比赛概括：

\(\mathrm{sum}=100+50+80\)

我的 AK 啊呜呜呜呜。

T1 【NOIP2016提高A组模拟9.2】积木：

题目大意：

每秒中有一定的概率浮现幻象，持续 \(x\) 秒的幻象将产生 \(x^2\) 的幻象值。在 \(N\) 秒内期望产生的幻象值是多少。

思路：

麻了，原题。

T2 【NOIP2016提高A组模拟9.4】树上摩托：

题目大意：

把树去掉任意条边，拆成若干棵新树，求让每棵新树的节点数相同的方案数。

思路：

两个性质：

1. 只要新树大小能整除于以当前结点为根的树的大小能够，他就可以拆。
2. 对于每一种新树大小，它只能有一个方案。

代码：

纪中机子日常爆栈。

const int N = 1e6 + 10;

inline ll Read()
{
	ll x = 0, f = 1;
	char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = -f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
	return x * f;
}

int n;
int head[N], tot;
struct edge
{
	int to, nxt;
}e[N << 1];

void add(int u, int v)
{
	e[++tot] = (edge){v, head[u]}, head[u] = tot;
}

int siz[N], fa[N];
int q[N], H, T;
inline void bfs(int s)
{
	H = 1, T = 0;
	q[++T] = s;
	fa[1] = 0;
	while(H <= T)
	{
		int u = q[H++];
		siz[u] = 1;
		for (int i = head[u]; i; i = e[i].nxt)
		{
			int v = e[i].to;
			if (~fa[v]) continue;
			fa[v] = u;
			q[++T] = v;
		}
	} 
	for (int i = T; i; i--)
	{
		int u = fa[q[i]], v = q[i];
		siz[u] += siz[v];
	}
}

int ans;

int main()
{
	memset(fa, -1, sizeof fa);
	n = Read();
	for (int i = 1; i < n; i++)
	{
		int u = Read(), v = Read();
		add(u, v), add(v, u);
	}
	bfs(1);
	for (int i = 1; i * i <= n; i++)
	{
		if (n % i) continue;
		int cnt1 = 0, cnt2 = 0;
		for (int j = 1; j <= n; j++)
			cnt1 += siz[j] % i == 0,
			cnt2 += siz[j] % (n / i) == 0;
		ans += (cnt1 == n / i) + (cnt2 == i);
		if(i * i == n) ans--;
	}
	printf ("%d\n", ans);
	return 0;
}

T3 【GDOI2017模拟9.4】矩阵：

题目大意：

思路：

先用把 \(\mathrm{mina}\times\mathrm{minb}\) 的子矩阵全部存进堆里。然后每次取出最优的，并把其向右向下延申的矩阵再存进堆。相同的矩阵用 hash 维护。

这个找第 \(k\) 大（小）的想法很像【Luogu P2048】[NOI2010] 超级钢琴和【Luogu P5283】[十二省联考2019]异或粽子。

代码：

#define num(i,j) (i-1) * m + j

inline ll Read()
{
	ll x = 0, f = 1;
	char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = -f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
	return x * f;
}

int n, m, mina, minb, k;
ll sum[N][N];

struct Matrix
{
	int x1, y1, x2, y2; ll val;
	Matrix(int X1, int Y1, int X2, int Y2)
	{
		x1 = X1, y1 = Y1, x2 = X2, y2 = Y2;
		val = sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
	}
	bool operator < (const Matrix &a) const
	{
		return val > a.val;
	}
};

priority_queue<Matrix> q;
map<pair<int, int>, int> hash;

int main()
{
	n = Read(), m = Read(), mina = Read(), minb = Read(), k = Read();
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + Read();
	
	for (int i = mina; i <= n; i++)
		for (int j = minb; j <= m; j++)
			q.push(Matrix(i - mina + 1, j - minb + 1, i, j));
	
	for (int i = 1; i < k; i++)
	{
		if (q.empty())
		{
			puts("-1");
			return 0;
		}
		Matrix tmp = q.top(); q.pop();
		int x1 = tmp.x1, y1 = tmp.y1, x2 = tmp.x2, y2 = tmp.y2;
		if(x2 < n && !hash[make_pair(num(x1, y1), num(x2 + 1, y2))]) 
			q.push(Matrix(x1, y1, x2 + 1, y2)),
			hash[make_pair(num(x1, y1), num(x2 + 1, y2))] = 1;
			
		if(y2 < m && !hash[make_pair(num(x1, y1), num(x2, y2 + 1))]) 
			q.push(Matrix(x1, y1, x2, y2 + 1)),
			hash[make_pair(num(x1, y1), num(x2, y2 + 1))] = 1;
	}
	if (q.empty())
	{
		puts("-1");
		return 0;
	}
	printf ("%lld\n", q.top().val);
	return 0;
}

标签：12,夏纪,Read,2021.8,ll,while,新树,&&,getchar
来源： https://www.cnblogs.com/GJY-JURUO/p/15132939.html