Perfect numbers: Difference between revisions
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sum += i |
sum += i |
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return sum == n</python> |
return sum == n</python> |
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Using functional form |
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<python>def perf(n): |
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return n == sum( i for i in xrange(1,n) if n % i == 0)</python> |
Revision as of 19:44, 23 August 2008
You are encouraged to solve this task according to the task description, using any language you may know.
Write a function which says whether a number is perfect.
A number is perfect if the sum of its factors is equal to twice the number. An equivalent condition is that n is perfect if the sum of n's factors that are less than n is equal to n.
BASIC
<qbasic>FUNCTION perf(n) sum = 0 for i = 1 to n - 1 IF n MOD i = 0 THEN sum = sum + i END IF NEXT i IF sum = n THEN perf = 1 ELSE perf = 0 END IF END FUNCTION</qbasic>
Java
<java>public static boolean perf(int n){ int sum= 0; for(int i= 1;i < n;i++){ if(n % i == 0){ sum+= i; } } return sum == n; }</java> Or for arbitrary precision: <java>import java.math.BigInteger;
public static boolean perf(BigInteger n){ BigInteger sum= BigInteger.ZERO; for(BigInteger i= BigInteger.ONE; i.compareTo(n) < 0;i=i.add(BigInteger.ONE)){ if(n.mod(i).compareTo(BigInteger.ZERO) == 0){ sum= sum.add(i); } } return sum.compareTo(n) == 0; }</java>
Python
<python>def perf(n):
sum = 0 for i in xrange(1,n): if n % i == 0: sum += i return sum == n</python>
Using functional form <python>def perf(n):
return n == sum( i for i in xrange(1,n) if n % i == 0)</python>