Project Euler Problem 21 Amicable numbers

Amicable numbers

Problem 21

Let d(n) be defined as the sum of proper divisors ofn (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 andb 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.

C++：

#include <iostream>
#include <cstring>

using namespace std;

#define MAXN 10000

int sum[MAXN+1];

void maketable(int n)
{
    memset(sum, 0, sizeof(sum));
    sum[1] = 0;

    int i=2, j;
    while(i<=n) {
        sum[i]++;
        j = i + i;      /* j=ki, k>1 */
        while(j <= n) {
            sum[j] += i;
            j += i;
        }
        i++;
    }
}

int main()
{
    maketable(MAXN);

    int allsum = 0;
    for(int i=1; i<=MAXN; i++)
        if(sum[i] <= MAXN && i < sum[i] && sum[sum[i]] == i)
            allsum += i + sum[i];

    cout << allsum << endl;

    return 0;
}

参考链接：I00039 亲密数