2022 International Collegiate Programming Contest, Jinan Site MKAEDGC

2022 International Collegiate Programming Contest, Jinan Site

目录

• 2022 International Collegiate Programming Contest, Jinan Site
  • VP概况
  • M - Best Carry Player
  • K - Stack Sort
  • A - Tower
  • E - Identical Parity
  • D - Frozen Scoreboard
  • G - Quick Sort
  • C - DFS Order 2

VP概况

没有罚时的情况下拿下签到，E因为打的表太小，猜错了结论，后面让队友读完大模拟来写，也很顺利，看榜发现后面的题不可做，补题的时候发现也不是1小时可以写出来的，下次比赛按照打满的思路来

M - Best Carry Player

推导得知运算顺序不影响答案，直接模拟

int n;	
void solve()
{
	cin>>n;
	vector<int> a, b, res;
	int x;	cin>>x;
	int ans = 0;
	while(x)
	{
		a.push_back(x % 10);
		x /= 10;
	}
	for(int i = 2; i <= n; i++)
	{
		cin>>x;
		b.clear();
		res.clear();
 
		while(x)
		{
			b.push_back(x % 10);
			x /= 10;
		}
 
		int len1 = a.size(), len2 = b.size();
		int add = 0;
		for(int j = 0; j < max(len1, len2); j++)
		{
			int t1 = 0, t2 = 0;
			if(j < len1)
				t1 = a[j];
			if(j < len2)
				t2 = b[j];
			int s = t1 + t2 + add;
			// cout<<s<<'\n';
			res.push_back(s % 10);
			// cout<<i<<" "<<j<<" "<<add<<'\n';
			add = s / 10;
			if(add == 1)
				ans++;
		}
		if(add)
			res.push_back(add);
		swap(a, res);
		// a = res;
	}
	cout<<ans<<'\n';
}

K - Stack Sort

看有多少个相邻段在数值上和数组下标上是连续的

const int N = 5e5 + 10;
 
int n, a[N], pos[N];
void solve()
{       
    cin>>n;
    int res = 0;
    for(int i = 1; i <= n; i++)
    {
        cin>>a[i];
        pos[a[i]] = i;
    }
    for(int i = n; i >= 1; i--)
    {
        int j = i;
        while(j >= 2 && pos[j - 1] >= pos[j])
            j--;
        res++;
        i = j;
    }
    cout<<res<<'\n';
 
    return;
}

A - Tower

如果所有数字都要变成 \(x\) ，那么就是一个中位数问题了，其区别在于除 \(2\) 的操作，除 \(2\) 代表可以有更多的中位数选择，那么可能得中位数数量有 \(O(N \log V)\) 种，然后去判断每个数到达 \(x\) 的最小操作次数，总共时间复杂度 \(N^2 \log V \times (\log V + \log N)\) ，这里包括排序的复杂度

int n, m, a[510];
vector<int> b;
 
ll check(int x)
{
	ll ans = 0;
	vector<int> c;
	// cout<<"X : ";
	for(int i = 1; i <= n; i++)
	{
		if(a[i] <= x)
		{
			c.push_back(x - a[i]);
			// cout<<x - a[i]<<"  ";
			continue;
		}
		int y = a[i];
		int cost = a[i] - x;
		int t = 0;
		while(y)
		{
			cost = min(abs(y - x) + t, cost);
			y /= 2;
			t++;
		}
		c.push_back(cost);
	}
	sort(c.begin(), c.end());
	reverse(c.begin(), c.end());
	for(int i = m; i < n; i++)
		ans += c[i];
	return ans;
}
void solve()
{
	b.clear();
	cin>>n>>m;
	for(int i = 1; i <= n; i++)
	{
		cin>>a[i];
		int x = a[i];
		while(x)
		{
			b.push_back(x);
			x /= 2;
		}	
	}
	b.push_back(0);
	sort(b.begin(), b.end());
	b.erase(unique(b.begin(), b.end()), b.end());
	ll res = 1e18;
	for(auto it : b)
		res = min(check(it), res);
	cout<<res<<'\n';
}

E - Identical Parity

我是打表找规律做的，队内的两位数学手没做出来，我先打表打出来了

规律如下，表中 \(1\) 代表 YES在 \(k\) 为奇数的时候，YES 情况成等差数列分布

\(k = 1, 10\)

\(k = 3, 111010\)

\(k = 5, 11111011100010\)

\(\dots\)

\(1\) 和 \(0\)的数量分部刚好等于 \(k + 1\) ，就很好的去做这个规律了

k:    1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25
n : 1  1
n : 2  0 1
n : 3  0 1 1
n : 4  0 1 1 1
n : 5  0 1 1 1 1
n : 6  0 1 0 1 1 1
n : 7  0 1 1 1 1 1 1
n : 8  0 1 0 1 1 1 1 1
n : 9  0 1 0 1 1 1 1 1 1
n : 10 0 1 0 1 0 1 1 1 1 1
n : 11 0 1 0 1 1 1 1 1 1 1 1
n : 12 0 1 0 1 1 1 1 1 1 1 1 1
n : 13 0 1 0 1 1 1 1 1 1 1 1 1 1
n : 14 0 1 0 1 0 1 0 1 1 1 1 1 1 1
n : 15 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1
n : 16 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1
n : 17 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 18 0 1 0 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1
n : 19 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 20 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 21 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 22 0 1 0 1 0 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1
n : 23 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 24 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 25 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

打表代码（丢失了）

AC代码：

typedef long long ll;
void solve()
{
	ll n, k;
	cin>>n>>k;
	if(n == 1)
	{
		cout<<"YES\n";
		return;
	}
	if(k == 1)
	{
		cout<<"NO\n";
		return;
	}
	if(k % 2 == 0)
	{
		cout<<"YES\n";
		return;
	}
	ll m = (n + 1) / 2;
	ll x = k;
	ll t = (n - (k - 1))  / (x + 1), r = (n - (k - 1)) % (x + 1);
	if(r != 0)	t++;
	if(r == 0) r = x + 1;
	ll d = 1ll + (t - 1) * 2;
	// cout<<"t: "<<t<<" m: "<<m<<" r:  "<<r<<"  d:  "<<d<<"\n";
	if(t <= m && r <= x + 1 - d)
		cout<<"YES\n";
	else
		cout<<"NO\n";
}

D - Frozen Scoreboard

大模拟

对于要求的总时间先减去已经确定的时间，即是封榜前AC的题的罚时，再枚举 \(S\) 其二进制下第 \(i\) 位为 \(1\) 代表通过第 \(i - 1\) 题，对于状态 \(S\) 封榜后通过的题，总时间减去\(240\) 的基础时间，再用罚时消耗总时间，剩下的时间再用封榜后的60min消耗

并不难，细节多而已，代码的注释更清楚

array<int, 3> a[15];
array<int, 4> res[15];
 
int n, m;
void solve()
{
	int ac, time;	cin>>ac>>time;
	for(int i = 0; i < m; i++)
	{
		char opt;	cin>>opt;
		if(opt == '.')	
		{
			res[i] = {0, 0, 0, 0};
			a[i] = {0, 0, 0};
		}
		else if(opt == '+')	
		{
			int x, y;	cin>>x>>opt>>y;
			time -= ((x - 1) * 20 + y);
			a[i] = {1, x, y};
			res[i] = {1, x, y, 0};
		}
		else if(opt == '?')
		{
			int x, y;	cin>>x>>y;
			a[i] = {2, x, y};
			res[i] = {2, 0, 0, 0};
		}	
		else if(opt == '-')
		{
			int x;	cin>>x;
			a[i] = {3, 0, x};
			res[i] = {3, 0, x, 0};
		}
	}
	// cout<<time<<'\n';
	for(int S = 0; S < (1 << m); S++)
	{
		bool ok = true;
		int cnt = 0;
		for(int i = 0; i < m; i++)
		{
			if((a[i][0] == 0 || a[i][0] == 3) && ((S >> i) & 1) == 0)
				ok = false;
			if(a[i][0] == 1 && ((S >> i) & 1) == 0)
				ok = false;
		}
		for(int i = 0; i < m; i++)
		{
			if(a[i][0] == 1)	cnt++;
			else if(a[i][0] == 2 && ((S >> i) & 1) == 1)	cnt++;
		}
 
		if(!ok || ac != cnt)	continue;
		int T = time;
		//  过题时间   罚时次数
		for(int i = 0; i < m; i++)
			if(((S >> i) & 1) && a[i][0] == 2)
			{
				T -= 240;
				T -= (a[i][2] - a[i][1]) * 20;
			}
		if(T < 0)	continue;
 
		// T 基础罚时
 
		
		// 优先消耗封榜的罚时 20 倍数
		// cout<<T<<'\n';
		for(int i = 0; i < m; i++)
			if(((S >> i) & 1) && a[i][0] == 2)
			{
				int c = min(a[i][1] - 1, T / 20);
				T -= c * 20; 
			}		
 
		// 消耗罚时时间
		for(int i = 0; i < m; i++)
			if(((S >> i) & 1) && a[i][0] == 2)
				T -= min(59, T);
 
 
		if(T == 0)	//输出答案
		{
			T = time;
			for(int i = 0; i < m; i++)
				if(((S >> i) & 1) && a[i][0] == 2)
				{
					T -= 240;
					T -= (a[i][2] - a[i][1]) * 20;
				}			
 
			// T 基础罚时
 
			// 优先消耗封榜的罚时 20 倍数
			for(int i = 0; i < m; i++)
				if(((S >> i) & 1) && a[i][0] == 2)
				{
					int c = min(a[i][1] - 1, T / 20);
					T -= c * 20; 
					res[i] = {1, a[i][2] - a[i][1] + c + 1, 240, 0};
					//           封榜前次数         封榜后的第几次
				}	
			// 消耗罚时时间
			for(int i = 0; i < m; i++)
				if(((S >> i) & 1) && a[i][0] == 2)
				{
					res[i][0] = 1;
					res[i][2] += min(59, T);
					T -= min(59, T);
				}
			cout<<"Yes\n";
			for(int i = 0; i < m; i++)
			{
				if(a[i][0] == 0)
					cout<<".\n";
				else if(a[i][0] == 1)
					cout<<"+ "<<res[i][1]<<"/"<<res[i][2]<<'\n';
				else if(a[i][0] == 2 && ((S >> i) & 1) == 1)
					cout<<"+ "<<res[i][1]<<"/"<<res[i][2]<<'\n';
				else if(a[i][0] == 2 && ((S >> i) & 1) == 0)
					cout<<"- "<<a[i][2]<<"\n";
				else 
					cout<<"- "<<a[i][2]<<"\n";
			}	
			return;
		}
 
	}
	cout<<"No\n";
 
}

G - Quick Sort

因为递归次数只有 \(\log n\) 次，则交换次数最多 \(n \log n\) 次，所以可以直接做

每个数字只会出现 \(1\) 次，用数据结构维护区间最大值，区间最小值即可

细节要注意的地方：par函数内，找到的下标不在范围内，那么我们就要找更大或更小的，为什么呢？因为 \(pivot\) swap了，导致 \(\geq\) 或 \(\leq\) \(pivot\)不在范围内

const int N = 5e5 + 10;
 
int n, a[N], res;
struct segtree
{
    int w1, w2;
}seg[N * 4];
 
void update(int id)
{
    seg[id].w1 = max(seg[id * 2].w1, seg[id * 2 + 1].w1);
    seg[id].w2 = min(seg[id * 2].w2, seg[id * 2 + 1].w2);
}
 
void build(int id, int l, int r)
{
    seg[id] = {0, 0};
    if(l == r)
    {
        seg[id] = {a[l], a[l]};
        return;
    }
    int mid = (l + r) >> 1;
    build(id * 2, l, mid);
    build(id * 2 + 1, mid + 1, r);
    update(id);
}
 
void change(int id, int l, int r, int pos)
{
    if(l == r)
    {
        seg[id] = {a[l], a[l]};
        return;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid)
        change(id * 2, l, mid, pos);
    else
        change(id * 2 + 1, mid + 1, r, pos);
    update(id);
}
 
int query1(int id, int l, int r, int val)
{
    if(l == r)
        return l;
    int mid = (l + r) >> 1;
    if(seg[id * 2].w1 >= val)
        return query1(id * 2, l, mid, val);
    else
        return query1(id * 2 + 1, mid + 1, r, val);
}
 
int query2(int id, int l, int r, int val)
{
    if(l == r)
        return l;
    int mid = (l + r) >> 1;
    if(seg[id * 2 + 1].w2 <= val)
        return query2(id * 2 + 1, mid + 1, r, val);
    else
        return query2(id * 2, l, mid, val);
}
 
int par(int l, int r)
{
    int tl = l - 1, tr = r + 1;
    int val = a[(l + r) / 2];
    while(1)
    {
        int p1 = query1(1, 1, n, val);
        if(p1 <= tl)    p1 = query1(1, 1, n, val + 1);
        int p2 = query2(1, 1, n, val);
        if(p2 >= tr)    p2 = query2(1, 1, n, val - 1);
        if(p1 >= p2)
            return p2;
        swap(a[p1], a[p2]);
        change(1, 1, n, p1);        change(1, 1, n, p2);
        res++;
        tl = p1, tr = p2;
    }   
}
void quicksort(int l, int r)
{
    if(l >= r || l < 0 || r < 0)  return;
    int p = par(l, r);
    quicksort(l, p);
    quicksort(p + 1, r);
}
 
void solve()
{       
    cin>>n;
    for(int i = 1; i <= n; i++)
        cin>>a[i];
    res = 0;
    build(1, 1, n);
    quicksort(1, n);
    cout<<res<<'\n';
    return;
}

C - DFS Order 2

回退背包，第一次见耶

2022 ICPC 济南站 C (回退背包) - 严格鸽的文章 - 知乎

typedef long long ll;
 
const int mod = 998244353;
const int N = 500 + 10;
ll qmi(ll a, ll b, ll mod)
{
    ll ans = 1 % mod;
    while(b)
    {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}
vector<int> e[N];
int n;
ll son[N], sz[N], fac[N];
ll f[N][N], g[N][N], h[N];
 
ll dfs1(int u, int from)
{
    son[u] = 0;
    sz[u] = 1;
    ll res = 1;
    for(auto v : e[u])
    {
        if(v == from)   continue;
        res = (res * dfs1(v, u)) % mod;
        sz[u] += sz[v];
        son[u]++;
    }
    res = (res * fac[son[u]]) % mod;
    return res;
}
 
 
void dfs(int u, int from)
{
    memset(g, 0, sizeof g);
    g[0][0] = 1;
    for(auto v : e[u])
    {
        if(v == from)   continue;
        for(int i = son[u]; i >= 1; i--)
            for(int j = sz[u]; j >= sz[v]; j--)
                g[i][j] = (g[i][j] + g[i - 1][j - sz[v]]) % mod;
    }
 
    for(auto v : e[u])
    {
        if(v == from)   continue;
        for(int i = 1; i <= son[u]; i++)
            for(int j = sz[v]; j <= sz[u]; j++)
                g[i][j] = (g[i][j] - g[i - 1][j - sz[v]] + mod) % mod;
 
        memset(h, 0, sizeof h);
        for(int i = 0; i <= son[u] - 1; i++)
            for(int k = 0; k <= sz[u]; k++)
                h[k + 1] = (h[k + 1] + (fac[i] * fac[son[u] - 1 - i] % mod) * g[i][k]) % mod;
        
        for(int i = 1; i <= n; i++)
            for(int k = 1; k <= n; k++)
                if(i + k <= n)
                    f[v][i + k] = (f[v][i + k] + f[u][i] * h[k]) % mod;
 
        for(int i = son[u]; i >= 1; i--)
            for(int j = sz[u]; j >= sz[v]; j--)
                g[i][j] = (g[i][j] + g[i - 1][j - sz[v]]) % mod;                
    }
    for(auto v : e[u])
    {
        if(v == from)   continue;
        dfs(v, u);
    }
}
 
void solve()
{       
    cin>>n;
    fac[0] = 1;
    for(int i = 1; i < N; i++)    fac[i] = (fac[i - 1] * i) % mod;
    for(int i = 1; i <= n - 1; i++)
    {
        int u, v;   cin>>u>>v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    f[1][1] = dfs1(1, 0);
    dfs(1, 0);
    for(int i = 1; i <= n; i++)
    {
        ll sum = 0;
        for(int k = 1; k <= n; k++) sum += f[i][k];
        sum %= mod;
        ll inv = qmi(sum, mod - 2, mod);
 
 
        for(int k = 1; k <= n; k++)
            cout<<(f[i][k] * f[1][1] % mod) * inv % mod<<" ";
        cout<<'\n';
    }
    return;
}