The authors should've mentioned these in the question. Like if q is 0 then you should print 10 but if it is any number between 1 to 9 then you should print that number only, i.e. 2 for 2, 4 for 4 not 14 for 4.

How is it POSSIBLE to find 42 test case?
It seems like everything should work fine..

150 : 556

Edited by author 14.06.2020 17:05

#include<bits/stdc++.h>
using namespace std;
bool isPrime(int n)
{
    if(n!=2&&n%2==0)return false;
    if(n<2)return false;
    for(int i=3;i<=sqrt(n);i+=2)
    {
        if(n%i==0)return false;
    }
    return true;
}
int main()
{
    int n;
    cin>>n;
    int x=9,r;
    if(n==1)cout<<1<<endl;
    else if(n==0)cout<<10<<endl;
else{
    vector<int>v;
    while(1)
    {
        if(n%x==0)
        {
           if(isPrime(n)==true)
           {
            if(n>9){cout<<-1<<endl;return 0;}
           }
           n=n/x;
           if(x==1)break;
           v.push_back(x);
        }
        else
        {
            x--;
        }

    }
    for(int i=v.size()-1;i>=0;i--)
    {
        cout<<v[i];
    }
    cout<<endl;
}
}

The output needs to be sorted in ascending order(!)

so for test number 5: 45

the correct answer is 59 (5*9)

and 95 (9*5) will be rejected.

Hope this helps someone.

What is the mistake?
n = int(input())
a = []
for i in range(1, n+1):
    if n % i == 0:
        a.append(i)
if n == 1:
    print("1")
    exit(0)
if n == 0:
    print("10")
    exit()
if len(a) > 2:
    for i in range(len(a) - 1):
        if a[i] * a[i + 1] == n or a[i]*a[i] == n:
            if a[i] * a[i + 1] == n:
                s = "".join(sorted(str(a[i])+str(a[i+1])))
                print(s)
            else:
                print(a[i], a[i], sep="")
else:
    print("-1")

Whats wrong with anything here?
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using ui=unsigned int;
int main() {
    //ios::sync_with_stdio(0);
    //cin.tie(0);
    ll n;cin>>n;
    if(!n){
        cout<<10;
        return 0;
    }
    else if(n<10){
        cout<<n;
        return 0;
    }
    vector<int>v;
    ll l=n;
    for(bool q=1;q;){
        q=0;
        for(ll i=9;i>1;i--){
            if(l%i==0){
                l/=i;
                v.push_back(i);
                q=1;

            }
        }
    }
    v.push_back(l);
    sort(v.begin(),v.end());
    if(v[0]==1)for(int i=1;i<v.size();i++)cout<<v[i];
    else cout<<-1;
    return 0;
}

if u have wa 44 just remove one in the cyclic factorization, if there is one.
<=sqrt() + 1   -- bad
<=sqrt()       -- good

I don't know the problem

use std::io;
use std::io::BufReader;
use std::io::BufRead;
fn minimal_product_digit(n: i64) -> i64{
    if n == 0 {return 10};
    if n < 10 { return n};

    let f : Vec<i64> = vec![2,3,5,7];
    let mut aux = n;
    let mut ret : Vec<i64> = Vec::new();
    for i in f{

        if  aux % i== 0 {

            loop{
                aux/=i;
                 ret.push(i);
                if aux % i != 0 {
                    break;
                }
            }
        }
    };

    if ret.is_empty() || aux > 1 {
        return -1;
    }

    match ret.iter().fold(0,|acc,elem| if *elem == 3{acc + 1} else {acc}) {

        1 => {

            let index = ret .iter().position(|&x| x == 3).unwrap();

            if index > 0{

                ret.remove(index);

                let k = ret.get_mut(index - 1).unwrap();

                *k = 6;

                ret.sort();

            }
        }

        _ => ()

    }
    let mut v : Vec<i64> = Vec::new();
    for i in ret{

        match i {

            2 | 3=>

                match v.last_mut(){

                    None => v.push(i),
                    Some(last) if i ==3 && *last == 2 => v.push(i),
                    Some(last) => {
                        let valor = *last;
                        if (valor * i) < 10{
                            *last = valor*i
                        }
                        else{
                            v.push(i);
                        }
                    }
                }
            _ => v.push(i)
        }
    }

        v.sort();
        v.iter().fold(0, |acc, elem| acc * 10 + elem)

}

fn main() {

    let mut line = String::new();
    let mut stdin = BufReader::new(io::stdin());

  stdin.read_line(&mut line).unwrap();

    let lim = line.trim().parse::<i64>().unwrap();

    println!("{:?}",minimal_product_digit(lim));

}

can anyone tell me what is test case 13?

i dont know why my code is failing test 4.can any body give me a series of test cases where this kind of program may fail. please help

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
    int Num = 0, ans = -2;

    cin >> Num;

    vector <int> fact;

    int n = 2;

    while (Num > 1)
    {
        if (n > 9) {
            ans = -1;
            break;
        }
        while (Num % n == 0) {
            Num /= n;
            fact.push_back(n);
        }
        n++;
    }

    sort(fact.begin(), fact.end());

    if (n < 10 & ans != -1)
    {
        ans = 0;
        for (int i = 0; i < fact.size(); i++) {
            //ans += fact.at(i) * (i*10);
            cout << fact[i];
        }
    }


    //cout << ans;
}

Hey! I solved this problem and here are some test that you may check for you solution:

0->10
1->1
2->2
3->3
4->4
5->5
6->6
7->7
8->8
9->9
10->25
11->-1
12->26
13->-1
14->27
15->35
16->28
17->-1
18->29
19->-1
20->45
21->37
22->-1
23->-1
24->38
25->55
26->-1
27->39
28->47
29->-1
30->56
31->-1
32->48
33->-1
34->-1
35->57
36->49
37->-1
38->-1
39->-1
40->58
41->-1
42->67
43->-1
44->-1
45->59
46->-1
47->-1
48->68
49->77
50->255
51->-1
52->-1
53->-1
54->69
55->-1
56->78
57->-1
58->-1
59->-1
60->256
61->-1
62->-1
63->79
64->88
65->-1
66->-1
67->-1
68->-1
69->-1
70->257
71->-1
72->89
73->-1
74->-1
75->355
76->-1
77->-1
78->-1
79->-1
80->258
81->99
82->-1
83->-1
84->267
85->-1
86->-1
87->-1
88->-1
89->-1
90->259
91->-1
92->-1
93->-1
94->-1
95->-1
96->268
97->-1
98->277
99->-1
100->455

Good Luck!

Edited by author 17.08.2020 15:23

Edited by author 17.08.2020 15:23

Edited by author 17.08.2020 15:23

what will the output if

input = 1 or 0 and why?

This problem can be solved using greedy technic.
Method is to divide the given number from 9 to 2. Each time we will continue to divide and update the number as we go. We will keep dividing N with the same number utill its impossible to do so.And for each division we will save the number which was divisible by given number. And by "save" i mean pushing the number in a stack or a in a vector or an array. If you are using other than stack you might print out the numbers in a reverse manner.

And when we are getting "-1"?
I already wrote that, we keep updating the number. so after we pass the 9 to 2 loop, we should be left with a value of 1 in the variable N ( given number ). Incase N!=1 we are just printing "-1".

For the case N=1 or N=0? for the case 1 you should just print "1" and for the case 0 print "10", we can not print 01 since this refers to number, 1.

If you have problem understanding what i have just written or for better understanding, give this a go : https://ideone.com/709cog

Edited by author 21.02.2020 00:34

Edited by author 21.02.2020 00:38

Edited by author 21.02.2020 00:41

I have tried everything.

Here are some testcases from my algorithm:
0: 10
1: 1
2: 2
3: 3
4: 4
5: 5
6: 6
7: 7
8: 8
9: 9
10: 25
12: 26
14: 27
15: 35
16: 28
18: 29
20: 45
21: 37
24: 38
25: 55
27: 39
28: 47
30: 56

Some big numbers:
16859136: 267778888
16875000: 3555555589
16934400: 355778889
16941456: 277777789
17006112: 48999999
17010000: 555567899
17146080: 256778999
17150000: 1555557778
17203200: 155788888
17210368: 177777888
17222625: 355579999
17280000: 555568889
17287200: 455777789
17294403: 177777777
17360406: 67799999
17364375: 555577799
17418240: 256788899
17496000: 355588999
17500000: 4555555578
17503290: 257777999
17561600: 455777888
17578125: 5555555559
17635968: 67889999
17640000: 555577889
17647350: 556777777
17694720: 256888889
17714700: 255699999
17718750: 2555555799

long long int minimoentero3(long int num)
{
     long long int nf=result;
     return nf;
}


int main()
{
    long int n;
    cin >> n;

    cout<<minimoentero3(n); <- Here the result is sent.
}

Edited by author 26.10.2017 18:28

Factorize (in 1..9 figures) the N and then reduce the resulting number of figures.

Edited by moderator 23.08.2020 01:12