Digit cube | Hackerearth practice probelm solution

   Problem:-

You are provided an array $A$ of size $N$ that contains non-negative integers. Your task is to determine whether the number that is formed by selecting the last digit of all the N numbers is divisible by $10$.

Note: View the sample explanation section for more clarification.

Input format

• First line: A single integer $N$ denoting the size of array $A$
• Second line: $N$ space-separated integers.

Output format

If the number is divisible by $10$, then print $Yes$. Otherwise, print $No$.

Constraints
$$\begin{gathered} 1\le N\le {10}^5 \\ 0\le A[i]\le {10}^5 \end{gathered}$$

Sample Input

5
85 25 65 21 84

Sample Output

No

Time Limit: 1

Memory Limit: 256

Source Limit:

Explanation

Last digit of $85$ is $5$.
Last digit of $25$ is $5$.
Last digit of $65$ is $5$.
Last digit of $21$ is $1$.
Last digit of $84$ is $4$.
Therefore the number formed is $55514$ which is not divisible by $10$.

Code:-

  Here I am going to give you two solution first one is on the basis of C language and second one is on the basis of c++ language which you can submit in c++14 and c++17 also

Solution 1 ( C language):-

#include <stdio.h>

#define MAX_N 3e6 + 14

#define LOG 50

#define MAX_S 136

int sum_of_digits(long long n) {

int sum = 0;

while (n > 0) {

sum += n % 10;

n = n / 10;

}

return sum;

}

int nxt[MAX_S][LOG];

int main() {

int t,i,j;

long long n,k;

for (j = 0; j < LOG; ++j) {

for (i = 1; i < MAX_S; ++i) {

if (j)

nxt[i][j] = nxt[nxt[i][j - 1]][j - 1];

else

nxt[i][0] = sum_of_digits(i * i * i);

}

}

scanf("%d", &t);

while (t--) {

scanf("%lld", &n);

scanf("%lld", &k);

k--;

n = sum_of_digits(n);

for (i = LOG - 1; i >= 0; --i)

if (k >> i & 1)

n = nxt[n][i];

printf("%lld \n", n*n*n);

}

}

Solution 2 ( C++ language):-

 This solution is based on the c++ language and you can submit ib c++14 and c++17 also.
In this solution first three lines of the main function is only for the decreasing the time of execution of the program..

ios_base::sync_with_stdio(false);

cin.tie(NULL);

cout.tie(NULL);

This is your choice that you want to use this or not but in some cases the code may take more time in execution and that time we need it .

#include <bits/stdc++.h>

#define ll long long

#define pb push_back

#define c(p) cout<<p<<"\n"

#define sz(a) (ll) (a.size())

using namespace std;

const ll N = 500005;

const ll mod= 1e9 + 7;

vector<ll> v[150];

ll f(ll n)

{

    ll c=0;

    while(n>0)

    {

        c+=n%10;

        n/=10;

    }

    return c;

}

void solve()

{

    ll x=0,y=0,c=0,ans=0;

ll n,m,k;

cin>>n>>k;

assert(n>=1 and n<=1e15);

assert(k>=1 and k<=1e15);

    x=f(n);

m=sz(v[x]);

    

    if(m>k)

    {

        c(v[x][k-1]);

        return;

    }

    if(m==1)

    {

        c(v[x][0]);

        return;

    }

    k -= (m-2);

    k%=2;

    c(v[x][m-k-1]);

}

signed main() {

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    ll n,x,c;

    for(ll i=1;i<140;++i)

    {

        n=i*i*i;

        c=0;

        std::set<ll> s;

        s.insert(n);

        v[i].pb(n);

        while(true)

        {

            x=f(n);

            v[i].pb(x*x*x);

            if(s.find(x*x*x)!=s.end())

            {

                break;

            }

            s.insert(x*x*x);

            n= x*x*x;

        }

    }

    int T;

    cin>>T;

    assert(T>=1 and T<=1e6);

    while(T--)

    {

        solve();

    }       

    return 0;                   

}

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