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C Program for FIND EVEN & ODD NO.S IN THE ARRAY
#include stdio.h>
#include string.h>
#include conio.h>
void main()
{
int i=0,j=0, l_e[100],l_o[100];
int a[100],odd=0,even=0,sum_even=0,sum_odd=0;
float avg_e=0,avg_o=0;
for(i=0;i<100;i++)
a[i]=0;
l_e[i]=0;
l_o[i]=0;
}
i=0; clrscr();
printf("\nENTER THE LIMIT OF ELEMENTS: ");
scanf("%d",&j);
for(i=0;i { printf("\nENTER THE NO %d: ",i+1); scanf("%d",&a[i]); if(a[i]%2==0) { sum_even=sum_even+a[i]; even++; avg_e=sum_even/even; l_e[i]=a[i]; } else { sum_odd=sum_odd+a[i]; odd++; avg_o=sum_odd/odd; l_o[i]=a[i]; } } i=0; printf("\nTHE EVEN NO.S ARE %d",even); printf("\nTHE ODD NO.S ARE %d",odd); printf("\nTHE TOTAL OF EVEN NO.S ARE %d",sum_even); printf("\nTHE TOTAL OF ODD NO.S ARE %d",sum_odd); printf("\nTHE AVERAGE OF ALL EVEN NO.S IS %f",avg_e); printf("\nTHE AVERAGE OF ALL ODD NO.S IS %f",avg_o); getch(); } OUTPUT : ENTER THE NO 1 1 : 2 ENTER THE NO 1 2 : 6 ENTER THE NO 1 3 : 3 ENTER THE NO 2 1 : 4 ENTER THE NO 2 2 : 7 ENTER THE NO 2 3 : 8 ENTER THE NO 3 1 : 9 ENTER THE NO 3 2 : 1 ENTER THE NO 3 3 : 2 NO ARE IN ASCENDING ORDER 1 2 2 3 4 6 7 8 9
printf("\nENTER THE NO %d: ",i+1);
scanf("%d",&a[i]);
if(a[i]%2==0)
sum_even=sum_even+a[i];
even++;
avg_e=sum_even/even;
l_e[i]=a[i];
else
sum_odd=sum_odd+a[i];
odd++;
avg_o=sum_odd/odd;
l_o[i]=a[i];
} i=0;
printf("\nTHE EVEN NO.S ARE %d",even);
printf("\nTHE ODD NO.S ARE %d",odd);
printf("\nTHE TOTAL OF EVEN NO.S ARE %d",sum_even);
printf("\nTHE TOTAL OF ODD NO.S ARE %d",sum_odd);
printf("\nTHE AVERAGE OF ALL EVEN NO.S IS %f",avg_e);
printf("\nTHE AVERAGE OF ALL ODD NO.S IS %f",avg_o);
getch();
OUTPUT :
ENTER THE NO 1 1 : 2
ENTER THE NO 1 2 : 6
ENTER THE NO 1 3 : 3
ENTER THE NO 2 1 : 4
ENTER THE NO 2 2 : 7
ENTER THE NO 2 3 : 8
ENTER THE NO 3 1 : 9
ENTER THE NO 3 2 : 1
ENTER THE NO 3 3 : 2
NO ARE IN ASCENDING ORDER
1 2 2
3 4 6
7 8 9
O v e r l o a d i n g U n a r y Op e r a t o r s Us i n g F r i e n d F u n c t i o n cl a ss s i g n { i n t a ,
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