava
#include<iostream.h>
#include<conio.h>
void main()
{
int n,r,s=0;
cout<<"enter the no.";
cin>>n;
int b=n;
while(n>0)
{
r=n%10;
n=n/10;
s=s+(r*r*r);
}
if(b==s)
{
cout<<"the no. is armstrong"<<"\n";
}
else
{
cout<<"the no. is not a armstrong no.";
}
}
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / ashish ranjan
#include<iostream.h>
#include<conio.h>
void main()
{
int sum=0,n,m;
clrscr();
cin>>n;
m=n;
while(n>0)
{
m=n%10;
sum=sum+(m*m*m);
n=n/10;
}
if(m==sum)
{
cout<<"the given number is amstrong";
}
else
{
cout<<"the given number is not amstrong";
}
getch();
}
| Is This Answer Correct ? | 13 Yes | 10 No |
Answer / sarang deshpande
#include<iostream.h>
#include<conio.h>
void main()
{
int n,n1,sum,rem;
clrscr();
cout<<" \n Enter a num ";
cin>>n;
n1=n;
sum=0;
while(n>0)
{
rem=n%10;
sum=sum+(rem*rem*rem);
n=n/10;
}
if(n1==sum)
{
cout<<"\n The number is armstrong "<<n1;
}
else
{
cout<<"\n The number is not armstrong "<<n1;
}
getch();
}
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / tapojit roy
#include<stdio.h>
#include<conio.h>
void main()
{
int n,m,x,sum=0;
clrscr();
printf ("the armstrong no. from 0 to 500 are\n");
for (n=1;n<=500;n++){
m=n;
while (m>0)
{
x=m%10;
sum=sum+(x*x*x);
m=m/10;
}
if (sum==n)
{
printf("%d\n",n);
}
sum=0;
}
getch();
}
| Is This Answer Correct ? | 29 Yes | 28 No |
Answer / ramesh
#include<iostream.h>
#include<conio.h>
void main()
{
int sum=0,n,m;
clrscr();
cin>>n;
m=n;
while(n>0)
{
m=n%10;
sum=sum+(m*m*m);
n=n/10;
}
if(m==sum)
{
cout<<"the given number is amstrong";
}
else
{
cout<<"the given number is not amstrong";
}
getch();
}
| Is This Answer Correct ? | 4 Yes | 3 No |
Answer / hitesh
Armstrong Program in C++
Armstrong number is a number that is the sum of its own digits each raised to the power of the number of digits is equal to the number itself.
For example 153 is armstrong number, 132 is not prime number. <a href="http://www.tutorial4us.com/cpp-program/cpp-armstrong-program" rel="nofollow">Armstrong program in c++</a> is very simple and easy to write.
| Is This Answer Correct ? | 0 Yes | 0 No |
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