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A PROGRAM TO FIND SUM & AVERAGE OF EVEN & ODD INTEGERS BY IDENTIFYING
anubala331
August 2014
*A PROGRAM TO FIND SUM & AVERAGE OF EVEN & ODD INTEGERS BY IDENTIFYING*/
#include<stdio.h>
int main()
{
int counto=0;
int ger,count=0,sume=0,sumo=0,avge=0,avgo=0;
do
{
printf("enter the integer other then 0 to count");
scanf("%d",&ger);
if((ger%2)==0)/
/if integer is even
{
count++;/
/COUNT IS ++MENTED TO COUNT EVEN VALUE
printf("even intgers entered untill now=%d\n",count);
sume=sume+ger;/
/EVEN INTEGER IS ADDED TO SUME
avge=sume/count;
//SUM IS DIVIDED BY NO. OF TERMS ADDED TO GET AVERAGE VALUE
printf("avg value of even numbrs.=%d\n",avge);
}
else if(ger%2 != 0)
//if integer is odd
{
counto=counto+1;/
/EVERYTIME ODD INTEGER ADDED IS COUNTED
printf("odd intgers entered untill now=%d\n",counto);
sumo=sumo+ger;//
ODD INTEGER IS ADDED TO SUMO EVERYTIME //
avgo=sumo/counto;
//SUM OF ODD INGER/NO. OF ODD TERMS ..TO GET AVERAGE VALUE.//
printf("avg value for odd nums .=%d\n",avgo);
}
}while(ger != 0);/
/IT WILL FIND AVERAGE VALUE OF BOTH EVEN & ODD INTEGER UNTILL 0 VALUE OF INTEGER IS ENTERED//
return 0;
}
anubala331
August 2014
[root@localhost]/home/c-asignments/chdr# ./a.out
enter the integer other then 0 to count4
even intgers entered untill now=1
avg value of even numbrs.=4
enter the integer other then 0 to count5
odd intgers entered untill now=1
avg value for odd nums .=5
enter the integer other then 0 to count6
even intgers entered untill now=2
avg value of even numbrs.=5
enter the integer other then 0 to count7
odd intgers entered untill now=2
avg value for odd nums .=6
enter the integer other then 0 to count534
even intgers entered untill now=3
avg value of even numbrs.=181
enter the integer other then 0 to count3
odd intgers entered untill now=3
avg value for odd nums .=5
enter the integer other then 0 to count
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