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.
计算28123以内不能表示成两个abundant数的和 的数的和。。。有点绕
#include <iostream>
#include <map>
using namespace std;
bool abundant(int a)
{
int res = 0;
for (int i = 1; i < a; i++)
{
if (a%i == 0)
res += i;
}
if (res>a)
return true;
else
return false;
}
int main()
{
map<int, int>num;
map<int, int>ab;
for (int i = 1; i <= 28123; i++)
{
if (abundant(i))
ab[i] = 1;
}
//cout << ab.size() << endl;
map<int, int>::iterator it1;
map<int, int>::iterator it2;
for (it1 = ab.begin(); it1 != ab.end(); it1++)
{
//cout << it1->first << endl;
for (it2 = ab.begin(); it2 != ab.end(); it2++)
{
num[it1->first + it2->first] = 1;
}
}
map<int, int>::iterator it3;
long long res = 0; //可以表示成两个abundant数和的数的和
for (it3 = num.begin(); it3 != num.end(); it3++)
{
if (it3->first <= 28123 && num[it3->first]!=0)
res += it3->first;
}
int total = (28123 + 1) * 28123 / 2;
cout << total-res << endl;
system("pause");
return 0;
}
速度有点慢但结果是正确的