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.
#include <iostream>
#include <cmath>
#include <stdio.h>
#include <fstream>
#include <string>
#include <time.h>
usingnamespace 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;i<size && abundant[i]<=x;i++) {
for (int j=0;j<size && abundant[j]<=x;j++) {
if (abundant[i]+abundant[j]==x) {
returntrue;
}
}
}
returnfalse;
}
int main() {
clock_t start, end;
start = clock();
int sum=0;
int count=0;
int count2=0;
for (int i=0;i<28124;i++) {
if (d(i)>i) {
abundant[count]=i;
count++;
count2++;
}
}
size=count2;
cout<<size<<endl;
for (int i=0;i<28124;i++) {
if (!summable(i)) {
sum=sum+i;
}
}
cout<<sum<<endl;
end= clock();
printf("\nTook %f seconds\n", (double)(end-start)/CLOCKS_PER_SEC);
system("pause");
return 0;
}
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 ?