# Did I fakesolve?

So I’m doing this problem here, and my code is as follows:

#include <iostream>
#include <vector>
#include <map>
#include <algorithm>

using std::cout;
using std::endl;
using std::vector;
using std::pair;

/**
* https://oj.uz/problem/view/NOI18_knapsack
* 15 5
* 4 12 1
* 2 1 1
* 10 4 1
* 1 1 1
* 2 2 1 should output 15
*/
int main() {
int limit;
int type_num;
std::cin >> limit >> type_num;

std::map<int, vector<pair<int, int>>> by_weight;
for (int t = 0; t < type_num; t++) {
int value;
int weight;
int amt;
std::cin >> value >> weight >> amt;
if (weight <= limit && 0 < amt) {
by_weight[weight].push_back({value, amt});
}
}

vector<vector<long long>> best(by_weight.size() + 1,
vector<long long>(limit + 1, INT32_MIN));
best[0][0] = 0;
int at = 1;
for (const auto& [w, items] : by_weight) {
std::sort(items.begin(), items.end(), std::greater<pair<int, int>>());
for (int i = 0; i <= limit; i++) {
best[at][i] = best[at - 1][i];
int copies = 0;
int type_at = 0;
int curr_used = 0;
long long profit = 0;
while ((copies + 1) * w <= i && type_at < items.size()) {
copies++;
profit += items[type_at].first;
if (best[at - 1][i - copies * w] != INT32_MIN) {
best[at][i] = std::max(
best[at][i],
best[at - 1][i - copies * w] + profit
);
}

curr_used++;
if (curr_used == items[type_at].second) {
curr_used = 0;
type_at++;
}
}
}
at++;
}
cout << *std::max_element(std::begin(best.back()),
std::end(best.back())) << endl;
}


It basically groups the items by weight, and then does the standard knapsack DP. However, even though its complexity shouldn’t allow it to pass, I still get all test cases on this. Does anyone know why?

Looks fine to me. What do you think the time complexity is?

It should be O(W^2\cdot \text{something}), and given that W^2 is already 4 million, I don’t think anything above a factor of 10 would work for something.

…bump?

An item of weight i contributes an additive factor of \frac{S^2}{i} to the runtime. Summing over all i from 1 to S gives S^2\cdot \left(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{S}\right)\approx S^2\log S.

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