SPOILER

opt(i,L) is the number of combinations of length L t...

same for me even if I set matrix_dist[I][I]=1. thank you!

Do not add spaces at the end of each test case and add new line also for the last test case

Why don't we update (instead of adding) a pair in the priority queue, if it already exists ?

In the helper code provided by HackerRank, if you do not return a long long int, you will fail the test 11 : this is why you need to change not only the return type of the function : journeyToMoon , but also the variable that stores the result in the helper code into :

long long int result = journeyToMoon(n, astronaut);

My solution in c++

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
 
#define watch(x) cout << (#x) << " is " << (x) << endl
 
using namespace std;
 
struct mypair{
    int a;
    int b;
};
 
bool mycmp(mypair p1, mypair p2){
    return (p1.a+p1.b)>(p2.a+p2.b);
}
int main() {
ios::sync_with_stdio(false); // makes cin and cout as fast and scanf printf
#ifndef ONLINE_JUDGE

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
 
#define watch(x) cout << (#x) << " is " << (x) << endl
 
 
using namespace std;
int main() {
 
ios::sync_with_stdio(false); // makes cin and cout as fast and scanf printf
#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
#endif
    int c;
    int test_n = 1;
    cin>>c;
    while(c--){
        int n,k,b,t;

I think the problem is the choice of the parameter : taking the greatest of tips is not the optimal choice. For example consider this case : A = [10,1 ], B = [11 ,8], X=1, Y=1 : the optimal choice is 8+11 = 19, while you algorithm : 11+1=12

This is my solution following the useful hint of Navkrishna Raghav

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
 
#define watch(x) cout << (#x) << " is " << (x) << endl
 
using namespace std;
 
struct job{
    long int a;
    long int b;
    long long int d;
};
 
bool cmp1(job j1, job j2){
    if (j1.d == j2.d)
        return j1.a<j2.a;
    return  j1.d>j2.d ;
}
 
class cmp2{

thank you !