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답안 #518670

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
518670 2022-01-24T11:58:40 Z Jeff12345121 Simfonija (COCI19_simfonija) C++14
44 / 110
 1000 ms 3996 KB 
#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];
    }

    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 k2 = k,indRight = n + 1,indLeft = 0,bonus = 0;
        while(k2--) {
            int opt1 = ( (indRight > sep2) ? (v[indRight - 1] + x) : -inf );
            int opt2 = ( (indLeft < sep1) ? ( -v[indLeft + 1] - x ): -inf );
            if (opt1 > opt2) {
                indRight--;
                bonus += opt1;
            } else {
                indLeft++;
                bonus += opt2;
            }
        }

        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";
}

Subtask #1
11.0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Correct 26 ms 204 KB Output is correct
2 Correct 27 ms 204 KB Output is correct
3 Correct 31 ms 204 KB Output is correct

Subtask #1
11.0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Correct 89 ms 3692 KB Output is correct
2 Correct 92 ms 3996 KB Output is correct
3 Correct 89 ms 3908 KB Output is correct

Subtask #1
11.0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Correct 93 ms 3736 KB Output is correct
2 Correct 89 ms 3908 KB Output is correct
3 Correct 87 ms 3916 KB Output is correct

Subtask #4
11.0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Correct 90 ms 3900 KB Output is correct
2 Correct 95 ms 3828 KB Output is correct
3 Correct 89 ms 3908 KB Output is correct

Subtask #5
0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1086 ms 3748 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #6
0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1065 ms 3788 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #7
0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1044 ms 3744 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #8
0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1070 ms 3748 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #9
0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1037 ms 3816 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #10
0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1049 ms 3780 KB Time limit exceeded
2 Halted 0 ms 0 KB -