HackerEarth Beautiful numbers problem solution

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.

#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';
    }
}