poj1528---(number theory)

Test instructions: A number in addition to all factors other than itself, the resulting number if less than itself, output deficient, otherwise output abundant, if equal, output perfect

The core of the problem: to ask for all the factors other than itself

int sum=1, I;             for (i=2; i<n;i++)           {               if(n%i==0)                   sum+ =i;           }

A number if divided by less than his number starting from 2, if n%i==0, stating that can n be divisible by I, I was one of his factors, sum+=i, again let I++,i to n-1 end

#include <stdio.h>#include<stdlib.h>intMain () {intN; printf ("Perfection output\n");  while(SCANF ("%d", &n)! =EOF) {        if(n==0) {printf ("END of output\n");  Break; }        Else if(n==1) printf ("%5d deficient\n", N); Else        {           intsum=1, I;  for(i=2; i<n;i++)           {               if(n%i==0) Sum+=i; }           if(sum<n) printf ("%5d deficient\n", N); Else if(sum==n) printf ("%5d perfect\n", N); Elseprintf"%5d abundant\n", N); }    }}

poj1528---(number theory)