A. Mixed Messages

dp[i][j]表示前i位，选择\(j\)个移动到一起的最小花费。

#include<bits/stdc++.h>

using namespace std;

#define int long long
constexpr int inf = 1E9;


int32_t main() {
    int n;
    string s;
    cin >> n >> s;
    const string t = "spbsu";
    array<int, 6> dp;
    dp.fill(inf), dp[0] = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 4; j >= 0; j--)
            if (t[j] == s[i])
                dp[j + 1] = min(dp[j + 1], dp[j] + (j < 2 ? -1 : j > 2 ? 1 : 0) * i);
    }
    cout << dp[5] - 6 << "\n";
    return 0;
}

B. I Flipped The Calendar...

直接模拟这个过程

#include <bits/stdc++.h>

using namespace std;

set<int> s;
int m[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

// 1791 61

int32_t main() {
    for (int i = 2000; i <= 2037; i += 4) s.insert(i);
    for (int i = 2000; i >= 1970; i -= 4) s.insert(i);
    int Y = 1970, Date = 3;
    int y;
    cin >> y;
    for (int i = 1970; i < y; i++) {
        Date = (Date + 365 + s.count(i)) % 7;
    }
    int res = 0;
    for (int i = 1, M, ans; i <= 12; i++) {
        M = m[i];
        if (i == 2) M += s.count(y);

        ans = 1, Date = (Date + 1) % 7;
        for (int j = 2; j <= M; j++, Date = (Date + 1) % 7) {
            if( Date == 6 ) cout << "\n";
            if (Date == 0) ans++;

        }
        res += ans;
    }
    cout << res << "\n";
    return 0;
}

I. Password

#include <bits/stdc++.h>

using namespace std;

#define int long long


int32_t main() {
    int n;
    cin >> n;
    if (n & 1) {
        int m = (n + 1) / 2;
        cout << (m + 2) / 3 * 2 - 1;
    } else {
        int m = n / 2;
        cout << (m + 2) / 3 * 2;
    }
    cout << " " << n << "\n";
    return 0;
}

J. Transport Pluses

可以注意到的是，任意两个加号是一定相交的，所以最多只用经过两个加号，这样的话我们只需要枚举中间经过那些加号即可。

#include<bits/stdc++.h>

using namespace std;

//#define int long long

constexpr int inf = 1E9;

using vi = vector<int>;


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int n, t;
    cin >> n >> t;

    int xh, yh;
    cin >> xh >> yh;

    int xe, ye;
    cin >> xe >> ye;

    vi x(n), y(n);

    for (int i = 0; i < n; i++)
        cin >> x[i] >> y[i];

    double ans = sqrt((xh - xe) * (xh - xe) + (yh - ye) * (yh - ye));
    vector<array<int, 3>> path;
    path.push_back({0, xe, ye});

    for (int i = 0, cost; i < n; i++)
        for (int j = 0; j < n; j++) {
            if (i == j) {
                cost = min(abs(xh - x[i]), abs(yh - y[i])) + t + min(abs(xe - x[i]), abs(ye - y[i]));
                if (cost >= ans) continue;

                ans = cost, path.clear();
                if (abs(xh - x[i]) < abs(yh - y[i]))
                    path.push_back({0, x[i], yh});
                else
                    path.push_back({0, xh, y[i]});
                if (abs(xe - x[i]) < abs(ye - y[i]))
                    path.push_back({i + 1, x[i], ye});
                else
                    path.push_back({i + 1, xe, y[i]});
                path.push_back({0, xe, ye});
            } else {
                cost = min(abs(xh - x[i]), abs(yh - y[i])) + t + t + min(abs(xe - x[j]), abs(ye - y[j]));
                if (cost >= ans) continue;
                ans = cost, path.clear();
                if (abs(xh - x[i]) < abs(yh - y[i]))
                    path.push_back({0, x[i], yh});
                else
                    path.push_back({0, xh, y[i]});
                path.push_back({i + 1, x[i], y[j]});
                if (abs(xe - x[j]) < abs(ye - y[j]))
                    path.push_back({j + 1, x[j], ye});
                else
                    path.push_back({j + 1, xe, y[j]});
                path.push_back({0, xe, ye});
            }
        }
    cout << fixed << setprecision(10) << ans << "\n" << path.size() << "\n";
    for (auto [a, b, c]: path)
        cout << a << " " << b << " " << c << "\n";
    return 0;
}