# 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

$1\le N\le {10}^{5}\phantom{\rule{0ex}{0ex}}0\le A\left[i\right]\le {10}^{5}$

Sample Input
`585 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] = 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;
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][k1]);
return;
}
if(m==1)
{
c(v[x]);
return;
}
k -= (m2);
k%=2;
c(v[x][mk1]);
}
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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