 |
| Looping Program practice |
P3. Program to print the series of natural no. from 100 to 1.
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i;
printf( “ series
of natural no. from 100 to 1 \n “) ;
for ( i= 100 ; i
>= 1 ; i -- )
{
printf( “ %d “, i
);
}
getch();
}
P4. Program to print the series of even natural no. up to 100 without if condition .
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i ;
printf( “ series of
even no. upto 100 are : “);
for ( i =2 ; i
<= 100 ; i = i + 2)
{
printf( “ %d “, i
);
}
getch();
}
P5. If we have to write the above program by if condition we have :-
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i ;
printf( “ series of
even no. upto 100 are : “);
for ( i =1 ; i
<= 100 ; i = i ++ )
{
if ( i % 2 == 0 )
printf( “ %d\n “, i
);
}
getch();
}
P6. Program to print the series of odd natural no. from 100 to 50 using if condition .
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i;
printf( “ series of
odd natural no from 100 to 50 \n “ );
for ( i =100 ; i
>= 50 ; i -- )
{
if ( i % 2 == 1
)
printf( “ %d \n
“, i ) ;
}
getch();
}
P7. Program to print the series of those natural nos. that are divisible by 2 & 4 upto 100.
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i;
printf( “ natural
no divisible by 2 & 4 \n “ );
for ( i =1 ; i
<= 100 ; i ++ )
{
if ( i % 2 == 0
&& i % 4 == 0 )
printf( “ %d \n
“, i ) ;
}
getch();
}
P8. Program to find sum of even and odd natural no. upto hundred.
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i , esum = 0 ,
osum = 0;
for ( i =1; i <=
100 ; i ++ )
{
if ( i % 2 == 0
)
esum = esum + i
;
else
osum = osum +i
;
}
printf( “ sum of
even upto 100 = %d \n“ , esum );
printf( “ sum of
odd upto 100 = %d \n “, osum );
getch();
}
P9. Program to input any no. after that print all factors.
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i , n ;
printf( “ enter any
no. “ );
scanf ( “ %d “,
&n );
printf( “ all
factors are \n “, );
for ( i =1 ; i
<= n ; i ++ )
{
if ( n % i == 0
)
printf( “ %d \n
“, i ) ;
}
getch();
}
P10. Program to input any no. after that find sum of all factors.
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i , n , fsum =
0 ;
printf( “ enter any
no. “ );
scanf ( “ %d “,
&n );
for ( i =1 ; i
<= n ; i ++ )
{
if ( n % i == 0
)
fsum = fsum + i
;
}
printf( “ sum of all
factors = %d “ , fsum );
getch();
}
P11. Program to input any no. after that check no. is perfect or not .
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i , n , sum =0
;
printf( “ enter any
no. “ );
scanf ( “ %d “,
&n );
for ( i =1 ; i
<= n ; i ++ )
{
if ( n % i == 0
)
sum = sum + i ;
}
if ( sum == n )
printf( “ perfect
no. “ );
else
printf( “ not
perfect no. “ );
getch();
}
P12. Program to input any no. after that check no. is prime or not .
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i , n , count =
0 ;
printf( “ enter any
no. “ );
scanf ( “ %d “,
&n );
printf( “ all
factors are \n “, );
for ( i =1 ; i
<= n ; i ++ )
{
if ( n % i == 0
)
count = count +
1 ;
}
if ( count == 2 )
printf( “ Prime no.
“);
else
printf( “ not prime
no. “ );
getch();
}
P13. Program to input two nos. after that check no. is amicable or not .
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int a , b , i ,
sum1 = 0 , sum2 = 0 ;
printf( “ enter any
two no. “ );
scanf ( “ %d %d “,
&a, &b );
for ( i =1 ; i
<= a ; i ++ )
{
if ( a % i == 0
)
sum1 = sum1 + i
;
}
for ( i = 1 ; i <
b ; i ++ )
{
if ( b % i == 0 )
sum2 = sum2 + i ;
}
if ( sum1 = b
&& sum2 = a )
printf( “ Amicable
no. “ );
else
printf( “ not
amicable no. “ ) ;
getch();
}
No comments:
Post a Comment