370

A-C

A

void solve() {
    int l, r;cin >> l >> r;
    if (l == r) {
        cout << "Invalid\n";
    } else if (l) {
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }
}

B

int a[110][110];
void solve() {
    int n;cin >> n;
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= i;j++) {
            cin >> a[i][j];
        }
    }
    int res = a[1][1];
    for (int i = 2;i <= n;i++) {
        if (i >= res) {
            res = a[i][res];
        } else {
            res = a[res][i];
        }
    }
    cout << res << '\n';
}

C

void solve() {
    string s, t;cin >> s >> t;
    vector<string> ans;
    for (int i = 0;i < s.size();i++) {
        if (s[i] != t[i]) {
            if (s[i] > t[i]) {
                s[i] = t[i];
                ans.push_back(s);
            }
        }
    }
    for (int i = s.size() - 1;i >= 0;i--) {
        if (s[i] != t[i]) {
            s[i] = t[i];
            ans.push_back(s);
        }
    }
    cout << ans.size() << '\n';
    for (auto x : ans)cout << x << '\n';
}

D - Cross Explosion

投放炸弹，若当前位置无炸弹则向四周散开直到遇到没有被炸过的位置为止。求最后剩几个位置没有被炸过

Solution

有想到这种方向，但还是没写出来。

一个新的思路！直接枚举两边的端点然后二分找到相应下标即可。

void solve() {
    int n, m, q;cin >> n >> m >> q;
    vector<set<int>> g1(n + 1), g2(m + 1);
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= m;j++) {
            g1[i].insert(j);g2[j].insert(i);
        }
    }
    auto erase = [&](int i, int j) {
        g1[i].erase(j);g2[j].erase(i);
        };
    while (q--) {
        int x, y;cin >> x >> y;
        if (g1[x].count(y)) {
            erase(x, y);continue;
        }

        // 上
        {
            auto it = g2[y].lower_bound(x);
            if (it != begin(g2[y])) erase(*prev(it), y);
        }
        // 下
        {
            auto it = g2[y].lower_bound(x);
            if (it != end(g2[y])) erase(*it, y);
        }
        // 左
        {
            auto it = g1[x].lower_bound(y);
            if (it != begin(g1[x])) erase(x, *prev(it));
        }
        // 右
        {
            auto it = g1[x].lower_bound(y);
            if (it != end(g1[x])) erase(x, *it);
        }
    }
    int ans = 0;
    for (int i = 1;i <= n;i++)ans += g1[i].size();
    cout << ans << '\n';
}

后面的题目暂时不管了