# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
923182 | LucaLucaM | Stove (JOI18_stove) | C++17 | 16 ms | 4320 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <iostream>
#include <vector>
#include <algorithm>
#include <cassert>
#include <cstring>
#warning That's the baby, that's not my baby
typedef long long ll;
const int NMAX = 1e6;
int a[NMAX + 1];
/**
dp[i][j] =def= timpul minim daca rezolv primele i, termin opresc la momentul a[i] + 1 si folosesc j
dp[i][j] = 1 + min(dp[i - 1][j] + t[i] - t[i - 1], dp[i - 1][j - 1])
defapt nu pot doar sa scot cele mai mari k diferente sau cv?
**/
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(0);
int n, k;
std::cin >> n >> k;
a[0] = -1;
for (int i = 1; i <= n; i++) {
std::cin >> a[i];
}
std::vector<int> diff;
k--; /// ca sa pornesc primu
for (int i = 2; i <= n; i++) {
diff.push_back(a[i] - a[i - 1] - 1);
}
std::sort(diff.begin(), diff.end());
std::reverse(diff.begin(), diff.end());
int answer = a[n] - a[1] + 1;
for (int i = 0; i < k; i++) {
answer -= diff[i];
}
std::cout << answer;
return 0;
}
Compilation message (stderr)
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