#include <bits/stdc++.h>
#define int long long
using namespace std;
#ifdef LOCAL
ifstream in("in.in");
ofstream out("out.out");
#else
#define in cin
#define out cout
#endif
const int nmax = 100005, inf = (1LL << 60);
int n, k;
vector<int> a, b, v, s;
const int MAX_X = 2000005;
int sum(int l, int r) {
return s[r] - s[l - 1];
}
int sol = inf;
void solve() {
int sep1 = 0, sep2 = n + 1;
/// sep1 is the biggest point so that v[sep1] <= -x, so that number behind sep1
/// is just the number of values that will have an x straight up added to them.
/// sep2 is the smallest number such that v[sep2] >= 0, so that they will all get
/// x added to them
/// between sep1 and sep2, sum will get added number of numbers that are there * x -their sum
/// we also get the top numbers from all of these.
s[0] = 0;
for (int i = 1; i <= n; i++) {
s[i] = s[i - 1] + v[i];
}
auto f = [&](int nr, int x) {
///returns the max that you can take out if you take out the last x values
///from right.
int Left = k - nr,cursum = 0,Right = n - nr + 1;
if (nr > 0) {
cursum += sum(Right, n) + nr * x;
}
if (Left > 0) {
cursum += -sum(1, Left) - Left * x;
}
return cursum;
};
for (int x = MAX_X; x >= 0; x--) {
while (sep1 <= n && v[sep1 + 1] <= -x) {
sep1++;
}
while (sep2 > 1 && v[sep2 - 1] >= 0) {
sep2--;
}
int curSum = 0;
if (sep2 <= n + 1) {
curSum += sum(sep2, n) + (n - sep2 + 1) * x; /// right
}
if (sep1 >= 1) { // there are SOME elements smaller than -x
curSum += -sum(1, sep1) - sep1 * x;
}
if (sep1 < sep2 - 1) {
curSum += (sep2 - sep1 - 1) * x + sum(sep1 + 1, sep2 - 1);
}
int bonus = 0;
for (int nr = 0; nr <= k; nr++) {
bonus = max(bonus, f(nr , x));
}
sol = min(sol, curSum - bonus);
}
}
int32_t main() {
in >> n >> k;
a.resize(n + 1);
b.resize(n + 1);
v.resize(n + 1);
s.resize(n + 1);
for (int i = 1; i <= n; i++) {
in >> a[i];
}
for (int i = 1; i <= n; i++) {
in >> b[i];
}
for (int i = 1; i <= n; i++) {
v[i] = a[i] - b[i];
}
sort(v.begin() + 1, v.end());
solve();
for (int i = 1; i <= n; i++) {
v[i] *= -1;
}
sort(v.begin() + 1, v.end());
solve();
out << sol << "\n";
}
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Correct |
25 ms |
204 KB |
Output is correct |
2 |
Correct |
21 ms |
272 KB |
Output is correct |
3 |
Correct |
20 ms |
284 KB |
Output is correct |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Correct |
103 ms |
3404 KB |
Output is correct |
2 |
Correct |
91 ms |
4780 KB |
Output is correct |
3 |
Correct |
89 ms |
4812 KB |
Output is correct |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Correct |
87 ms |
3424 KB |
Output is correct |
2 |
Correct |
84 ms |
4812 KB |
Output is correct |
3 |
Correct |
85 ms |
4840 KB |
Output is correct |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Correct |
87 ms |
3404 KB |
Output is correct |
2 |
Correct |
83 ms |
4848 KB |
Output is correct |
3 |
Correct |
83 ms |
4804 KB |
Output is correct |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Execution timed out |
1079 ms |
3404 KB |
Time limit exceeded |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Execution timed out |
1092 ms |
3404 KB |
Time limit exceeded |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Execution timed out |
1078 ms |
3404 KB |
Time limit exceeded |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Execution timed out |
1096 ms |
3404 KB |
Time limit exceeded |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Execution timed out |
1083 ms |
3404 KB |
Time limit exceeded |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Execution timed out |
1100 ms |
3404 KB |
Time limit exceeded |
2 |
Halted |
0 ms |
0 KB |
- |