Take an integer n (n >= 0) and a digit d (0 <= d <= 9) as an integer. Square all numbers k (0 <= k <= n)between 0 and n. Count the numbers of digits d used in the writing of all the k**2. Call nb_dig (or nbDig or ...) the function taking n and d as parameters and returning this count.
#Examples:
n = 10, d = 1, the k*k are 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
We are using the digit 1 in 1, 16, 81, 100. The total count is then 4.
nb_dig(25, 1):
the numbers of interest are
1, 4, 9, 10, 11, 12, 13, 14, 19, 21 which squared are 1, 16, 81, 100, 121, 144, 169, 196, 361, 441
so there are 11 digits `1` for the squares of numbers between 0 and 25.
Note that 121 has twice the digit 1.
function nbDig(n, d) {
// your code
var arr=[];
var num=0;
for(var i=0;i<=n;i++){
arr.push(i*i);
}
arr=arr.join("").split("");
var arrs=arr.filter(function(x,index){
return x==d
})
return arrs.length;
}
测试：nbDig(5750, 0) 4700
最佳答案：
function nbDig(n, d) { var res=0; for (var g=0;g<=n;g++){ var square=(g*g+"").split(""); square.forEach((s)=>s==d?res++:null) }return res; }
原文：http://www.cnblogs.com/chengnanbei/p/7233156.html