All in a Single File

Trivial Solution

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import math

count = 0
n = 0

for x in xrange(100000):
    n += x + x + 1
    z = n
    for y in xrange(int(math.log10(n)) + 1):
        last = z / 10
        if z % 10 < last % 10:
            count -= 1
            break
        z = last
    count += 1

print count

Takes about 0.13 secs

49! Another square!

Here is my solution:

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d=0
for x in range(1,(10**5+1)):
    n=0
    s=str(x**2)
    for i in range(0,(len(s)-1)):
        if s[i]<=s[i+1]:
            n=n+1
    if n==(len(s)-1):
        d=d+1

print(d)

Trying to make one-liners is sometimes fun.

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sum((int("".join(list(sorted(str(n**2))))) == n**2) for n in range(1, 100000))
# prints 49

Explanation:

n**2 is our perfect square. str(n**2) turns it into a string , which is also an iterable (with elements being individual characters), so sorted(str(n**2)) sorts the characters in the element. list(sorted(...)) forces the result to a list (instead of a sorted type). "".join(list(...)) glues the characters together, so that the result is a string with the characters sorted, and int("".join(...)) turns it back into an int .

In short, int("".join(...)) sorts the digits of our perfect square. We compare it with the original perfect square, checking if they are equal; if they are, then the digits of the perfect square are listed in non-decreasing order, which is what we want to find.

Now, int(...) for n in range(1, 100000) iterates it for all perfect squares from $1^2$ to $99999^2$ . Every number that we seek will give a True , which counts as a 1 , and every number that we don't seek will give a False , which counts as a 0 . Thus sum(...) gives the count of the True values, or in other words the number of perfect squares with sorted digits. This is what we want.

\( // ConsoleApplication111.cpp : Defines the entry point for the console application. //

#include "stdafx.h"

#include<iostream>

using namespace std;

int jeff(int y){

if \(y == 0\)return 1;

if \(y % 10 >= \(y/10\) % 10\) return jeff\(y / 10\);

if \(y % 10 < \(y/10\) % 10\) return 0;

}

int main(){

int b = 0;

for \(int i = 0; i < 100000; i\+\+\)\{

    if \(jeff\(i×i\) == 1\)\{

        cout << i << "\\n";

        b\+\+;

    \}



\}

cout << b << "/";

} }\)