周六训练-1018

C Cut

我的质因数分解不是在杨老师这里学的，所以我写挂了，杨老师好闪拜谢杨老师

#include <iostream>
#define int long long

using namespace std;

const int MaxN = 1e5 + 10;

int f[MaxN][20], mn[MaxN][20], nxt[MaxN], pri[MaxN], a[MaxN], p[MaxN], tot, n, q;
bool vis[MaxN];

int qmn(int x, int len, int res = 1e9) {
	for (int i = 19; i >= 0; i--) {
		if (len & (1 << i)) {
			res = min(res, mn[x][i]), x += (1 << i);
		}
	}
	return res;
}

int GO(int x, int len) {
	for (int i = 19; i >= 0; i--) {
		if (len & (1 << i)) {
			x = f[x][i];
		}
	}
	return x;
}

signed main() {
	ios::sync_with_stdio(0), cin.tie(0);
	cin >> n >> q;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = 1; i < MaxN; i++) {
		nxt[i] = n + 1;
	}
	for (int i = n; i >= 1; i--) {
		p[i] = n + 1;
		for (int j = 2; j * j <= a[i]; j++) {
			if (a[i] % j == 0) {
				p[i] = min(p[i], nxt[j]), nxt[j] = i;
				while (a[i] % j == 0) {
					a[i] /= j; 
				}
			}
		}
		if (a[i] > 1) p[i] = min(p[i], nxt[a[i]]), nxt[a[i]] = i;
		mn[i][0] = p[i];
	}
	for (int i = 1; i <= 19; i++) {
		for (int j = 1; j + (1 << i) - 1 <= n; j++) {
			mn[j][i] = min(mn[j][i - 1], mn[j + (1 << (i - 1))][i - 1]);
		}
	}
	for (int i = 1; i <= n; i++) {
		f[i][0] = qmn(i, p[i] - 1 - i + 1);
	}
	f[n + 1][0] = n + 1;
	for (int i = 1; i <= 19; i++) {
		for (int j = 1; j <= n + 1; j++) {
			f[j][i] = f[f[j][i - 1]][i - 1];
		}
	}
	for (int u, v, l, r; q; q--) {
		cin >> u >> v, l = 0, r = MaxN - 1;
		while (l < r) {
			int mid = l + r >> 1;
			GO(u, mid) > v ? r = mid : l = mid + 1; 
		}
		cout << l << '\n';
	}
	return 0;
}

D Distinctive Roots in a Tree

我被误导了，我以为是用相等的颜色就将路径标记，但是由于这个会有多个相等的所以不能做。

我们从整体到个体来思考，我们考虑单个点，他如果儿子的某一个子树内有和它相同的，那么这个子树外面的都要标记为不可能。

如果当前点子树外有和它相同的，那么子树内的都要标记为不可能。

#include <iostream>
#include <vector>
#include <map>

using namespace std;

const int MaxN = 2e5 + 10;

int dfn[MaxN], sz[MaxN], sum[MaxN], d[MaxN], id, a[MaxN], n;
vector<int> g[MaxN];
map<int, int> cnt, tot;

void DFS(int x, int fa = 0) {
	dfn[x] = ++id;
	int pre = cnt[a[x]];
	cnt[a[x]]++, sz[x] = 1;
	for (int i : g[x]) {
		if (i == fa) continue;
		int tmp = cnt[a[x]];
		DFS(i, x); 
		sz[x] += sz[i];
		if (tmp != cnt[a[x]]) {
			d[1]++, d[dfn[i]]--, d[dfn[i] + sz[i]]++;
		}
	}
	if (cnt[a[x]] - pre < tot[a[x]]) {
		d[dfn[x]]++, d[dfn[x] + sz[x]]--;
	}
}

int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i], tot[a[i]]++;
	}
	for (int i = 1, u, v; i < n; i++) {
		cin >> u >> v;
		g[u].push_back(v), g[v].push_back(u);
	}
	DFS(1);
	for (int i = 1; i <= n; i++) {
		d[i] += d[i - 1];
		d[0] += !d[i];
	}
	cout << d[0] << '\n';
	return 0;
}