HackerEarth Maximum Sum problem solution

In this HackerEarth Maximum Sum problem solution, we have given an array A of N integers. Now, you have to output the sum of unique values of the maximum subarray sum of all the possible subarrays of the given array A.

HackerEarth Maximum Sum problem solution.

#include <bits/stdc++.h>

using namespace std;

const int N = 5E3 + 5;
long long a[N];
set<long long> st;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    for(int i = 0; i < n; i ++)
        cin >> a[i];
    long long msf, cur;
    for(int i = 0; i < n; i ++) {
        for(int j = i; j < n; j ++) {
            if(j == i) {
                msf = a[j];
                cur = a[j];
                st.insert(a[j]);
            } else {
                cur = max(a[j], cur + a[j]);
                msf = max(msf, cur);
                st.insert(msf);
            }
        }
    }
    long long ans = 0;
    for(auto i : st)
        ans += i;
    cout << ans;
    return 0;
}

Second solution

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define endl "\n"
#define eps 0.00000001
LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
LL dp[50001];
set<LL> s;
int a[5001];
int main()
    {
        ios_base::sync_with_stdio(0);
        int n;
        cin >> n;
        for(int i = 1; i <= n; i++)
            cin >> a[i];
        for(int i = 1; i <= n; i++){
            LL mx = -1000000000;
            for(int j = i; j <= n; j++){
                if(dp[j - 1] < 0){
                    dp[j] = a[j];
                }
                else{
                    dp[j] = dp[j - 1] + a[j];
                }
                mx = max(mx , dp[j]);
                s.insert(mx);
            }
            for(int j = i; j <= n; j++){
                dp[j] = 0;
            }
        }
        LL ans = 0;
        for(LL i: s){
            ans += i;
        }
        cout << ans;
    }