The 14th Blue Bridge Cup Competition Software Competition Provincial Competition (C/C++ Group B)

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At present, except for questions B and F that have not been filled, the rest of the questions have been updated, and they can all be AC ​​after the unofficial data test. Welcome to exchange

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Question A. Date Statistics

1. Description of the topic

  Xiaolan now has an array with a length of 100, and the value of each element in the array is within the
range of 0 to 9. The elements in the array are as follows from left to right:

5 6 8 6 9 1 6 1 2 4 9 1 9 8 2 3 6 4 7 7 5 9 5 0 3 8 7 5 8 1 5 8 6 1 8 3 0 3 7 9 2
7 0 5 8 8 5 7 0 9 9 1 9 4 4 6 8 6 3 3 8 5 1 6 3 4 6 7 0 7 8 2 7 6 8 9 5 6 5 6 1 4 0 1
0 0 9 4 8 0 9 1 2 8 5 0 2 5 3 3

Now he wants to find some subsequences from this array that satisfy the following conditions:

• 1. The length of the subsequence is $8$；
• 2. . This subsequence can form a yyyymmdd yyyymmdd in subscript order$Date\; in\; yyyymmdd$ format, and
     the date is required to be2023 2023The date of a day in 2023, such as$20230902$$20230902，20231223$。$yyyy$ yyyy$\_$
     $yyyy$ represents the year,$mm$ means month,$dd$ represents the number of days, and when the length of the month or the number of days
     is only one digit, a leading zero is required to be added.

  Please help Xiaolan calculate how many different dates in 2023 can be found according to the above conditions.
You only need to count once for the same date.

2. Problem-solving ideas

  Consider the enumeration of the eight-layer cycle. In the middle, you need to cut branches to speed up the search steps. It is not recommended to write dfs, or you will write badly in the examination room like me. Note that the answer needs to be deduplicated. The answer is 235.

3. Template code

  Temporary change

Item B.01 Entropy of Strings

1. Description of the topic

  I have never studied mathematics.

2. Problem-solving ideas

3. Template code

Question C. Smelting Metals

1. Description of the topic

  Xiaolan has a magic furnace for ooming common $O$ smelted into a special metal$x$ . This
furnace has a property called conversion rate$V$，$V$ is a positive integer, which means consume$V$ common
metal$O$ happens to be able to smelt a special metal$X$ , when common metalOOThe number of$O$$At\; V$ ,
smelting cannot be continued.
  Now gives$N$ smelting records, each record contains two integers$A$ and$B$ , which means that AAhas been invested this time$A$ common metal$O$ , and finally smelted$B$ special metal$x$ . Each record is independentOO
that was not consumed last time$O$ will not be added to the next smelting.
  According to this$N$ smelting records, please guess the conversion rate$What\; are\; the\; minimum\; and\; maximum\; values\; of\; V$ , and the title ensures that there is no unsolvable situation in the evaluation data.

2. Problem-solving ideas

  If you have seen the example, it is obvious that the two upper and lower bounds of the answer can be directly dichotomized. Because the structure of the formula is $\frac{A}{C}=B$。$A$ is unchanged, we first consider the smallest$C$ , because it is necessary to guarantee the BBof all formulas$B$ are unchanged, if$C$ is too small, obviously there will be a certain group of$B$ increases, so it is necessary to ensure that each group is consistenta[i]/x <= b[i]. Conversely consider seeking the largest$C$ , if$C$ is too big, obviously there will be a certain group of$B$ becomes smaller, and it is necessary to ensure that each group is consistenta[i]/x >= B.

3. Template code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> PII;
#define pb(s) push_back(s);
#define SZ(s) ((int)s.size());
#define ms(s,x) memset(s, x, sizeof(s))
#define all(s) s.begin(),s.end()
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const int N = 200010;

int n;
int a[N], b[N];
bool check(LL x) {
    
    
	for (int i = 1; i <= n; ++i) {
    
    
		if (a[i] / x > b[i]) return false;
	}
	return true;
}
bool check2(LL x) {
    
    
	for (int i = 1; i <= n; ++i) {
    
    
		if (a[i] / x < b[i]) return false;
	}
	return true;
}
void solve()
{
    
    
	cin >> n;
	for (int i = 1; i <= n; ++i) cin >> a[i] >> b[i];
	LL l = 1, r = 1e9;
	while (l < r) {
    
    
		LL mid = l + r >> 1;
		if (check(mid)) r = mid;
		else l = mid + 1;
	}
	int s = r;
	l = 1, r = 1e9;
	while (l < r) {
    
    
		LL mid = l + r + 1 >> 1;
		if (check2(mid)) l = mid;
		else r = mid - 1;
	}
	cout << s << " " << r << '\n';
}
int main()
{
    
    
	ios_base :: sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t = 1;
	while (t--)
	{
    
    
		solve();
	}
	return 0;
}

Question D. Airplane Landing

1. Description of the topic

   $N$ planes are about to land at an airport with only one runway. of whichiiAirplane$i$$Ti$ arrives at the sky above the airport at any time, and when it arrives, its remaining fuel can continue to circle$Di$ units of time, that is, it
can be at$The\; landing\; starts\; at\; Ti$ time, and can be landed at$Ti+Di$ starts to land at the moment. The landing process requires$Li$
unit time.
  As soon as one plane finishes landing, another plane can immediately begin landing at the same time, but not
before the previous plane has finished landing.
  Please judge$Whether\; all\; N$ planes can land safely.

2. Problem-solving ideas

   see NNThe maximum$N$10 is ,Tand the maximum is10N. Considering the full arrangement and enumerating all the falling situations, as long as there is one matching situation, the complexity of calculation is roughly$10!\times 10\times 10$ equals3e8, the theory can pass.

3. Template code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> PII;
#define pb(s) push_back(s);
#define SZ(s) ((int)s.size());
#define ms(s,x) memset(s, x, sizeof(s))
#define all(s) s.begin(),s.end()
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const int N = 200010;


int n;
int a[N], b[N], c[N];
void solve()
{
    
    
	cin >> n;
	for (int i = 0; i < n; ++i) cin >> a[i] >> b[i] >> c[i];
	std::vector<int> d(n);
	auto check = [&]() {
    
    
		int v = 0;
		for (int i = 0; i < n; ++i) {
    
    
			int x = d[i];
			if (i == 0) {
    
    
				v = a[x] + c[x];
			} else {
    
    
				if (a[x] + b[x] < v) return false;
				v = max(v, a[x]) + c[x];
			}
		}
		return true;
	};
	iota(all(d), 0);
	bool f = false;
	do {
    
    
		if (check()) {
    
    
			f = true;
			break;
		}
	} while (next_permutation(all(d)));
	if (f) cout << "YES" << '\n';
	else cout << "NO" << '\n';
}
int main()
{
    
    
	ios_base :: sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t = 1;
	cin >> t;
	while (t--)
	{
    
    
		solve();
	}
	return 0;
}

Question E. Solitaire sequence

1. Description of the topic

   For a length KKInteger sequence of$K$$A_1,A_2,...,A_K$, which we call the Solitaire sequence if and only if $A_i$The first digit of is exactly equal to $A_{i-1}$The last digit of $(2\le i\le K)$ .
   eg$12,23,35,56,61,11$ is the sequence of Solitaire;$12,23,34,56$ is not a Solitaire sequence, because56 56The first digit of$56$$The\; last\; digit\; of\; 34$ . All lengths are1 1The integer sequence of$1\; is\; a\; Solitaire\; sequence.$
   Now given a length$N$ -like sequence$A_1,A_2,...,A_N$, please calculate how many numbers are deleted from it at least, so that the remaining sequence is a Solitaire sequence?

2. Problem-solving ideas

   When I finished reading the questions in the examination room, I felt a little strange, but found that the thinking was relatively simple. First of all, we only need to pay attention to the first digit and the last digit of a number, which is defined as $f\left[i\right]$ is the numberiithe length of the longest Solitaire sequence ending in$i$$The\; range\; of\; i$ is$[0,9]$ . For each number, set its first digit to$a$ , the number at the end is$b$ , then there is a transfer equation:
$$f[b]=max(f[b],f[a]+1)$$
   Finally at$f\left[0\right],f\left[1\right]...f\left[9\right]$ take a maximum$ans$ , the answer is$n-ans$。

3. Template code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> PII;
#define pb(s) push_back(s);
#define SZ(s) ((int)s.size());
#define ms(s,x) memset(s, x, sizeof(s))
#define all(s) s.begin(),s.end()
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const int N = 200010;

int n;
int a[N], b[N];
int f[10];
void solve()
{
    
    
	cin >> n;
	for (int i = 1; i <= n; ++i) {
    
    
		int x;
		cin >> x;
		b[i] = x % 10;
		string s = to_string(x);
		a[i] = s[0] - '0';
	}
	for (int i = 1; i <= n; ++i) {
    
    
		f[b[i]] = max(f[b[i]], f[a[i]] + 1);
	}
	int ans = 0;
	for (int i = 0; i <= 9; ++i) ans = max(ans, f[i]);
	cout << n - ans << '\n';
}
int main()
{
    
    
	ios_base :: sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t = 1;
	while (t--)
	{
    
    
		solve();
	}
	return 0;
}

Question F. Number of islands

1. Description of the topic

Temporary update, I feel bad to write

2. Problem-solving ideas

3. Template code

Question G. Substring shorthand

1. Description of the topic

   A very new shorthand method is popular in the circle of programmers: for a string, only the first and last characters are reserved, and all characters between the first and last characters are replaced by the length of this part. For example $interandtion-alization$ is abbreviated as$i18n$，$Kuberne\; tes$ (note that the hyphen is not part of the string) is abbreviated as$K$$K8s,Lanqiao$ abbreviated as$L5o$ et al.
   In this question, we stipulate that the length is greater than or equal to$All\; strings\; of\; K$ can use this shorthand method (the length is less than$K$ strings are not eligible for this shorthand).
   Given a string$S$ and two characters$c1$ and$c2$ , please calculate$How\; many\; S$ have$c1$ starts withc 2 c2Can the substring at the end of$c$$2\; use\; this\; shorthand?$

2. Problem-solving ideas

   It feels even more strange to put this Gquestion , a prefix and template question. Suppose the subscript is $The\; character\; of\; i$ is$c1$ , then we only need to count in the interval$[i+k-1,n]$ how many$c2$ is enough, prefix and preprocess$c2$ characters, just add the answer directly, note that the answer will explodeint, complexity$O\left(n\right)$

3. Template code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> PII;
#define pb(s) push_back(s);
#define SZ(s) ((int)s.size());
#define ms(s,x) memset(s, x, sizeof(s))
#define all(s) s.begin(),s.end()
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const int N = 500010;

int k;
string s;
char c1, c2;
int a[N];
void solve()
{
    
    
	cin >> k >> s >> c1 >> c2;
	int n = s.size();
	s = '?' + s;
	for (int i = 1; i <= n; ++i) {
    
    
		a[i] = (s[i] == c2);
		a[i] += a[i - 1];
	}
	LL ans = 0;
	for (int i = 1; i + k - 2 < n; ++i) {
    
    
		if (s[i] == c1) ans += a[n] - a[i + k - 2];
	}
	cout << ans << '\n';
}
int main()
{
    
    
	ios_base :: sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t = 1;
	while (t--)
	{
    
    
		solve();
	}
	return 0;
}

Question H. Integer Deletion

1. Description of the topic

   Given a length NNInteger sequence of$N$$A_1,A_2,...,A_N$. You have to repeat the following operations $K$ times:
   Each time select the smallest integer in the sequence (if there are more than one minimum value, select the first one), and delete it. And add the deleted value to the integer adjacent to it.
   OutputKKSequence after$K\; operations.$

2. Problem-solving ideas

   It seems to be a relatively typical topic. Use the priority queue to maintain, store the value and subscript, and then use an array to cntaccumulate the sum of each subscript. When the smallest value pops up and the subscript iis if it cnt[i]is not equal at this time 0, it means it The actual value needs to be added cnt[i]. We will increase it and put it back into the priority column. Note that it needs to be cleared cnt[i]. If it cnt[i]is equal to 0at this time, then we have successfully popped the current smallest element. At this time, we need to increase the value of the previous element and the next element. We need to simulate a linked list to record who the front and back elements of each element are, indicating that the subscript is pre[i]the iupper Who is an element ne[i]indicates who is the next element iwhose is, and n-kthe loop ends when the number of elements in the heap is exhausted. It is not difficult to imagine that the number of entry and exit of heap elements is linear.

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3. Template code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> PII;
#define pb(s) push_back(s);
#define SZ(s) ((int)s.size());
#define ms(s,x) memset(s, x, sizeof(s))
#define all(s) s.begin(),s.end()
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const int N = 500010;

int n, k;
int pre[N], ne[N];
LL cnt[N];
void solve()
{
    
    
	cin >> n >> k;
	priority_queue<pair<LL, int>, vector<pair<LL, int>>, greater<pair<LL, int>> >q;
	for (int i = 1; i <= n; ++i) {
    
    
		LL v;
		cin >> v;
		q.push({
    
    v, i});
		pre[i] = i - 1;
		ne[i] = i + 1;
	}
	int g = n - k;
	while (q.size() > g) {
    
    
		auto p = q.top(); q.pop();
		LL v = p.first, ix = p.second;
		if (cnt[ix]) {
    
    
			q.push({
    
    v + cnt[ix], ix});
			cnt[ix] = 0;
		} else {
    
    
			int l = pre[ix], r = ne[ix];
			cnt[l] += v;
			cnt[r] += v;
			ne[l] = r;
			pre[r] = l;
		}
	}
	std::vector<LL> a(n + 1);
	for (int i = 0; i < g; ++i) {
    
    
		auto p = q.top(); q.pop();
		a[p.second] = p.first + cnt[p.second];
	}
	for (int i = 1; i <= n; ++i) {
    
    
		if (a[i]) cout << a[i] << " ";
	}
}
int main()
{
    
    
	ios_base :: sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t = 1;
	while (t--)
	{
    
    
		solve();
	}
	return 0;
}

Question I. Scenic Tourism

1. Description of the topic

   A scenic spot has a total of $N$ attractions, numbered$1$ to$N.$ _ There are N − 1 N − 1among the attractions$N-A$ two-way shuttle bus line is connected to form a tree structure. Traveling between scenic spots can only be done by these shuttle buses, which will take a certain amount of time. Xiao Ming is a senior tour guide of this scenic spot. He takes guests to visit KK
   in a fixed order every day.$K$ attractions:$A_1,A_2,...,A_K$. Today, due to time constraints, Xiao Ming decided to skip one of the scenic spots and only take tourists to visit $K-1$ attraction. Specifically, if Xiao Ming chooses to skip$A_i$, $A_1,A_2,...,A_{i-1},A_{i+1},...,A_K,(1\le i\le K)$ .
   Please tell me about any$Ai$ , calculate how much time Xiao Ming needs to spend on the shuttle bus between attractions if he skips this attraction?

2. Problem-solving ideas

   $For\; the\; LCA$ template question (the problem is that the competition cannot write the board), you must first consider finding the distance between any two points on the tree. set point$u,$The nearest common ancestor of$v$$z$ , define$f\left[i\right]$ is point$i$ to the root node, then there is a formula:
$$dist(u,v)=f[u]+f[v]-2\ast f\left[z\right]$$
   first calculate the distance that needs to be walked when not skipping any pointansand the distance between any two adjacent jump points. Suppose we skip four points are$a,b,c,d$ . When the skipped point is$a$ , we only need toanssubtract$a$ to$The\; distance\; of\; b$ , when the skipped point is$d$ , we only need toanssubtract$c$ to$The\; distance\; of\; d$ , this is the case of the first and last two points.
   So what if the skipped point is the middle point? For example, what we skip is$b$ , then you need toanssubtract$a$ to$b$ and$b$ to$The\; distance\; of\; c$ , and also need to add$a$ to$c$ , the rest of the intermediate points are treated in the same way.

3. Template code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> PII;
#define pb(s) push_back(s);
#define SZ(s) ((int)s.size());
#define ms(s,x) memset(s, x, sizeof(s))
#define all(s) s.begin(),s.end()
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const int N = 200010;

int n, m;
std::vector<PII> e[N];
int depth[N], fa[N][32];
LL f[N];
int root;
void bfs(int root)
{
    
    
	memset(depth, 0x3f, sizeof depth);
	depth[0] = 0, depth[root] = 1;
	queue<int> q;
	q.push(root);
	while (!q.empty()) {
    
    
		auto t = q.front();
		q.pop();
		for (auto [j, c] : e[t]) {
    
    
			if (depth[j] > depth[t] + 1) {
    
    
				depth[j] = depth[t] + 1;
				q.push(j);
				fa[j][0] = t;
				for (int k = 1; k <= 15; k++) {
    
    
					fa[j][k] = fa[fa[j][k - 1]][k - 1];
				}
			}
		}
	}
}
void dfs(int u, int fa) {
    
    
	for (auto [v, c] : e[u]) {
    
    
		if (v == fa) continue;
		f[v] = f[u] + c;
		dfs(v, u);
	}
}
int lca(int a, int b) {
    
    
	if (depth[a] < depth[b]) swap(a, b);
	for (int k = 15; k >= 0; k--) {
    
    
		if (depth[fa[a][k]] >= depth[b]) {
    
    
			a = fa[a][k];
		}
	}
	if (a == b) return a;
	for (int k = 15; k >= 0; --k) {
    
    
		if (fa[a][k] != fa[b][k]) {
    
    
			a = fa[a][k];
			b = fa[b][k];
		}
	}
	return fa[a][0];
}
void solve()
{
    
    
	cin >> n >> m;
	for (int i = 0; i < n - 1; ++i) {
    
    
		int u, v , c;
		cin >> u >> v >> c;
		e[u].push_back({
    
    v, c});
		e[v].push_back({
    
    u, c});
	}
	bfs(1);
	dfs(1, -1);
	std::vector<LL> g(m + 1), w(m + 1);
	for (int i = 1; i <= m; ++i) cin >> g[i];
	LL ans = 0;
	for (int i = 1; i < m; ++i) {
    
    
		int u = g[i], v = g[i + 1];
		int z = lca(u, v);
		w[i] = f[u] + f[v] - 2 * f[z];
		ans += w[i];
	}
	for (int i = 1; i <= m; ++i) {
    
    
		if (i == 1) cout << ans - w[1] << " ";
		else if (i == m) cout << ans - w[m - 1] << " ";
		else {
    
    
			LL res = ans - w[i] - w[i - 1];
			int u = g[i - 1], v = g[i + 1];
			int z = lca(u, v);
			res += f[u] + f[v] - 2 * f[z];
			cout << res << " ";
		}
	}
}
int main()
{
    
    
	ios_base :: sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t = 1;
	while (t--)
	{
    
    
		solve();
	}
	return 0;
}

Question J. Cutting Trees

1. Description of the topic

   given a tree by $A\; tree\; composed\; of\; n$ nodes and$m$ non-repeating random number pairs$(a1,b1),(a_2,b_2),...,(a_m,b_m)$ , among which$ai$ are different from each other,$b_i$different from each other, $a_i,b_j(1\le i,j\le m)$ .
   Xiao Ming wants to know whether it is possible to choose a tree edge to cut, so that for each$(ai,bi)$ full$ai$ sum$bi$ is not connected, if possible, output the number of the edge that should be broken (the number is from$1$ ), otherwise output$-1$。

2. Problem-solving ideas

LCA LCA    Again$LCA$ template question, but I have never done it before (I can’t write$LCA)$ . Consider a pair of unordered pairs of points$x$和$y$ , if we cut off an edge to make these two points disconnected, then this edge must be from$x$到yyA point on the$y$$x$到yyThe edge weights of the$y$1 path are added . We can usetree difference for this operation. for$m$ unordered numbers operate in this way for us, and finally if the weight of a certain edge is$m$ means that it meets the conditions. We select the side with the largest number that meets the conditions as the answer. If there is no weight, the value is$The\; edge\; of\; m$ means that there is no solution.

3. Template code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> PII;
#define pb(s) push_back(s);
#define SZ(s) ((int)s.size());
#define ms(s,x) memset(s, x, sizeof(s))
#define all(s) s.begin(),s.end()
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const int N = 200010;

int n, m;
std::vector<int> e[N];
int depth[N], fa[N][32];
int f[N];
int root;
int ans;
map<PII, int> mp;
void bfs(int root)
{
    
    
	ms(depth, 0x3f);
	depth[0] = 0, depth[root] = 1;
	queue<int> q;
	q.push(root);
	while (!q.empty()) {
    
    
		auto t = q.front();
		q.pop();
		for (int j : e[t]) {
    
    
			if (depth[j] > depth[t] + 1) {
    
    
				depth[j] = depth[t] + 1;
				q.push(j);
				fa[j][0] = t;
				for (int k = 1; k <= 15; k++) {
    
    
					fa[j][k] = fa[fa[j][k - 1]][k - 1];
				}
			}
		}
	}
}
int lca(int a, int b) {
    
    
	if (depth[a] < depth[b]) swap(a, b);
	for (int k = 15; k >= 0; k--) {
    
    
		if (depth[fa[a][k]] >= depth[b]) {
    
    
			a = fa[a][k];
		}
	}
	if (a == b) return a;
	for (int k = 15; k >= 0; --k) {
    
    
		if (fa[a][k] != fa[b][k]) {
    
    
			a = fa[a][k];
			b = fa[b][k];
		}
	}
	return fa[a][0];
}
int dfs(int u, int fa) {
    
    
	int res = f[u];
	for (auto v : e[u]) {
    
    
		if (v == fa) continue;
		int g = dfs(v, u);
		if (g == m) {
    
    
			ans = max(ans, mp[ {
    
    v, u}]);
		}
		res += g;
	}
	return res;
}
void solve()
{
    
    
	cin >> n >> m;
	for (int i = 0; i < n - 1; ++i) {
    
    
		int u, v;
		cin >> u >> v;
		mp[ {
    
    u, v}] = mp[ {
    
    v, u}] = i + 1;
		e[u].push_back(v);
		e[v].push_back(u);
	}
	bfs(1);
	for (int i = 0; i < m; ++i) {
    
    
		int u, v;
		cin >> u >> v;
		int z = lca(u, v);
		f[u]++;
		f[v]++;
		f[z] -= 2;
	}
	dfs(1, -1);
	cout << (ans == 0 ? -1 : ans) << '\n';
}
int main()
{
    
    
	ios_base :: sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t = 1;
	while (t--)
	{
    
    
		solve();
	}
	return 0;
}