How do I solve this assignment

Question

1.00/5 (2 votes)

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Given a natural number k (1≤k≤18), we need to find the smallest square number with k digits, where each digit is also a square number.

Input:

The number of test cases t (0<t<10^3).
For each test case, input a natural number k.
Output:

Each line corresponds to a test case and prints the found number. If no number is found, print -1.
Example:
Input:
2
1
3

Output:
1
100

What I have tried:

#include<iostream>
#include<cmath>
using namespace std;

int scp(int n){
    int can = sqrt(n);
    if (can * can == n)
        return 1;
    else
        return 0;
}

long long find(int k){
    if (k > 18){
        return -1;
    }

    long long socantim = 0;
    for (int i = 0; i < k; i++){
        int khoitao = 1;
        while (!scp(khoitao)){
            khoitao++;
        }
        socantim = socantim * 10 + khoitao;
    }

    if (socantim == 0){
        return -1;
    }

    return socantim;
}

int main(){
    int t;
    cin >> t;
    while (t--){
        int a;
        cin >> a;
        cout << find(a) << endl;
    }
    return 0;
}

Posted 17-Jul-23 21:54pm

B20DCVT339-Trần Anh Tuấn

Updated 18-Jul-23 0:54am

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Comments

Patrice T 18-Jul-23 3:57am

What is the problem with your code ?

B20DCVT339-Trần Anh Tuấn 18-Jul-23 4:05am

i think i thought wrong algorithm

B20DCVT339-Trần Anh Tuấn 18-Jul-23 4:05am

my code is not running properly testcase

Patrice T 18-Jul-23 4:21am

Use Improve question to update your question.
So that everyone can pay attention to this information. anf give details.

Andre Oosthuizen 18-Jul-23 4:22am

You need to be more specific, "not running properly" does not tell us anything. Where does it error, what is the error code? What output are getting now compared to what is expected?

Patrice T 18-Jul-23 4:23am

By the way, for me the answer for 1 digit should be 0.

3 solutions

Solution 1

While we are more than willing to help those that are stuck, that doesn't mean that we are here to do it all for you! We can't do all the work, you are either getting paid for this, or it's part of your grades and it wouldn't be at all fair for us to do it all for you.

So we need you to do the work, and we will help you when you get stuck. That doesn't mean we will give you a step by step solution you can hand in!
Start by explaining where you are at the moment, and what the next step in the process is. Then tell us what you have tried to get that next step working, and what happened when you did.

Start by reading the assignment carefully, and working out what you need to know - then try to do it manually. Remember that "each digit is also a square number", and that an 18 digit number won't fit in an int at all ...

So, it's going to be up to you.
Fortunately, you have a tool available to you which will help you find out what is going on: the debugger. How you use it depends on your compiler system, but a quick Google for the name of your IDE and "debugger" should give you the info you need.

Put a breakpoint on the first line in the function, and run your code through the debugger. Then look at your code, and at your data and work out what should happen manually. Then single step each line checking that what you expected to happen is exactly what did. When it isn't, that's when you have a problem, and you can back-track (or run it again and look more closely) to find out why.

Sorry, but we can't do that for you - time for you to learn a new (and very, very useful) skill: debugging!

Posted 17-Jul-23 22:44pm

OriginalGriff

Solution 3

Input:
2
1
3

Output:
1
100

I fear the assignment is wrong.
I don't see any reason why 0 is not the answer for k=1.

Posted 18-Jul-23 0:54am

Patrice T

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CPallini · Accepted Answer · 2023-07-17T22:46:00

Solution 2

Hint 1: if k is odd then 10^k-1 is the solution.
Then, for even values of k:
Hint 2: you need a 64-bit integer to hold number having up to 18 digits.
Hint 3: you may solve it by construction, generating the squares.

Posted 17-Jul-23 22:46pm

CPallini

Comments

Patrice T 18-Jul-23 18:26pm

+5

CPallini 19-Jul-23 0:34am

Thank you.