Amicable numbers -- Javascript 实现

Posted 2020-09-16 zsychanpin

问题描写叙述：

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.

实现：

(function(){

var factor = function (n){
var arr = new Array();
for(var i = 1;i < n; i++)
{
if(n%i == 0 && arr.indexOf(i) == -1)
arr.push(i);
}
return arr;
}

var sumArr = function(arr){
var sum = 0 ;

for(var i = 0 ; i < arr.length; i++)
sum += arr[i];

return sum ;
}


for(var i = 2;i< 10000; i++)
{
var r1 = sumArr(factor(i));
var r2 = sumArr(factor(r1));

if(i == r2 && i != r1)
console.log("num1 : " + i + ", num2 : " + r1);
}

})();