Equilateral numbers

input/code.cpp: In function 'int main()':
input/code.cpp:23:17: warning: overflow in conversion from 'double' to 'int' changes value from '1.0e+18' to '2147483647' [-Woverflow]
   23 |         dp[i] = 1e18;
      |                 ^~~~
input/code.cpp:19:8: warning: unused variable 'm' [-Wunused-variable]
   19 |     ll m = 0;
      |        ^

#include<iostream>
#include<vector>
#include <cmath>
#include <map>
using namespace std;

typedef long long ll;

#define lim 101010

int dp[lim];
map<ll, int> ek;

ll e(ll k){
    return k*(k+1)/2;
}
int main()
{
    ll m = 0;
    dp[0] = 0;
    for(ll i = 1; i< lim; ++i){
        ek[e(i)] = 1;
        dp[i] = 1e18;
        for (ll j =0; e(j) <= i; j++) {
            dp[i] = min(dp[i-e(j)] + 1, dp[i]);
        }
        // m  = max(dp[i], m);
    }
    for(ll i = 1; i< 10000; ++i){
        for(ll j = 1; j< 1000; ++j){
            ek[e(i) + e(j)] = 2;
        }
    }

    ll n;
    cin >> n;

    int res = 3;
    for(ll i = 1; e(i) <= n; ++i){
        if(e(i) == n){
            cout << "1\n";
            return 0;
        }
        if(ek[n - e(i)]){
            res = min(res, ek[n-e(i)]+1);
        }
        if(n - e(i) < lim){
            res = min(res, dp[n-e(i)]+1);
        }
    }
    cout << res << endl;
}

Test 1

Verdict: TIME LIMIT EXCEEDED

Test 2

Verdict: TIME LIMIT EXCEEDED

Test 3

Verdict: TIME LIMIT EXCEEDED

Test 4

Verdict: TIME LIMIT EXCEEDED

Test 5

Verdict: TIME LIMIT EXCEEDED

Test 6

Verdict: TIME LIMIT EXCEEDED

Test 7

Verdict: TIME LIMIT EXCEEDED

Test 8

Verdict: TIME LIMIT EXCEEDED

Test 9

Verdict: TIME LIMIT EXCEEDED

Test 10

Verdict: TIME LIMIT EXCEEDED

Test 11

Verdict: TIME LIMIT EXCEEDED

Test 12

Verdict: TIME LIMIT EXCEEDED