A题 ： 超级闪光牛可乐

点：不会出现输出-1的情况，直接逮一个输出1000次即可

void solve() {
    int x; cin >> x;
    int n; cin >> n;
    int val; char c;
    while (n -- ) {
        cin >> c >> val;
    }
    for (int i = 1; i <= 1000; i ++ ) cout << c;
    cout << endl;
     
}

B题 ： 人列计算机

点： getline进行输入

#include <bits/stdc++.h>
using namespace std;
int main() {
    string a[6];
    for (int i = 1; i <= 5; i ++ ) getline(cin, a[i]);
    int v1 = a[2][0] - '0', v2 = a[4][0] - '0', v3 = a[3][0] - '0';
    if (a[3][5] == 38) {
        cout << (v1 & v2) << endl;
    }else if (a[3][5] == '=') {
        cout << (v1 | v2) << endl;
    }else {
        cout << !v3 << endl;
    }
    return 0;
}

C题 ： 时间管理大师

点：全化成分钟

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n; cin >> n;
    set<int> s;
    while (n -- ) {
        int h, m;
        cin >> h >> m;
        int now = h * 60 + m;
        s.insert(now - 1), s.insert(now - 3), s.insert(now - 5);
    }
    cout << s.size() << endl;
    for (int t : s) cout << t / 60 << ' ' << t % 60 << endl;
    return 0;
}

D题 ： 我不是大富翁

点： 背包

#include <bits/stdc++.h>
using namespace std;
const int N = 5010;
int st[N], backup[N];
int main() {
    int n, m; cin >> n >> m;
    st[1] = backup[1] = 1;
    for (int i = 1; i <= m; i ++ ) {
        int t; cin >> t;
        t %= n;
        for (int j = 1; j <= n; j ++ ) {
            if (st[j] == i) {
                int x, y;
                x = j - t;
                while (x < 1) x += n;
                y = j + t;
                while (y > n) y -= n;
                backup[x] = backup[y] = i + 1;
            }
        }
        memcpy(st, backup, sizeof st);
    }
    if (st[1] == m + 1) cout << "YES" << endl;
    else cout << "NO" << endl;
    return 0;
}

E题 ： 多重映射

点： 正难则反

#include <bits/stdc++.h>
using namespace std;
inline void solve() {
    int n, m; cin >> n >> m;
    vector<int> in(n), l(n), r(n);
    map<int, int> fa; 
    for (int i = 0; i < n; i ++ ) cin >> in[i];
    for (int i = 0; i < m; i ++ ) cin >> l[i] >> r[i];
    for (int i = m - 1; i >= 0; i -- ) {
        if (fa.count(r[i])) {
            fa[l[i]] = fa[r[i]];
        }else {
            fa[l[i]] = r[i];
        }
    }
    for (int i = 0; i < n; i ++ ) {
        cout << (fa[in[i]] ? fa[in[i]] : in[i]) << " \n"[i == n - 1];
    }
}
int main() {
    int _; cin >> _;
    while (_ -- ) solve();
    return 0;
}

其他