在某网站看到一道js题，觉得有点意思
Some numbers have funny properties. For example:
89 --> 8¹ + 9² = 89 * 1
695 --> 6² + 9³ + 5⁴= 1390 = 695 * 2
46288 --> 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51
Given a positive integer n written as abcd... (a, b, c, d... being digits) and a positive integer p we want to find a positive integer k, if it exists, such as the sum of the digits of n taken to the successive powers of p is equal to k * n. In other words:
Is there an integer k such as : (a ^ p + b ^ (p+1) + c ^(p+2) + d ^ (p+3) + ...) = n * k
If it is the case we will return k, if not return -1.
Note: n, p will always be given as strictly positive integers.
digPow(89, 1) should return 1 since 8¹ + 9² = 89 = 89 * 1
digPow(92, 1) should return -1 since there is no k such as 9¹ + 2² equals 92 * k
digPow(695, 2) should return 2 since 6² + 9³ + 5⁴= 1390 = 695 * 2
digPow(46288, 3) should return 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 236
<!doctype html> <html lang="en"> <head> <meta charset="UTF-8"> <title>Document</title> </head> <body> <h1 onclick="digPow(46288,3)">Math</h1> <script> function digPow(n, p){ // ... var len = (n.toString()).length; var newString = []; var intNum = 0; for(var i=0;i<len;i++){ newString[i] = Math.pow(parseInt(nString[i]),p); p++; intNum += parseInt(newString[i]); } if(intNum%n ==0 ){ var multi = parseInt(intNum/n); return multi; }else{ return -1; } } </script> </body> </html>
有更好的方法欢迎推荐！