Codeforces Round #684 (Div. 2)ABC1C2

Posted 2020-12-21 kanoon

比赛链接：https://codeforces.com/contest/1440

A. Buy the String

题解

枚举字符串中 (0) 或 (1) 的个数即可。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
        int n, c0, c1, h;
        cin >> n >> c0 >> c1 >> h;
        string s;
        cin >> s;
        int cnt[2] = {};
        for (char c : s) ++cnt[c - ‘0‘];
        int ans = INT_MAX;
        for (int i = 0; i <= n; i++) {
            ans = min(ans, i * c0 + (n - i) * c1 + h * abs(i - cnt[0]));
        }
        cout << ans << "
";
    }
    return 0;
}

B. Sum of Medians

题解

贪心，先把较小的数填入每个数组的前 (lceil frac{n}{2} ceil - 1) 个元素，然后依次填完每个数组即可。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
        int n, k;
        cin >> n >> k;
        vector<vector<int>> a(k);
        for (int i = 0, j = 0; i < n * k; i++) {
            int x;
            cin >> x;
            a[j].push_back(x);
            if (a[j].size() + 1 == (n + 1) / 2) ++j;
            if (a[j].size() == n) ++j;
            if (j == k) j = 0;
        }
        long long ans = 0;
        for (int i = 0; i < k; i++) {
            ans += a[i][(n + 1) / 2 - 1];
        }
        cout << ans << "
";
    }
    return 0;
}

C1. Binary Table (Easy Version)

题解

依次操作每个 (2 imes 2) 方阵即可，方阵中 (1) 的个数的规律为：1 -> 2 -> 3 -> 0 。

代码

见C2。

C2. Binary Table (Hard Version)

题解

依次把所有 (1) 都挤到右下角的 (2 imes 2) 的方阵中，然后操作一下该方阵即可。

证明

除右下角的方阵外最多操作 (n imes m - 4) 次，右下角的方阵最多操作 (3) 次，所以最多操作 (n imes m - 1) 次。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        vector<string> MP(n);
        for (auto &x : MP) cin >> x;
        vector<pair<int, int>> v;
        auto op = [&](int x, int y) {
            v.emplace_back(x, y);
            MP[x][y] = MP[x][y] == ‘0‘ ? ‘1‘ : ‘0‘;
        };
        for (int i = 0; i < n - 2; i++) {
            for (int j = 0; j < m; j++) {
                if (MP[i][j] == ‘1‘) {
                    op(i, j);
                    op(i + 1, j);
                    if (j == m - 1) op(i + 1, j - 1);
                    else op(i + 1, j + 1);
                }
            }
        }
        for (int j = 0; j + 2 < m; j++) {
            for (int i = n - 2; i < n; i++) {
                if (MP[i][j] == ‘1‘) {
                    op(i, j);
                    op(i, j + 1);
                    if (i == n - 2) op(i + 1, j + 1);
                    if (i == n - 1) op(i - 1, j + 1);
                }
            }
        }
        auto cal = [&](int x, int y) {
            string s;
            s += MP[x][y];
            s += MP[x][y + 1];
            s += MP[x + 1][y];
            s += MP[x + 1][y + 1];
            for (int tot_1 = count(s.begin(), s.end(), ‘1‘); tot_1 != 0; ) {
                if (tot_1 == 1) {
                    int cnt_0 = 0;
                    for (int i = 0; i < 4; i++) {
                        if (s[i] == ‘1‘) {
                            v.emplace_back(x + (i >= 2), y + (i == 1 or i == 3));
                            s[i] = ‘0‘;
                        } else if (cnt_0 < 2) {
                            v.emplace_back(x + (i >= 2), y + (i == 1 or i == 3));
                            s[i] = ‘1‘;
                            ++cnt_0;
                        }
                    }
                } else if (tot_1 == 2) {
                    int cnt_1 = 0;
                    for (int i = 0; i < 4; i++) {
                        if (s[i] == ‘1‘) {
                            if (cnt_1 < 1) {
                                v.emplace_back(x + (i >= 2), y + (i == 1 or i == 3));
                                s[i] = ‘0‘;
                                ++cnt_1;
                            }
                        } else {
                            v.emplace_back(x + (i >= 2), y + (i == 1 or i == 3));
                            s[i] = ‘1‘;
                        }
                    }
                } else if (tot_1 == 3) {
                    for (int i = 0; i < 4; i++) {
                        if (s[i] == ‘1‘) {
                            v.emplace_back(x + (i >= 2), y + (i == 1 or i == 3));
                            s[i] = ‘0‘;
                        }
                    }
                } else if (tot_1 == 4) {
                    int cnt_1 = 0;
                    for (int i = 0; i < 4; i++) {
                        if (s[i] == ‘1‘) {
                            v.emplace_back(x + (i >= 2), y + (i == 1 or i == 3));
                            s[i] = ‘0‘;
                            if (++cnt_1 == 3) break;
                        }
                    }
                }
                tot_1 = count(s.begin(), s.end(), ‘1‘);
            }
            for (int i = 0; i < 4; i++) {
                int nx = x + (i >= 2);
                int ny = y + (i == 1 or i == 3);
                MP[nx][ny] = s[i];
            }
        };
        cal(n - 2, m - 2);
        cout << v.size() / 3 << "
";
        int cnt = 0;
        for (auto [x, y] : v) {
            cout << x + 1 << ‘ ‘ << y + 1 << ‘ ‘;
            if (++cnt % 3 == 0) cout << "
";
        }
    }
    return 0;
}