A

B

C

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>

using namespace std;

const int N = 2e6 + 10;

int read() {
	int res = 0; bool sym = 0; char ch = getchar();
	while (!isdigit(ch)) sym |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) res = (res << 3) + (res << 1) + (ch ^ 48), ch = getchar();
	return sym ? -res : res;
}

int n, vis[N];

int main() {
	int T = read();
	while (T--) {
		n = read(); int ans = 0;
		if (n == 1) {printf("1\n"); continue;}
		for (int i = 1; i < n; i++) {
			int x = read(), y = read(); vis[x]++; vis[y]++;
		}
		for (int i = 1; i <= n; i++) {
			if (vis[i] == 1) ans++;
		}
		for (int i = 1; i <= n; i++) vis[i] = 0;
		printf("%d\n", ans);
	}
}

D

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>

using namespace std;

const int N = 2e6 + 10;

int read() {
	int res = 0; bool sym = 0; char ch = getchar();
	while (!isdigit(ch)) sym |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) res = (res << 3) + (res << 1) + (ch ^ 48), ch = getchar();
	return sym ? -res : res;
}

int n, ans[N], a[N];

void dfs(int x, int cnt, int now, int l, int r, int need) { // 当前数x，cnt个1，目前枚举位置now，左右边界l，r
	if (now == r + 1) {
		if (cnt == need) ans[++n] = x; return;
	}
	if (cnt < need) dfs(x + (1 << now - 1), cnt + 1, now + 1, l, r, need);
	dfs(x, cnt, now + 1, l, r, need);
}

int main() {
	int T = read();
	ans[1] = 2; ans[2] = 12; n = 2;
	for (int i = 4; i <= 30; i++) { // 最大0容量
		if (1 << i > 1e9) break; int x = 1 << i;
		for (int j = 2; j <= i + 1 >> 1; j++) { // 后缀0的数量
			x += (1 << j);
			dfs(x, 0, j + 2, j + 2, i, j - 2);
			x -= (1 << j);
		}
	}
	sort(ans + 1, ans + n + 1);
	while (T--) {
		int l = read(), r = read();
		int x = lower_bound(ans + 1, ans + n + 1, l) - ans;
		if (ans[x] <= r) printf("%d\n", ans[x]); else printf("-1\n");
	}
}

E

F

G

H

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>

using namespace std;

const int N = 2e6 + 10, mod = 20220911;

int read() {
	int res = 0; bool sym = 0; char ch = getchar();
	while (!isdigit(ch)) sym |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) res = (res << 3) + (res << 1) + (ch ^ 48), ch = getchar();
	return sym ? -res : res;
}

string st;

int n, s[N], top;

int main() {
	cin >> st; top = 1;
	while (st != "fin") {
		if (st[0] == 'a') {cin >> st; continue;}
		if (st[0] == 'r') {top++; cin >> st; continue;}
		if (st[0] == 'l') s[top]++;
		if (st[0] == 'f') {
			int x = read();
			s[top] = 1ll * s[top] * x % mod; top--;
			s[top] = 1ll * (s[top] + s[top + 1]) % mod;
			s[top + 1] = 0;
			cin >> st;
		}
		cin >> st;
		// for (int i = 1; i <= top; i++) printf("%d ", s[i]); printf("\n");
	}
	// printf("%d\n", top);
	printf("%d", s[1]);
}

I

J

K

L