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.

我们直接编写函数d(n)，如果d(d(n))=n,并且n!=d(n)那么n就是符合要求的数

我们直接暴力搜索

#include

<

iostream

>

#include

<

cmath

>

using

namespace

std;

int

d(

int

tmp){

int

sum

=

0

;

int

i;

for

(i

=

1

; i

<

sqrt(tmp); i

++

){

if

(tmp

%

i

==

0

){

sum

=

sum

+

i

+

tmp

/

i;

}

}

if

(i

*

i

==

tmp){

sum

+=

i;

}

return

sum

-

tmp;

}

int

main(){

int

sum

=

0

;

for

(

int

i

=

1

; i

<

10000

; i

++

){

if

(i

==

d(d(i))

&&

i

!=

d(i)){

sum

+=

i;

}

}

cout

<<

sum

<<

endl;

return

0

;

}