AtCoder Beginner Contest 359（ABCDEFG题）视频讲解

A - Count Takahashi

Problem Statement

You are given $N$ strings.
The $i$ -th string $S_i$ $\leq i \leq N)$ is either Takahashi or Aoki.
How many $i$ are there such that $S_i$ is equal to Takahashi?

Constraints

$\leq N \leq 100$
$N$ is an integer.
Each $S_i$ is Takahashi or Aoki. $\leq i \leq N)$

Input

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

$N$
$S_1$
$S_2$
$\vdots$
$S_N$

Output

Print the count of $i$ such that $S_i$ is equal to Takahashi as an integer in a single line.

Sample Input 1

3
Aoki
Takahashi
Takahashi

Sample Output 1

$S_2$ and $S_3$ are equal to Takahashi, while $S_1$ is not.
Therefore, print 2.

Sample Input 2

2
Aoki
Aoki

Sample Output 2

It is possible that no $S_i$ is equal to Takahashi.

Sample Input 3

20
Aoki
Takahashi
Takahashi
Aoki
Aoki
Aoki
Aoki
Takahashi
Aoki
Aoki
Aoki
Takahashi
Takahashi
Aoki
Takahashi
Aoki
Aoki
Aoki
Aoki
Takahashi

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;

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

	int n;
	cin >> n;

	int res = 0;
	for (int i = 1; i <= n; i ++) {
		string s;
		cin >> s;
		if (s == "Takahashi") res ++;
	}
	cout << res << endl;

	return 0;
}

There are $2 N$ people standing in a row, and the person at the $i$ -th position from the left is wearing clothes of color $A_i$ . Here, the clothes have $N$ colors from $1$ to $N$ , and exactly two people are wearing clothes of each color.
Find how many of the integers $i=1,2,\ldots,N$ satisfy the following condition:
There is exactly one person between the two people wearing clothes of color $i$ .

Constraints

$\leq N \leq 100$
$\leq A_i \leq N$
Each integer from $1$ through $N$ appears exactly twice in $A$ .
All input values are integers.

Input

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

$N$
$A_1$ $A_2$ $\ldots$ $A_{2N}$

Output

Print the answer.

Sample Input 1

3
1 2 1 3 2 3

Sample Output 1

There are two values of $i$ that satisfy the condition: $1$ and $3$ .
In fact, the people wearing clothes of color $1$ are at the 1st and 3rd positions from the left, with exactly one person in between.

Sample Input 2

2
1 1 2 2

Sample Output 2

There may be no $i$ that satisfies the condition.

Sample Input 3

4
4 3 2 3 2 1 4 1

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;

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

	int n;
	cin >> n;

	std::vector<int> a(2 * n);
	for (int i = 0; i < 2 * n; i ++)
		cin >> a[i];

	int res = 0;
	for (int i = 1; i < 2 * n - 1; i ++)
		if (a[i - 1] == a[i + 1])
			res ++;

	cout << res << endl;

	return 0;
}

C - Tile Distance 2

Problem Statement

The coordinate plane is covered with $2\times1$ tiles. The tiles are laid out according to the following rules:
For an integer pair $(i, j)$ , the square ${i,j}=\lbrace(x,y)\mid i\leq x\leq i+1\wedge j\leq y\leq j+1\rbrace$ is contained in one tile.
When $i + j$ is even, $A _ {i,j}$ and $A _ {i + 1,j}$ are contained in the same tile.
Tiles include their boundaries, and no two different tiles share a positive area.
Near the origin, the tiles are laid out as follows:

Takahashi starts at the point $S _ x+0.5,S _ y+0.5)$ on the coordinate plane.
He can repeat the following move as many times as he likes:
Choose a direction (up, down, left, or right) and a positive integer $n$ . Move $n$ units in that direction.
Each time he enters a tile, he pays a toll of $1$ .
Find the minimum toll he must pay to reach the point $T _ x+0.5,T _ y+0.5)$ .

Constraints

$0\leq S _ x\leq2\times10 ^ {16}$
$0\leq S _ y\leq2\times10 ^ {16}$
$0\leq T _ x\leq2\times10 ^ {16}$
$0\leq T _ y\leq2\times10 ^ {16}$
All input values are integers.

Input

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

$S _ x$ $S _ y$
$T _ x$ $T _ y$

Output

Print the minimum toll Takahashi must pay.

Sample Input 1

5 0
2 5

Sample Output 1

For example, Takahashi can pay a toll of $5$ by moving as follows:

Move left by $1$ . Pay a toll of $0$ .
Move up by $1$ . Pay a toll of $1$ .
Move left by $1$ . Pay a toll of $0$ .
Move up by $3$ . Pay a toll of $3$ .
Move left by $1$ . Pay a toll of $0$ .
Move up by $1$ . Pay a toll of $1$ .
It is impossible to reduce the toll to $4$ or less, so print 5.

Sample Input 2

3 1
4 1

Sample Output 2

There are cases where no toll needs to be paid.

Sample Input 3

2552608206527595 5411232866732612
771856005518028 7206210729152763

Sample Output 3

1794977862420151

Note that the value to be output may exceed the range of a $32$ -bit integer.

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);

	int sx, sy, fx, fy;
	cin >> sx >> sy >> fx >> fy;

	int res = abs(sy - fy);
	if ((sx + sy & 1) && fx < sx || ((sx + sy) % 2 == 0) && fx > sx) {
		res += max(0ll, abs(sx - fx) - abs(sy - fy) >> 1);
	} else {
		res += max(0ll, abs(sx - fx) - abs(sy - fy) + 1 >> 1);
	}

	cout << res << endl;

	return 0;
}

D - Avoid K Palindrome

Problem Statement

You are given a string $S$ of length $N$ consisting of characters A, B, and ?.
You are also given a positive integer $K$ .
A string $T$ consisting of A and B is considered a good string if it satisfies the following condition:
No contiguous substring of length $K$ in $T$ is a palindrome.
Let $q$ be the number of ? characters in $S$ .
There are $2^q$ strings that can be obtained by replacing each ? in $S$ with either A or B. Find how many of these strings are good strings.
The count can be very large, so find it modulo $998244353$ .

Constraints

$\leq K \leq N \leq 1000$
$\leq 10$
$S$ is a string consisting of A, B, and ?.
The length of $S$ is $N$ .
$N$ and $K$ are integers.

Input

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

$N$ $K$
$S$

Output

Print the answer.

Sample Input 1

7 4
AB?A?BA

Sample Output 1

The given string has two ?s.
There are four strings obtained by replacing each ? with A or B:
ABAAABA
ABAABBA
ABBAABA
ABBABBA
Among these, the last three contain the contiguous substring ABBA of length 4, which is a palindrome, and thus are not good strings.
Therefore, you should print 1.

Sample Input 2

40 7
????????????????????????????????????????

Sample Output 2

116295436

Ensure to find the number of good strings modulo $998244353$ .

Sample Input 3

15 5
ABABA??????????

Sample Output 3

It is possible that there is no way to replace the ?s to obtain a good string.

Sample Input 4

40 8
?A?B??B?B?AA?A?B??B?A???B?BB?B???BA??BAA

Sample Output 4

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, M = 11, mod = 998244353;

int n, k;
string s;
int f[N][1 << M];

bool check(int x) {
	string a, b;
	for (int i = 0; i < k; i ++)
		a += char((x >> i & 1) ^ 48);
	b = a;
	reverse(b.begin(), b.end());
	return a != b;
}

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

	cin >> n >> k >> s;
	s = ' ' + s;

	std::vector<int> avl;
	avl.push_back(0);
	for (int i = 1; i <= k; i ++) {
		std::vector<int> ok;
		while (avl.size()) {
			int tmp = avl.back();
			avl.pop_back();
			if (s[i] == '?') {
				ok.push_back(tmp), ok.push_back(tmp | (1ll << i - 1));
			} else if (s[i] == 'B') {
				ok.push_back(tmp | (1ll << i - 1));
			} else {
				ok.push_back(tmp);
			}
		}
		avl = ok;
	}
	for (auto v : avl)
		if (check(v))
			f[k][v] = 1;
	for (int i = k + 1; i <= n; i ++)
		for (int j = 0; j < 1ll << k; j ++)
			if (s[i] == '?') {
				if (check(j) && check(j >> 1)) f[i][j >> 1] = (f[i][j >> 1] + f[i - 1][j]) % mod;
				if (check(j) && check((j >> 1) | (1ll << k - 1))) f[i][(j >> 1) | (1ll << k - 1)] = (f[i][(j >> 1) | (1ll << k - 1)] + f[i - 1][j]) % mod;
			} else if (s[i] == 'B') {
				if (check(j) && check((j >> 1) | (1ll << k - 1))) f[i][(j >> 1) | (1ll << k - 1)] = (f[i][(j >> 1) | (1ll << k - 1)] + f[i - 1][j]) % mod;
			} else {
				if (check(j) && check(j >> 1)) f[i][j >> 1] = (f[i][j >> 1] + f[i - 1][j]) % mod;
			}

	int res = 0;
	for (int i = 0; i < 1ll << k; i ++)
		res = (res + f[n][i]) % mod;

	cout << res << endl;

	return 0;
}

E - Water Tank

Problem Statement

You are given a sequence of positive integers of length $N$ : $2,\dotsc,H _ N)$ .
There is a sequence of non-negative integers of length $N + 1$ : $1,\dotsc,A _ N)$ . Initially, $1=\dotsb=A _ N=0$ .
Perform the following operations repeatedly on $A$ :

Increase the value of $A _ 0$ by $1$. For $i=1,2,\ldots,N$ in this order, perform the following operation: If $A _ {i-1}\gt A _ i$ and $A _ {i-1}\gt H _ i$, decrease the value of $A _ {i-1}$ by 1 and increase the value of $A _ i$ by $1$.

For each $i=1,2,\ldots,N$, find the number of operations before $A _ i>0$ holds for the first time. ## Constraints

$1\leq N\leq2\times10 ^ 5$
$1\leq H _ i\leq10 ^ 9\ (1\leq i\leq N)$
All input values are integers.

Input

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

$N$
$H _ 1$ $H _ 2$ $\dotsc$ $H _ N$

Output

Print the answers for $i=1,2,\ldots,N$ in a single line, separated by spaces.

Sample Input 1

5
3 1 4 1 5

Sample Output 1

4 5 13 14 26

The first five operations go as follows.
Here, each row corresponds to one operation, with the leftmost column representing step 1 and the others representing step 2.

From this diagram, $1\gt0$ holds for the first time after the 4th operation, and $2\gt0$ holds for the first time after the 5th operation.
Similarly, the answers for $A _ 3, A _ 4, A _ 5$ are $13, 14, 26$ , respectively.
Therefore, you should print 4 5 13 14 26.

Sample Input 2

6
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 2

1000000001 2000000001 3000000001 4000000001 5000000001 6000000001

Note that the values to be output may not fit within a $32$ -bit integer.

Sample Input 3

15
748 169 586 329 972 529 432 519 408 587 138 249 656 114 632

Sample Output 3

749 918 1921 2250 4861 5390 5822 6428 6836 7796 7934 8294 10109 10223 11373

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], to[N], f[N];

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

	cin >> n;

	stack<int> stk;
	for (int i = 1; i <= n; i ++) {
		cin >> a[i];
		while (stk.size() && a[stk.top()] < a[i]) stk.pop();
		if (stk.size()) to[i] = stk.top();
		stk.push(i);
	}

	for (int i = 1; i <= n; i ++)
		f[i] = f[to[i]] + (i - to[i]) * a[i], cout << f[i] + 1 << " ";

	return 0;
}

F - Tree Degree Optimization

Problem Statement

You are given a sequence of integers $A=(A_1,\ldots,A_N)$ . For a tree $T$ with $N$ vertices, define $f (T)$ as follows:
Let $d_i$ be the degree of vertex $i$ in $T$ . Then, $f(T)=\sum_{i=1}^N {d_i}^2 A_i$ .
Find the minimum possible value of $f (T)$ .
The constraints guarantee the answer to be less than $2^{63}$ .

Constraints

$2\leq N\leq 2\times 10^5$
$1\leq A_i \leq 10^9$
All input values are integers.

Input

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

$N$
$A_1$ $A_2$ $\ldots$ $A_N$

Output

Print the answer.

Sample Input 1

4
3 2 5 2

Sample Output 1

Consider a tree $T$ with an edge connecting vertices $1$ and $2$ , an edge connecting vertices $2$ and $4$ , and an edge connecting vertices $4$ and $3$ .
Then, $1^2\times 3 + 2^2\times 2 + 1^2\times 5 + 2^2\times 2 = 24$ . It can be proven that this is the minimum value of $f (T)$ .

Sample Input 2

3
4 3 2

Sample Output 2

Sample Input 3

7
10 5 10 2 10 13 15

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;
int a[N], d[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];
	priority_queue<PII, vector<PII>, greater<PII>> heap;
	for (int i = 1; i <= n; i ++)
		d[i] = 1, heap.push({a[i] * (2 * d[i] + 1), i});

	for (int i = 1; i <= n - 2; i ++) {
		auto tmp = heap.top(); heap.pop();
		d[tmp.se] ++, heap.push({a[tmp.se] * (2 * d[tmp.se] + 1), tmp.se});
	}
	int res = 0;
	for (int i = 1; i <= n; i ++) res += d[i] * d[i] * a[i];

	cout << res << endl;

	return 0;
}

G - Sum of Tree Distance

Problem Statement

You are given a tree with $N$ vertices. The $i$ -th edge connects vertices $u_i$ and $v_i$ bidirectionally.
Additionally, you are given an integer sequence $A=(A_1,\ldots,A_N)$ .
Here, define $f (i, j)$ as follows:
If $A_i = A_j$ , then $f (i, j)$ is the minimum number of edges you need to traverse to move from vertex $i$ to vertex $j$ . If $A_i \neq A_j$ , then $f (i, j) = 0$ .
Calculate the value of the following expression:

$\displaystyle \sum_{i=1}^{N-1}\sum_{j=i+1}^N f(i,j)$.
## Constraints

$\leq N \leq 2 \times 10^5$
$\leq u_i, v_i \leq N$
$\leq A_i \leq N$
The input graph is a tree.
All input values are integers.

Input

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

$N$
$u_1$ $v_1$
$\vdots$
$u_{N-1}$ $v_{N-1}$
$A_1$ $A_2$ $\ldots$ $A_N$

Output

Print the answer.

Sample Input 1

Sample Output 1

$f (1, 4) = 2, f (2, 3) = 2$ . For all other $KaTeX parse error: Expected 'EOF', got '&' at position 17: …, j\ (1 \leq i &̲lt; j \leq N)$ , we have $f (i, j) = 0$ , so the answer is $2 + 2 = 4$ .

Sample Input 2

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

Sample Output 2

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, B = 448;

int n;
int a[N], dep[N], idx[N], tot[N], cnt[N], acl[N][B + 10], id[N];
int f[N << 1][20], Euler[N << 1], pl[N], m, sub;
std::vector<int> g[N], pos[N], big;

void dfs(int u, int fa) {
	Euler[ ++ m] = u, pl[u] = m;
	for (auto v : g[u]) {
		if (v == fa) continue;
		dep[v] = dep[u] + 1;
		dfs(v, u), Euler[ ++ m] = u;
	}
}
void dfs2(int u, int fa) {
	if (id[a[u]]) acl[u][id[a[u]]] ++;
	for (auto v : g[u]) {
		if (v == fa) continue;
		dfs2(v, u);
		for (auto i : big) acl[u][i] += acl[v][i];
	}
	for (auto i : big) {
		int sum = 0, res = 0;
		for (auto v : g[u]) {
			if (v == fa) continue;
			res += acl[v][i] * sum, sum += acl[v][i];
		}
		sub += res * dep[u] * 2;
	}
	if (id[a[u]]) sub += (acl[u][id[a[u]]] - 1) * dep[u] * 2;
}
void build() {
	for (int j = 0; j <= log2(m); j ++)
		for (int i = 1; i + (1ll << j) - 1 <= m; i ++)
			if (!j) f[i][j] = i;
			else {
				if (dep[Euler[f[i][j - 1]]] < dep[Euler[f[i + (1ll << j - 1)][j - 1]]]) f[i][j] = f[i][j - 1];
				else f[i][j] = f[i + (1ll << j - 1)][j - 1];
			}
}
int query(int l, int r) {
	int k = log2(r - l + 1);
	if (dep[Euler[f[l][k]]] < dep[Euler[f[r - (1ll << k) + 1][k]]]) return f[l][k];
	return f[r - (1ll << k) + 1][k];
}
int lca(int a, int b) {
	if (pl[a] > pl[b]) swap(a, b);
	return Euler[query(pl[a], pl[b])];
}

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

	cin >> n;
	for (int i = 1; i < n; i ++) {
		int u, v;
		cin >> u >> v;
		g[u].push_back(v), g[v].push_back(u);
	}
	for (int i = 1; i <= n; i ++)
		cin >> a[i], pos[a[i]].push_back(i);
	dfs(1, -1), build();

	int res1 = 0, res2 = 0, tmp = 0;
	for (int i = 1; i <= n; i ++) {
		if (pos[a[i]].size() <= B) {
			for (int j = 0; j < idx[a[i]]; j ++)
				res2 += dep[lca(i, pos[a[i]][j])] * 2;
			idx[a[i]] ++;
		} else if (!id[a[i]]) id[a[i]] = ++ tmp, big.push_back(tmp);
		res1 += tot[a[i]] + cnt[a[i]] * dep[i];
		tot[a[i]] += dep[i], cnt[a[i]] ++;
	}
	dfs2(1, -1);
	cout << res1 - res2 - sub << endl;

	return 0;
}

视频题解

AtCoder Beginner Contest 359（A ~ G 题讲解）

最后祝大家早日