学步园

Amicable numbers

Problem 21

05 July 2002

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a  b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

 

Answer:

31626

 

void p21()
{
    long sum=0;
    for(int n=1; n<=10000; ++n)
    {
        int sum_1=1;
        for(int i=2;i<n;++i)
        {
            if(n%i==0)
                sum_1+=i;;
        }
        if(sum_1>=n || sum_1==1)
            continue;
        int sum_2=1;
        for(int i=2;i<sum_1;++i)
        {
            if(sum_1%i==0)
                sum_2+=i;
        }
        if(sum_2==n)
        {
            cout<<"n="<<n<<",sum_1="<<sum_1<<"sum_2="<<sum_2<<endl;
            sum=sum+n+sum_1;
        }
    }

    cout<<sum<<endl;
}