In this HackerEarth Beautiful numbers problem solution, Any number is called beautiful if it consists of 2N digits and the sum of the first N digits is equal to the sum of the last N digits. Your task is to find the count of beautiful numbers in the interval from L to R (including L and R).

Beautiful numbers do not have leading zeroes.


HackerEarth Beautiful numbers problem solution


HackerEarth Beautiful numbers problem solution.

#include<bits/stdc++.h>
using namespace std;

#define F first
#define S second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define vi vector<int>
#define all(x) x.begin(),x.end()
#define fix fixed<<setprecision(10)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define repb(i,b,a) for(int i=int(b);i>=int(a);i--)
#define FastIO ios_base::sync_with_stdio(0),cin.tie(0)

typedef double db;
typedef long long ll;

const int N=2e5+5;
const int mod=1e9+7;

vi sum[50][10],v;

void pre(){
rep(i,1,9999){
int t=i,s=0,d=0;
while(t){
s+=(t%10);
t/=10;
d++;
}
rep(j,d,4) sum[s][j].pb(i);
}
rep(i,1,9999){
int t=i,s=0,d=0;
while(t){
s+=(t%10);
t/=10;
d++;
}
for(int u:sum[s][d]){
string s=to_string(i);
string t=to_string(u);
while(t.size()<d) t="0"+t;
s+=t;
v.pb(stoi(s));
}
}
sort(all(v));
}

void solve(){
int l,r;
cin>>l>>r;
cout<<ub(all(v),r)-lb(all(v),l)<<'\n';
}

signed main(){
FastIO;
pre();
int t;
cin>>t;
while(t--) solve();
return 0;
}

Second solution

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int D = 5, MXS = 1000, N = 1e4;
vector<int> v[D][MXS], ans;

int main() {
ios::sync_with_stdio(0), cin.tie(0);
for (int i = 1; i < N; ++i) {
string t = to_string(i);
v[t.size()][accumulate(t.begin(), t.end(), 0) - t.size() * '0'].push_back(i);
}
for (int d = 1; d < D; ++d)
for (int s = 0; s < MXS; ++s)
for (auto l : v[d][s])
for (int dd = 0; dd <= d; ++dd)
for (auto r : v[dd][s])
ans.push_back(l * (int) pow(10, d) + r);
sort(ans.begin(), ans.end());
int t;
cin >> t;
while (t--) {
int l, r;
cin >> l >> r;
++r;
cout << lower_bound(ans.begin(), ans.end(), r) - lower_bound(ans.begin(), ans.end(), l) << '\n';
}
}