### code not working

here is what my code should do:
 ``` A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number. A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n. As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit. Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers. ```

(or take a look at http://projecteuler.net/problem=23
here is my code:
 ``12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758`` ``````#include #include #include #include #include #include using namespace std; int abundant[28124]; int size; int d(int n) { int divisors=0; for (int i=1;i<=sqrt(double(n));i++) { if (n%i==0) { divisors+=i; divisors+=n/i; } } return divisors-n; } bool summable(int x){ for (int i=0;ii) { abundant[count]=i; count++; count2++; } } size=count2; cout<

the problem is that the answer i get from the program is wrong. with this program i get around 3900000 but the answer should be around 4100000
can someone tell me what i am doing wrong ?
Last edited on
d() is incorrect. If the number is a perfect square, its square root gets added twice into the result.
Last edited on
thank you. it is solved now.
for someone who's interested i added
 ``123`` ``````if (floor(sqrt(double(n)))==sqrt(double(n))) { divisors=divisors-sqrt(double(n)); }``````

before the return in d()
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