简单数学，直接暴力模拟

void solve()
{       
    ll n;   cin>>n;
    ll s = 0, base = 1;
    while(s + base < n)
    {
        s += base;
        base++;
    }
    cout<<n - s<<'\n';
    return;
}

B - The Time

模拟时间

void solve()
{       
    char a,b,c,d,e;
    cin>>a>>b>>e>>c>>d;
    int t = (a - '0') * 10 * 60 + (b - '0') * 60 
    + (c - '0') * 10 + (d - '0');
    int add;    cin>>add;
    t += add;
    int h = t / 60;
    h %= 24;
    int m = t % 60;
    if(h < 10)  cout<<0<<h;
    else   cout<<h;
    cout<<":";
    if(m < 10)  cout<<0<<m;
    else   cout<<m;
    cout<<'\n';
    return;
}

C - Not Equal on a Segment

我的做法是对于\([l, r]\) 区间，其区间最大值和最小值都等于 \(x\) ，那么一定无解，相反的，只要最大值不等于 \(x\) ，或最小值不等于 \(x\) 那么一定有解，我们用 ST表处理区间最值，并维护位置即可

看了题解发现只要维护与自己相同前一个数的下标即可

typedef long long ll;

ll n, s;
 
bool check(ll x)
{
    ll t = x;
    ll sub = 0;
    while(t)
    {
        t /= 10;
        sub += 9;
    }
    if(x - sub >= s)
        return true;
    else return false;
}
void solve()
{       
    cin>>n>>s;
    ll l = 1, r = n + 2;
    while(l < r)
    { 
        ll mid = (l + r) >> 1;
        if(check(mid))  r = mid;
        else l = mid + 1;
    }
    ll res = 0;
    if(l <= n && check(l))    res = n - l + 1;
    // cout<<l<<" "<<res<<'\n';
 
    // cout<<l<<" "<<res<<'\n';
    if(l > n)   l = n + 1;
    
    for(ll i = s; i <= l - 1; i++)
    {
        ll t = i;
        ll sub = 0;
        while(t)
        {
            sub = sub + (t % 10);
            t /= 10;
        }
        if(i - sub >= s)    res++;
 
    }
    cout<<res<<'\n';
 
    return;
}

D - Optimal Number Permutation

观察到:

\(i = n\)时贡献为 \(0\)
其他贡献为 \((n - i) \times |y_i - x_i + i - n|\)
不难发现可以构造出 \(y_i - x_i = -(i - n)\) ，使得 \(|y_i - x_i + i - n| = 0\)
打表发现，\(1\) 的距离要是 \(n - 1\)，更多的 \(i\) 的距离要是 \(n - i\)
手动放一个\(n = 9\) 的情况，\(1,3,5,7,9,7,5,3,1,2,4,6,8,8,6,4,2,9\)， \(n = 10\) 的情况，\(1,3,5,7,9,9,7,5,3,1,2,4,6,8,10,8,6,4,2,10\)
发现了奇数，偶数分开放，\(n\) 来插空（取决于 \(n\) 的奇偶性）

int n;
void solve()
{       
    cin>>n;
    for(int i = 1; i < n; i += 2)
        cout<<i<<" ";
    if(n & 1)
        cout<<n<<" ";
    for(int i = ((n - 1) % 2 == 1 ? n - 1 : n - 2); i >= 1; i -= 2)
        cout<<i<<" ";
    for(int i = 2; i < n; i += 2)
        cout<<i<<" ";
    if(n % 2 == 0)
        cout<<n<<" ";
    for(int i = ((n - 1) % 2 == 0 ? n - 1 : n - 2); i >= 1; i -= 2)
        cout<<i<<" ";
    cout<<n;
    cout<<'\n';
    
    return;
}

E - Ants in Leaves

对于相同时间在相同节点排队的贡献是排队的蚂蚁数量

luogu的题解也说得很详细了

graph 1 --- 2 1 --- 3 1 --- 4 2 --- 5 2 --- 6 2 --- 7

const int N = 5e5 + 10;


int n, m, a[N], dep[N];
vector<int> e[N];

void dfs(int u, int from)
{
    dep[u] = dep[from] + 1;
    bool ok = false;
    for(auto v : e[u])
    {
        if(v == from)   continue;
        ok = true;
        dfs(v, u);
    }
    if(!ok)
        a[++m] = dep[u];
}
void solve()
{       
    cin>>n;
    for(int i = 1; i < n; i++)
    {
        int u, v;   cin>>u>>v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dep[1] = 0;
    int res = 0;
    for(auto v : e[1])
    {
        m = 0;
        dfs(v, 1);
        sort(a + 1, a + 1 + m);
        for(int i = 1; i <= m; i++) a[i] = max(a[i - 1] + 1, a[i]);
        res = max(a[m], res);
    }
    cout<<res<<'\n';
    return;
}