UVA Problem#382

Question:

http://uva.onlinejudge.org/external/3/382.html 

Solution: 

void perfection(int n){

int i , sum=0, m=n/2; 

for(i=1; i<=m; i++){ 

if(n%i==0)

sum += i; 

}

if(n>sum){ 

printf("DEFICIENT\n"); 

}

else if(n<sum){ 

printf("ABUNDANT\n"); 

}

else if(n == sum){

printf("PERFECT\n"); 

} 

}

  int main(){

....

.... 

}