AtCoder Beginner Contest 350 （ABCDEF题）视频讲解-Toy模板网

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A - Past ABCs

Problem Statement

You are given a string $S$ of length $6$ . It is guaranteed that the first three characters of $S$ are ABC and the last three characters are digits.
Determine if $S$ is the abbreviation of a contest held and concluded on AtCoder before the start of this contest.
Here, a string $T$ is “the abbreviation of a contest held and concluded on AtCoder before the start of this contest” if and only if it equals one of the following $348$ strings:
ABC001, ABC002, $\ldots$ , ABC314, ABC315, ABC317, ABC318, $\ldots$ , ABC348, ABC349.
Note that ABC316 is not included.

Constraints

$S$ is a string of length $6$ where the first three characters are ABC and the last three characters are digits.

Input

The input is given from Standard Input in the following format:

$S$

Output

If $S$ is the abbreviation of a contest held and concluded on AtCoder before the start of this contest, print Yes; otherwise, print No.

Sample Input 1

ABC349

Sample Output 1

Yes

ABC349 is the abbreviation of a contest held and concluded on AtCoder last week.

Sample Input 2

ABC350

Sample Output 2

No

ABC350 is this contest, which has not concluded yet.

Sample Input 3

ABC316

Sample Output 3

No

ABC316 was not held on AtCoder.

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	string s;
	cin >> s;

	s.erase(0, 3);

	if (stoi(s) < 350 && s != "316" && s != "000") cout << "Yes" << endl;
	else cout << "No" << endl;

	return 0;
}

B - Dentist Aoki

Problem Statement

Takahashi has $N$ teeth, one in each of the holes numbered $\dots, N$ .

Dentist Aoki will perform $Q$ treatments on these teeth and holes.

In the $i$ -th treatment, hole $T_i$ is treated as follows:
If there is a tooth in hole $T_i$ , remove the tooth from hole $T_i$ .
If there is no tooth in hole $T_i$ (i.e., the hole is empty), grow a tooth in hole $T_i$ .
After all treatments are completed, how many teeth does Takahashi have?

Constraints

All input values are integers.
$\le N, Q \le 1000$
$\le T_i \le N$

Input

The input is given from Standard Input in the following format:

$N$ $Q$
$T_1$ $T_2$ $\dots$ $T_Q$

Output

Print the number of teeth as an integer.

Sample Input 1

30 6
2 9 18 27 18 9

Sample Output 1

Initially, Takahashi has $30$ teeth, and Aoki performs six treatments.
In the first treatment, hole $2$ is treated. There is a tooth in hole $2$ , so it is removed.
In the second treatment, hole $9$ is treated. There is a tooth in hole $9$ , so it is removed.
In the third treatment, hole $18$ is treated. There is a tooth in hole $18$ , so it is removed.
In the fourth treatment, hole $27$ is treated. There is a tooth in hole $27$ , so it is removed.
In the fifth treatment, hole $18$ is treated. There is no tooth in hole $18$ , so a tooth is grown.
In the sixth treatment, hole $9$ is treated. There is no tooth in hole $9$ , so a tooth is grown.
The final count of teeth is $28$ .

Sample Input 2

1 7
1 1 1 1 1 1 1

Sample Output 2

Sample Input 3

9 20
9 5 1 2 2 2 8 9 2 1 6 2 6 5 8 7 8 5 9 8

Sample Output 3

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 1e3 + 10;

int n, q;
int cnt[N];

signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	cin >> n >> q;

	for (int i = 1; i <= n; i ++)
		cnt[i] = 1;
	while (q --) {
		int tooth;
		cin >> tooth;

		cnt[tooth] ^= 1;
	}

	int res = 0;
	for (int i = 1; i <= n; i ++)
		res += cnt[i];

	cout << res << endl;

	return 0;
}

C - Sort

Problem Statement

You are given a permutation $A=(A_1,\ldots,A_N)$ of $(1,2,\ldots,N)$ .

Transform $A$ into $(1,2,\ldots,N)$ by performing the following operation between $0$ and $N - 1$ times, inclusive:
Operation: Choose any pair of integers $(i, j)$ such that $KaTeX parse error: Expected 'EOF', got '&' at position 9: 1\leq i &̲lt; j \leq N$ . Swap the elements at the $i$ -th and $j$ -th positions of $A$ .
It can be proved that under the given constraints, it is always possible to transform $A$ into $(1,2,\ldots,N)$ .

Constraints

$\leq N \leq 2\times 10^5$
$(A_1,\ldots,A_N)$ is a permutation of $(1,2,\ldots,N)$ .
All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$
$A_1$ $\ldots$ $A_N$

Output

Let $K$ be the number of operations. Print $K + 1$ lines.

The first line should contain $K$ .

The $(l + 1)$ -th line ( $1\leq l \leq K$ ) should contain the integers $i$ and $j$ chosen for the $l$ -th operation, separated by a space.

Any output that satisfies the conditions in the problem statement will be considered correct.

Sample Input 1

5
3 4 1 2 5

Sample Output 1

2
1 3
2 4

The operations change the sequence as follows:
Initially, $A = (3, 4, 1, 2, 5)$ .
The first operation swaps the first and third elements, making $A = (1, 4, 3, 2, 5)$ .
The second operation swaps the second and fourth elements, making $A = (1, 2, 3, 4, 5)$ .
Other outputs such as the following are also considered correct:

Sample Input 2

4
1 2 3 4

Sample Output 2

Sample Input 3

3
3 1 2

Sample Output 3

2
1 2
2 3

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 2e5 + 10;

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

signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	cin >> n;

	for (int i = 1; i <= n; i ++)
		cin >> a[i], p[a[i]] = i;

	std::vector<PII> res;
	for (int i = 1; i <= n; i ++) {
		if (p[i] == i) continue;
		res.emplace_back(i, p[i]);
		int pos = p[i];
		swap(p[i], p[a[i]]), swap(a[i], a[pos]);
	}

	cout << res.size() << endl;
	for (auto v : res)
		cout << v.fi << " " << v.se << endl;

	return 0;
}

D - New Friends

Problem Statement

There is an SNS used by $N$ users, labeled with numbers from $1$ to $N$ .
In this SNS, two users can become friends with each other.

Friendship is bidirectional; if user X is a friend of user Y, user Y is always a friend of user X.
Currently, there are $M$ pairs of friendships on the SNS, with the $i$ -th pair consisting of users $A_i$ and $B_i$ .
Determine the maximum number of times the following operation can be performed:
Operation: Choose three users X, Y, and Z such that X and Y are friends, Y and Z are friends, but X and Z are not. Make X and Z friends.

Constraints

$\leq N \leq 2 \times 10^5$
$\leq M \leq 2 \times 10^5$
$KaTeX parse error: Expected 'EOF', got '&' at position 12: 1 \leq A_i &̲lt; B_i \leq N$
The pairs $A_i, B_i)$ are distinct.
All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$ $M$
$A_1$ $B_1$
$\vdots$
$A_M$ $B_M$

Output

Print the answer.

Sample Input 1

Sample Output 1

Three new friendships with a friend’s friend can occur as follows:
User $1$ becomes friends with user $3$ , who is a friend of their friend (user $2$ )
User $3$ becomes friends with user $4$ , who is a friend of their friend (user $1$ )
User $2$ becomes friends with user $4$ , who is a friend of their friend (user $1$ )
There will not be four or more new friendships.

Sample Input 2

3 0

Sample Output 2

If there are no initial friendships, no new friendships can occur.

Sample Input 3

Sample Output 3

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 2e5 + 10;

int n, m;
int p[N], sz[N], cnt[N];

int find(int x) {
	if (p[x] != x) p[x] = find(p[x]);
	return p[x];
}

signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	cin >> n >> m;

	for (int i = 1; i <= n; i ++)
		p[i] = i, sz[i] = 1;

	for (int i = 1; i <= m; i ++) {
		int u, v;
		cin >> u >> v;
		int f1 = find(u), f2 = find(v);
		if (f1 == f2) {
			cnt[f1] ++;
		} else {
			p[f1] = f2;
			sz[f2] += sz[f1];
			cnt[f2] += cnt[f1];
		}
	}

	int res = 0;
	for (int i = 1; i <= n; i ++)
		if (find(i) == i) {
			res += sz[i] * (sz[i] - 1) / 2 - sz[i] + 1 - cnt[i];
		}

	cout << res << endl;

	return 0;
}

E - Toward 0

Problem Statement

You are given an integer $N$ . You can perform the following two types of operations:
Pay $X$ yen to replace $N$ with $\displaystyle\left\lfloor\frac{N}{A}\right\rfloor$ .
Pay $Y$ yen to roll a die (dice) that shows an integer between $1$ and $6$ , inclusive, with equal probability. Let $b$ be the outcome of the die, and replace $N$ with $\displaystyle\left\lfloor\frac{N}{b}\right\rfloor$ .
Here, $\lfloor s \rfloor$ denotes the greatest integer less than or equal to $s$ . For example, $\lfloor 3 \rfloor=3$ and $\lfloor 2.5 \rfloor=2$ .
Determine the minimum expected cost paid before $N$ becomes $0$ when optimally choosing operations.

The outcome of the die in each operation is independent of other rolls, and the choice of operation can be made after observing the results of the previous operations.

Constraints

$\leq N \leq 10^{18}$
$\leq A \leq 6$
$\leq X, Y \leq 10^9$
All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$ $A$ $X$ $Y$

Output

Print the answer.

Your output will be considered correct if the absolute or relative error from the true answer is at most $10^{-6}$ .

Sample Input 1

3 2 10 20

Sample Output 1

20.000000000000000

The available operations are as follows:
Pay $10$ yen. Replace $N$ with $\displaystyle\left\lfloor\frac{N}{2}\right\rfloor$ .
Pay $20$ yen. Roll a die. Let $b$ be the outcome, and replace $N$ with $\displaystyle\left\lfloor\frac{N}{b}\right\rfloor$ .
The optimal strategy is to perform the first operation twice.

Sample Input 2

3 2 20 20

Sample Output 2

32.000000000000000

The available operations are as follows:
Pay $20$ yen. Replace $N$ with $\displaystyle\left\lfloor\frac{N}{2}\right\rfloor$ .
Pay $20$ yen. Roll a die. Let $b$ be the outcome, and replace $N$ with $\displaystyle\left\lfloor\frac{N}{b}\right\rfloor$ .
The optimal strategy is as follows:
First, perform the second operation to roll the die.
If the outcome is $4$ or greater, then $N$ becomes $0$ .
If the outcome is $2$ or $3$ , then $N$ becomes $1$ . Now, perform the first operation to make $N = 0$ .
If the outcome is $1$ , restart from the beginning.

Sample Input 3

314159265358979323 4 223606797 173205080

Sample Output 3

6418410657.7408381

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

int n, a, x, y;
unordered_map<int, double> f;

double DP(int m) {
	if (!m) return 0;
	if (f.count(m)) return f[m];

	double sum = y;
	for (int i = 2; i <= 6; i ++)
		sum += DP(m / i) + y;
	sum /= 5;
	f[m] = min(sum, DP(m / a) + x);

	return f[m];
}

signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	cin >> n >> a >> x >> y;

	printf("%.8lf\n", DP(n));

	return 0;
}

F - Transpose

Problem Statement

You are given a string $S_1 S_2 S_3 \dots S_{|S|}$ consisting of uppercase and lowercase English letters, (, and ).

The parentheses in the string $S$ are properly matched.
Repeat the following operation until no more operations can be performed:
First, select one pair of integers $(l, r)$ that satisfies all of the following conditions:
$KaTeX parse error: Expected 'EOF', got '&' at position 3: l &̲lt; r$
$S_l =$ (
$S_r =$ )
Each of the characters $S_{l+1}, S_{l+2}, \dots, S_{r-1}$ is an uppercase or lowercase English letter.
Let $\overline{S_{r-1} S_{r-2} \dots S_{l+1}}$ .
Here, $\overline{x}$ denotes the string obtained by toggling the case of each character in $x$ (uppercase to lowercase and vice versa).
Then, delete the $l$ -th through $r$ -th characters of $S$ and insert $T$ at the position where the deletion occurred.
Refer to the sample inputs and outputs for clarification.
It can be proved that it is possible to remove all (s and )s from the string using the above operations, and the final string is independent of how and in what order the operations are performed.

Determine the final string.

What does it mean that the parentheses in $S$ are properly matched? First, a correct parenthesis sequence is defined as follows: A correct parenthesis sequence is a string that satisfies one of the following conditions: It is an empty string. There exists a correct parenthesis sequence $A$, and the string is formed by concatenating (, $A$, ) in this order. There exist non-empty correct parenthesis sequences $A$ and $B$, and the string is formed by concatenating $A$ and $B$ in this order. The parentheses in $S$ are properly matched if and only if the (s and )s extracted from $S$ without changing the order form a correct parenthesis sequence. ## Constraints

$\le |S| \le 5 \times 10^5$
$S$ consists of uppercase and lowercase English letters, (, and ).
The parentheses in $S$ are properly matched.

Input

The input is given from Standard Input in the following format:

$S$

Output

Print the final string.

Sample Input 1

((A)y)x

Sample Output 1

YAx

Let us perform the operations for $S =$ ((A)y)x.
Choose $l = 2$ and $r = 4$ . The substring (A) is removed and replaced with a.
After this operation, $S =$ (ay)x.
Choose $l = 1$ and $r = 4$ . The substring (ay) is removed and replaced with YA.
After this operation, $S =$ YAx.
After removing the parentheses, the string becomes YAx, which should be printed.

Sample Input 2

((XYZ)n(X(y)Z))

Sample Output 2

XYZNXYZ

Let us perform the operations for $S =$ ((XYZ)n(X(y)Z)).
Choose $l = 10$ and $r = 12$ . The substring (y) is removed and replaced with Y.
After this operation, $S =$ ((XYZ)n(XYZ)).
Choose $l = 2$ and $r = 6$ . The substring (XYZ) is removed and replaced with zyx.
After this operation, $S =$ (zyxn(XYZ)).
Choose $l = 6$ and $r = 10$ . The substring (XYZ) is removed and replaced with zyx.
After this operation, $S =$ (zyxnzyx).
Choose $l = 1$ and $r = 9$ . The substring (zyxnzyx) is removed and replaced with XYZNXYZ.
After this operation, $S =$ XYZNXYZ.
After removing the parentheses, the string becomes XYZNXYZ, which should be printed.

Sample Input 3

(((()))(()))(())

Sample Output 3

The final outcome may be an empty string.

Sample Input 4

dF(qT(plC())NnnfR(GsdccC))PO()KjsiI((ysA)eWW)ve

Sample Output 4

dFGsdccCrFNNnplCtQPOKjsiIwwEysAve

Solution

后期补一下这题目的视频

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 5e5 + 10;

int n;
string s, tmp, ans = " ";
struct Splay {
    int s[2], p, val;
    int sz, flg;
    void init(int _val, int _p) { p = _p, val = _val, sz = 1; }
}tr[N];
int root, idx;

void pushup(int u) {
    tr[u].sz = tr[tr[u].s[0]].sz + 1 + tr[tr[u].s[1]].sz;
}
void pushdown(int u) {
    if (tr[u].flg) {
        swap(tr[u].s[0], tr[u].s[1]);
        tr[tr[u].s[0]].flg ^= 1;
        tr[tr[u].s[1]].flg ^= 1;
        tr[u].flg = 0;
    }
}
void rotate(int x) {
    int y = tr[x].p, z = tr[y].p;
    int k = tr[y].s[1] == x;
    tr[x].p = z, tr[z].s[tr[z].s[1] == y] = x;
    tr[y].s[k] = tr[x].s[k ^ 1], tr[tr[x].s[k ^ 1]].p = y;
    tr[x].s[k ^ 1] = y, tr[y].p = x;
    pushup(y), pushup(x);
}
void splay(int x, int k) {
    while (tr[x].p != k) {
        int y = tr[x].p, z = tr[y].p;
        if (z != k) {
            if ((tr[z].s[0] == y) ^ (tr[y].s[0] == x)) rotate(x);
            else rotate(y);
        }
        rotate(x);
    }
    if (!k) root = x;
}
void insert(int val) {
    int u = root, p = 0;
    while (u) p = u, u = tr[u].s[val > tr[u].val];
    u = ++ idx;
    if (p) tr[p].s[val > tr[p].val] = u;
    tr[u].init(val, p);
    splay(u, 0);
}
int get_k(int k) {
    int u = root;
    while (1) {
        pushdown(u);
        if (tr[tr[u].s[0]].sz >= k) u = tr[u].s[0];
        else if (tr[tr[u].s[0]].sz + 1 == k) return u;
        else k -= tr[tr[u].s[0]].sz + 1, u = tr[u].s[1];
    }
}
void output(int u) {
    pushdown(u);
    if (tr[u].s[0]) output(tr[u].s[0]);
    if (tr[u].val >= 1 && tr[u].val <= n) cout << ans[tr[u].val];
    if (tr[u].s[1]) output(tr[u].s[1]);
}

signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	cin >> s;
	n = s.size(), s = ' ' + s;

	int cnt = 0;
	std::vector<int> l;
	std::vector<PII> seg;
	for (int i = 1; i <= n; i ++) {
		if (s[i] == '(') {
			if (cnt & 1) {
				for (auto &v : tmp)
					if (v >= 'A' && v <= 'Z')
						v = v - 'A' + 'a';
					else
						v = v - 'a' + 'A';
			}
			ans += tmp, tmp.clear();
			l.emplace_back(ans.size());
			cnt ++;
		}
		else if (isalpha(s[i])) {
			tmp += s[i];
		} else {
			if (cnt & 1) {
				for (auto &v : tmp)
					if (v >= 'A' && v <= 'Z')
						v = v - 'A' + 'a';
					else
						v = v - 'a' + 'A';
			}
			ans += tmp, tmp.clear();
			seg.emplace_back(l.back(), ans.size() - 1);
			cnt --, l.pop_back();
		}
	}
	if (tmp.size()) ans += tmp;
	n = ans.size() - 1;

	// for (auto v : seg)
	// 	cout << v.fi << " " << v.se << endl;
	for (int i = 0; i <= n + 1; i ++)
		insert(i);

	for (auto v : seg) {
		int l = v.fi, r = v.se;
		l = get_k(l), r = get_k(r + 2);
		splay(r, 0), splay(l, r);
		tr[tr[l].s[1]].flg ^= 1;
	}

	output(root);

	return 0;
}