euler21

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.

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#include <stdio.h>

unsigned d(int num)
{
    int i, sum = 0;

    for (i=1; i < num/2+1; i++) {
        if (num%i == 0)
            sum += i;
    }

    return sum;
}

#define N 10000
int main(int argc, const char *argv[])
{
    int i, sum = 0;
    unsigned d1, d2;

    for (i=2; i<N; i++) {
        d1 = d(i);
        d2 = d(d1);
        if (d2 == i && i != d1 )
            sum += i;
    }

    printf("sum: %d\n", sum);

    return 0;
}

效率不高,还有很大的优化空间。