A题：简单判断三个字符是否一致

#include <iostream>
#include <string>
using namespace std;

int main() {
    string s;
    cin >> s;
    if (s[0] == s[1] && s[1] == s[2])
        cout << "Won" << endl;
    else
        cout << "Lost" << endl;
    return 0;
}

B题：遍历计算酒精含量，寻找酒醉点

#include <iostream>
using namespace std;
typedef long long ll;

int main() {
    int n, x;
    cin >> n >> x;
    int ans = -1;
    ll total = 0;
    for (int i = 0; i < n; i++) {
        int v, p;
        cin >> v >> p;
        total += v * p;
        if (ans == -1 && total > x * 100) {
            ans = i + 1;
        }
    }
    cout << ans << endl;
    return 0;
}

C题：对每个位置向左右扩展，找到第一个比当前值小的位置，计算区间面积取最大值

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAXN = 1000005;

int arr[MAXN], leftB[MAXN], rightB[MAXN];

int main() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> arr[i];
    vector<pair<int, int>> st;
    st.push_back({0, 0});
    for (int i = 1; i <= n; ++i) {
        while (st.back().first >= arr[i]) st.pop_back();
        leftB[i] = st.back().second;
        st.push_back({arr[i], i});
    }
    st.clear();
    st.push_back({0, n + 1});
    for (int i = n; i >= 1; --i) {
        while (st.back().first >= arr[i]) st.pop_back();
        rightB[i] = st.back().second;
        st.push_back({arr[i], i});
    }
    ll best = 0;
    for (int i = 1; i <= n; ++i) {
        best = max(best, (ll)arr[i] * (rightB[i] - leftB[i] - 1));
    }
    cout << best << endl;
    return 0;
}

D题：状态DP，记录当前结果为真和假的方案数

#include <iostream>
#include <string>
using namespace std;
typedef long long ll;
const int MAXN = 1000005;

ll trueCnt[MAXN], falseCnt[MAXN];

int main() {
    int n;
    cin >> n;
    trueCnt[0] = falseCnt[0] = 1;
    for (int i = 1; i <= n; ++i) {
        string op;
        cin >> op;
        if (op == "AND") {
            trueCnt[i] = 2 * trueCnt[i - 1] + falseCnt[i - 1];
            falseCnt[i] = falseCnt[i - 1];
        } else {
            trueCnt[i] = trueCnt[i - 1];
            falseCnt[i] = trueCnt[i - 1] + 2 * falseCnt[i - 1];
        }
    }
    cout << trueCnt[n] << endl;
    return 0;
}

E题：维护变换函数，快速回答多个点在不同操作后的坐标

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAXN = 1000005;

ll xs[MAXN], ys[MAXN], op1[MAXN], op2[MAXN];
pair<ll, ll> result[MAXN];

ll eval(ll idx, pair<ll, pair<ll, ll>> func) {
    ll val = (func.first == 0) ? xs[idx] : ys[idx];
    return val * func.second.first + func.second.second;
}

int main() {
    ll n, m, q;
    cin >> n;
    for (ll i = 1; i <= n; ++i) cin >> xs[i] >> ys[i];
    cin >> m;
    for (ll i = 1; i <= m; ++i) {
        cin >> op1[i];
        if (op1[i] >= 3) cin >> op2[i];
    }
    cin >> q;
    vector<pair<ll, pair<ll, ll>>> queries;
    for (ll i = 0; i < q; ++i) {
        ll a, b;
        cin >> a >> b;
        queries.push_back({a, {b, i}});
    }
    sort(queries.begin(), queries.end());
    pair<ll, pair<ll, ll>> fx = {0, {1, 0}};
    pair<ll, pair<ll, ll>> fy = {1, {1, 0}};
    ll idx = 0;
    for (ll i = 0; i <= m; ++i) {
        if (op1[i] == 1) {
            swap(fx, fy);
            fy.second.first *= -1;
            fy.second.second *= -1;
        } else if (op1[i] == 2) {
            swap(fx, fy);
            fx.second.first *= -1;
            fx.second.second *= -1;
        } else if (op1[i] == 3) {
            fx.second.first *= -1;
            fx.second.second = 2 * op2[i] - fx.second.second;
        } else if (op1[i] == 4) {
            fy.second.first *= -1;
            fy.second.second = 2 * op2[i] - fy.second.second;
        }
        while (idx < q && queries[idx].first == i) {
            ll id = queries[idx].second.second;
            ll pos = queries[idx].second.first;
            result[id].first = eval(pos, fx);
            result[id].second = eval(pos, fy);
            idx++;
        }
    }
    for (ll i = 0; i < q; ++i) {
        cout << result[i].first << ' ' << result[i].second << endl;
    }
    return 0;
}

F题：使用期望递推，处理特殊位置回退到起点的情况

#include <iostream>
#include <set>
#include <cmath>
#include <cstdio>
using namespace std;
const int MAXN = 1000005;
pair<double, double> dp[MAXN];

int main() {
    int n, m, k;
    cin >> n >> m >> k;
    set<int> trap;
    for (int i = 0; i < k; ++i) {
        int x;
        cin >> x;
        trap.insert(x);
    }
    dp[n] = {0, 0};
    pair<double, double> sum = {0, 0};
    for (int i = n - 1; i >= 0; --i) {
        sum.first += dp[i + 1].first;
        sum.second += dp[i + 1].second;
        sum.first -= dp[min(n, i + m + 1)].first;
        sum.second -= dp[min(n, i + m + 1)].second;
        if (trap.count(i)) {
            dp[i] = {1, 0};
        } else {
            dp[i].first = sum.first / m;
            dp[i].second = sum.second / m + 1;
        }
    }
    if (fabs(dp[0].first - 1.0) < 1e-12) {
        cout << -1 << endl;
    } else {
        printf("%.4f\n", dp[0].second / (1 - dp[0].first));
    }
    return 0;
}