总览

Complete

Solutions

A. Trippi Troppi

Solve

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#include <bits/stdc++.h>
using namespace std;
int main() {
    int t;
    cin >> t;
    string ans;
    string tmp;
    while (t--) {
        for (int i = 1; i <= 3; ++i) {
            cin >> tmp;
            ans += tmp[0];
        }
        cout << ans << '\n';
        ans = "";        
    }
    return 0;
}

B. Bobritto Bandito

Solve

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#include <bits/stdc++.h>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, m, l, r;
        cin >> n >> m >> l >> r;
        int sub_l = l, sub_r = r;
        while (sub_r - sub_l > m) {
            if (abs(sub_r) > abs (sub_l)) {
                sub_r--;
            } else {
                sub_l++;
            }
        }
        cout << sub_l << ' ' << sub_r << '\n';
    }
    return 0;
}

C. Brr Brrr Patapim

Solve

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#include <bits/stdc++.h>
using namespace std;
#define all(x) x.begin(), x.end()

int main() {
    int t;
    cin >> t;
    vector<int> ans;
    while (t--) {
        int n;
        cin >> n;
        ans.resize(2 * n, 0);
        unordered_set<int> s;

        for (int i = 0, temp; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                cin >> temp;
                if ((i + j + 1) < 2 * n) {
                    s.insert(temp);
                    ans[i + j  + 1] = temp;
                }
            }
        }
        for (int i = 1; i <= 2 * n; ++i) {
            if (s.find(i) == s.end()) {
                ans[0] = i;
            }
        }
        for_each(all(ans), [](int n) {
            cout << n << ' ';
        });
        cout << '\n';
    }
    return 0;
}

D. Tung Tung Sahur

Solve

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#include <bits/stdc++.h>
using namespace std;
void solve(int t) {
    while (t--) {
        string p, cmp;
        cin >> p >> cmp;
        int i = 0, j = 0;
        bool flag = true;
        // 使用 && 只在两个字符串都有字符时进入循环
        while (i < p.length() && j < cmp.length()) {
            int li = i, lj = j;

            while (i < p.length() && p[i] == p[li]) ++i;
            while (j < cmp.length() && cmp[j] == cmp[lj]) ++j;
            
            // 检查当前两组字符是否一致（字符必须相同）
            if (p[li] != cmp[lj]) {
                flag = false;
                break;
            }
            int cnt_p = i - li, cnt_cmp = j - lj;
            // 检查 cmp 的数量是否在合法范围内
            if (cnt_cmp < cnt_p || cnt_cmp > 2 * cnt_p) {
                flag = false;
                break;
            }
        }
        // 遍历结束后，如果两字符串未同时结束则说明不匹配
        if (i != p.length() || j != cmp.length()) {
            flag = false;
        }
        cout << (flag ? "YES" : "NO") << '\n';
    }
}
int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

    int t;
    cin >> t;
    solve(t);
    return 0;
}

E. Boneca Ambalabu

Solve

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#include <bits/stdc++.h>
using namespace std;

#define int long long
const int MAX_BITS = 31;
int t, n;

// 正解
void solve() {
    cin >> n;
    vector<int> a(n);
    for (auto &x : a) cin >> x;

    // 统计每一位上有多少个 1
    vector<int> bitCount(MAX_BITS, 0);
    for (int i = 0; i < n; ++i) {
        for (int b = 0; b < MAX_BITS; ++b) {
            if ((a[i] >> b) & 1) {
                bitCount[b]++;
            }
        }
    }

    int ans = 0;
    for (int i = 0; i < n; ++i) {
        int total = 0;
        for (int b = 0; b < MAX_BITS; ++b) {
            int cnt1 = bitCount[b];
            int cnt0 = n - cnt1;
            if ((a[i] >> b) & 1) {
                // 当前数该位为 1，与 cnt0 个 0 异或为 1
                total += cnt0 * (1LL << b);
            } else {
                // 当前数该位为 0，与 cnt1 个 1 异或为 1
                total += cnt1 * (1LL << b);
            }
        }
        ans = max(ans, total);
    }

    cout << ans << '\n';
}

// 暴力
void solve1() {
    cin >> n;
    vector<int> a(n);
    for (auto &x : a) cin >> x;

    int res = 0;
    for (int i = 0; i < n; ++i) {
        int cur = 0;
        for (int j = 0; j < n; ++j) {
            cur += (a[i] ^ a[j]);
        }
        res = max(res, cur);
    }
    cout << res << '\n';
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> t;
    while (t--) solve();
}