PROBLEM LINK:
Setter: Utkarsh Gupta
Tester: Manan Grover
Editorialist: Ajit Sharma Kasturi
DIFFICULTY:
CAKEWALK
PREREQUISITES:
Divisibility rules
PROBLEM:
We need to print an N digit odd number which is divisible by 3 but not divisibility by 9.
QUICK EXPLANATION:
- If N=1 we ouput 3 else we output 30000\dots03 where the total zeros are N-2.
EXPLANATION:
Recall that a number is divisible by 3 if the sum of the digits is divisible by 3 and a number is divisible by 9 if the sum of the digits is divisible by 9.
Now, to construct the number, there are many ways to go about it. One such way is as follows:
-
If N=1, we just output 3.
-
if N \gt 1, we can print a number where the first and last digits equal to 3 and the rest of the digits are equal to 0. The number constructed will be 3000\dots003 . Clearly this number is odd. Also the sum of the digits is 3+3=6 which is divisible by 3 but not divisible by 9. Hence, this number satisifies the required conditions.
TIME COMPLEXITY:
O(N) for each testcase.
SOLUTION:
Editorialist's solution
#include <bits/stdc++.h>
using namespace std;
int main()
{
int tests;
cin >> tests;
while (tests--)
{
int n;
cin >> n;
for (int i = 1; i <= n; i++)
{
if (i == 1 || i == n)
cout << 3;
else
cout << 0;
}
cout << endl;
}
return 0;
}
Setter's solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#include <chrono>
#include <random>
#define ll long long int
#define ull unsigned long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(i,n) for(ll i=0;i<n;i++)
#define loop(i,a,b) for(ll i=a;i<=b;i++)
#define vi vector <int>
#define vs vector <string>
#define vc vector <char>
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
#define max3(a,b,c) max(max(a,b),c)
#define min3(a,b,c) min(min(a,b),c)
#define deb(x) cerr<<#x<<' '<<'='<<' '<<x<<'\n'
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// ordered_set s ; s.order_of_key(val) no. of elements strictly less than val
// s.find_by_order(i) itertor to ith element (0 indexed)
typedef vector<vector<ll>> matrix;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
void solve()
{
int n;
cin>>n;
string s="";
int sum=0;
for(int i=1;i<=n;i++)
{
sum+=3;
s+='3';
}
if(sum%9==0)
s[0]='9';
cout<<s<<'\n';
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int T=1;
cin>>T;
int t=0;
while(t++<T)
{
//cout<<"Case #"<<t<<":"<<' ';
solve();
//cout<<'\n';
}
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Tester's solution
#include <bits/stdc++.h>
using namespace std;
int main(){
ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
int t;
cin>>t;
while(t--){
int n;
cin>>n;
for(int i = 0; i < n; i++){
if(i == 0 || i == n - 1){
cout<<3;
}else{
cout<<0;
}
}
cout<<"\n";
}
return 0;
}
Please comment below if you have any questions, alternate solutions, or suggestions. 
