Codeforces Round #715 (Div. 2) (A~D 补题记录)

补题链接：Here

经典手速场

1509A. Average Height

题意：要找出最大不平衡对序列

先输出奇数，然后输出偶数

void solve() {
    int n;
    cin >> n;
    vector<int> odd, even;
    for (int i = 0, x; i < n; ++i) {
        cin >> x;
        if (x & 1) odd.push_back(x);
        else
            even.push_back(x);
    }
    for (int x : odd) cout << x << " ";
    for (int x : even) cout << x << " ";
    cout << "
";
}

1509B. TMT Document

题意：给定一个 T-M字符串，求问是否能全拆分为 TMT 子序列

思路：

要能组成 TMT 就要是 T、M顺序一定并 cntT = 2 * cntM 和 (n \% 3== 0)

void solve() {
    int n;
    string s;
    cin >> n >> s;
    int ct = 0, cm = 0;
    bool f = true;
    for (int i = 0; f && i < n; ++i) {
        s[i] == 'T' ? ct++ : cm++;
        if (cm > ct || (ct > 2 * cm + n / 3 - cm)) f = false;
    }
    cout << (f && cm * 2 == ct && n % 3 == 0 ? "YES
" : "NO
");
}

1509C. The Sports Festival

题意：

学生会要参加接力赛，每位成员跑步速度为 (a_i) ，给定定义：

[d_i = max(a_1,a_2,dots,a_i) - min(a_1,a_2,dots,a_i) ]

求出最小的 (sum_{i = 1}^n d_i)

思路：

待补。

using ll = long long;
ll dp[2005][2005];
int n;
int A[2005];
void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> A[i];
    sort(A + 1, A + n + 1);
    for (int i = 1; i <= n; ++i)
        for (int j = i + 1; j <= n; ++j) dp[i][j] = 1e18;
    for (int i = 1; i <= n; ++i) dp[i][i] = 0;
    for (int len = 1; len < n; ++len) {
        for (int i = 1; i + len - 1 <= n; ++i) {
            int j = i + len - 1;
            if (j < n) dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + A[j + 1] - A[i]);
            if (i > 1) dp[i - 1][j] = min(dp[i - 1][j], dp[i][j] + A[j] - A[i - 1]);
        }
    }
    cout << dp[1][n] << '
';
}

另外一种写法

using ll = long long;
void solve() {
    int n;
    cin >> n;
    vector<ll> s(n);
    for (ll &x : s) cin >> x;
    sort(s.begin(), s.end());
    vector<ll> dp0(n), dp1(n);
    for (int k = 1; k < n; ++k) {
        for (int i = k; i < n; ++i)
            dp1[i] = min(dp0[i - 1], dp0[i]) + s[i] - s[i - k];
        swap(dp0, dp1);
    }
    cout << dp0[n - 1] << '
';
}

1508A/1509D. Binary Literature

题意：

在一场二进制小说写作比赛中，需要由三个长度为 (2 · n) 的字符串组成的 (3 · n) 长度的字符串（其中至少包括 (3) 个字符串的两个作为子序列）

先贴一下AC代码

void solve() {
    int n;
    string a, b, c;
    cin >> n >> a >> b >> c;
    int x = 0, y = 0, z = 0;
    for (int i = 0; i < 2 * n; ++i) {
        if (a[i] == '1') ++x;
        if (b[i] == '1') ++y;
        if (c[i] == '1') ++z;
    }
    if (x > y) swap(a, b), swap(x, y);
    if (y > z) swap(b, c), swap(y, z);
    if (x > y) swap(a, b), swap(x, y);
    char cc = '0';
    if (y > n) {
        cc = '1';
        swap(a, c), swap(x, z);
    }
    x = y      = 0;
    string ans = "";
    while (true) {
        while (x < 2 * n && a[x] != cc) ans += a[x], ++x;
        while (y < 2 * n && b[y] != cc) ans += b[y], ++y;
        if (x == 2 * n && y == 2 * n) break;
        ans += cc;
        if (x < 2 * n) x++;
        if (y < 2 * n) y++;
    }
    while (ans.size() < 3 * n) ans += '0';
    cout << ans << '
';
}