# Help me find testcases where my code for SOLVEMORE is failing

So, here’s my code for SOLVEMORE from STARTERS 78.
This is the basic idea of what I’m doing.
Take sum of ai and bi, store it in ci {si, i}. ci is a vector of pair<ll, int>, which stores sum and index.
Then, sort c.
Find cumulative sum for c.
Using std::upper_bound, find idx for which ci > k.
Now, we find if there is an ai, which when added to ci-1 <=k
We do this by adding ci.second, that is original indices involved, and are part of problems solved to a set.
We iterate over a, if there’s an a s.t c[idx-1] + a[i] <= k and i is not in set, then we increase our answer.

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

typedef long long ll;

int main() {
int t;
cin >> t;
while (t--) {
ll n, k, i;
cin >> n >> k;
vector<ll> a(n), b(n);
vector<pair<ll, int>> c(n);
for (i = 0; i < n; i++) cin >> a[i];
for (i = 0; i < n; i++) {
cin >> b[i];
c[i].first = a[i] + b[i];
c[i].second = i;
}
sort(c.begin(), c.end());
for (i = 1; i < n; i++) c[i].first += c[i - 1].first;
auto idx = distance(c.begin(),
upper_bound(c.begin(), c.end(), make_pair(k, INT_MIN)));
unordered_set<int> us;
for (i = 0; i < idx; i++) {
us.emplace(c[i].second);
}
ll ans = idx;
if (idx - 1 >= 0) {
for (i = 0; i < n; i++) {
if (!us.count(i) && a[i] + c[idx - 1].first <= k) {
ans++;
break;
}
}
}
std::cout << ans << endl;
}
return 0;
}

``````