A - Divisible

Problem Statement

You are given positive integers $N$ and $K$, and a sequence of length $N$, $A=(A_1,A_2,\dots ,A_N)$.
Extract all elements of $A$ that are multiples of $K$, divide them by $K$, and print the quotients.

Constraints

$1\le N,K\le 100$
KaTeX parse error: Expected 'EOF', got '&' at position 11: 1\leq A_1 &̲lt; A_2 < \l…
$A$ has at least one multiple of $K$.
All given numbers are integers.

Input

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

$N$ $K$
$A_1$ $A_2$ $\dots$ $A_N$

Output

Divide all elements of $A$ that are multiples of $K$ and print the quotients in ascending order with spaces in between.

Sample Input 1

5 2
2 5 6 7 10

Sample Output 1

1 3 5

The multiples of $2$ among the elements in $A$ are $2$, $6$, and $10$. Divide them by $2$ to get $1$, $3$, and $5$, and print them in ascending order with spaces in between.

Sample Input 2

3 1
3 4 7

Sample Output 2

3 4 7

Sample Input 3

5 10
50 51 54 60 65

Sample Output 3

5 6

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, k;
	cin >> n >> k;

	std::vector<int> a(n);
	for (int i = 0; i < n; i ++) {
		cin >> a[i];
		if (a[i] % k == 0)
			cout << a[i] / k << " ";
	}

	return 0;
}

B - Substring

Problem Statement

You are given a string $S$ consisting of lowercase English letters. How many different non-empty substrings does $S$ have?
A substring is a contiguous subsequence. For example, xxx is a substring of yxxxy but not of xxyxx.

Constraints

$S$ is a string of length between $1$ and $100$, inclusive, consisting of lowercase English letters.

Input

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

$S$

Output

Print the answer.

Sample Input 1

yay

Sample Output 1

5

$S$ has the following five different non-empty substrings:
a
y
ay
ya
yay

Sample Input 2

aababc

Sample Output 2

17

Sample Input 3

abracadabra

Sample Output 3

54

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;

	int n = s.size();
	set<string> res;
	for (int i = 0; i < n; i ++)
		for (int j = 1; j <= n - i; j ++)
			res.insert(s.substr(i, j));

	cout << res.size() << endl;

	return 0;
}

C - Ideal Holidays

Problem Statement

In the Kingdom of AtCoder, a week consists of $A+B$ days, with the first through $A$-th days being holidays and the $(A+1)$-th through $(A+B)$-th being weekdays.
Takahashi has $N$ plans, and the $i$-th plan is scheduled $D_i$ days later.
He has forgotten what day of the week it is today. Determine if it is possible for all of his $N$ plans to be scheduled on holidays.

Constraints

$1\le N\le 2\times 10^5$
$1\le A,B\le 10^9$
KaTeX parse error: Expected 'EOF', got '&' at position 10: 1\leq D_1&̲lt;D_2<\ldot…

Input

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

$N$ $A$ $B$
$D_1$ $D_2$ $\dots$ $D_N$

Output

Print Yes in a single line if it is possible for all of Takahashi’s $N$ plans to be scheduled on holidays, and No otherwise.

Sample Input 1

3 2 5
1 2 9

Sample Output 1

Yes

In this input, a week consists of seven days, with the first through second days being holidays and the third through seventh days being weekdays.
Let us assume today is the seventh day of the week. In this case, one day later would be the first day of the week, two days later would be the second day of the week, and nine days later would also be the second day of the week, making all plans scheduled on holidays. Therefore, it is possible for all of Takahashi’s $N$ plans to be scheduled on holidays.

Sample Input 2

2 5 10
10 15

Sample Output 2

No

Sample Input 3

4 347 347
347 700 705 710

Sample Output 3

Yes

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, a, b;
int d[N];

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

	cin >> n >> a >> b;

	int l1 = -1e18, r1 = 1e18, l2 = -1e18, r2 = 1e18;
	for (int i = 1; i <= n; i ++) {
		cin >> d[i];
		d[i] %= (a + b);
		int tl = (0 - d[i] + a + b) % (a + b), tr = (a - 1 - d[i] + a + b) % (a + b);
		if (tl > tr) {
			l1 = max(0ll, l1), r1 = min(r1, tr);
			l2 = max(l2, tl), r2 = min(a + b - 1, r2);
		} else
			l1 = max(l1, tl), r1 = min(r1, tr), l2 = max(l2, tl), r2 = min(r2, tr);
	}

	if (l1 <= r1 || l2 <= r2) cout << "Yes" << endl;
	else cout << "No" << endl;

	return 0;
}

D - Popcount and XOR

Problem Statement

You are given non-negative integers $a$, $b$, and $C$.
Determine if there is a pair of non-negative integers $(X,Y)$ that satisfies all of the following five conditions. If such a pair exists, print one.
KaTeX parse error: Expected 'EOF', got '&' at position 10: 0 \leq X &̲lt; 2^{60}
KaTeX parse error: Expected 'EOF', got '&' at position 10: 0 \leq Y &̲lt; 2^{60}
$\mathrm{popcount}(X)=a$
$\mathrm{popcount}(Y)=b$
$X\oplus Y=C$
Here, $\oplus$ denotes the bitwise XOR.
If multiple pairs $(X,Y)$ satisfy the conditions, you may print any of them.

What is popcount? For a non-negative integer $x$, the popcount of $x$ is the number of $1$s in the binary representation of $x$. More precisely, for a non-negative integer $x$ such that $\displaystyle x=\sum _ {i=0} ^ \infty b _ i2 ^ i\ (b _ i\in\lbrace0,1\rbrace)$, we have $\displaystyle\operatorname{popcount}(x)=\sum _ {i=0} ^ \infty b _ i$. For example, $13$ in binary is 1101, so $\operatorname{popcount}(13)=3$. What is bitwise XOR? For non-negative integers $x, y$, the bitwise exclusive OR $x \oplus y$ is defined as follows. The $2^k$'s place $\ (k\geq0)$ in the binary representation of $x \oplus y$ is $1$ if exactly one of the $2^k$'s places $\ (k\geq0)$ in the binary representations of $x$ and $y$ is $1$, and $0$ otherwise. For example, $9$ and $3$ in binary are 1001 and 0011, respectively, so $9 \oplus 3 = 10$ (in binary, 1010). #### Constraints

$0\le a\le 60$
$0\le b\le 60$
KaTeX parse error: Expected 'EOF', got '&' at position 10: 0 \leq C &̲lt; 2^{60}
All input values are integers.

Input

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

$a$ $b$ $C$

Output

If there is a pair of non-negative integers that satisfies the conditions, choose one such pair $(X,Y)$ and print $X$ and $Y$ in this order, with a space in between.
If no such pair exists, print -1.

Sample Input 1

3 4 7

Sample Output 1

28 27

The pair $(X,Y)=(28,27)$ satisfies the conditions.
Here, $X$ and $Y$ in binary are 11100 and 11011, respectively.
$X$ in binary is 11100, so $\mathrm{popcount}(X)=3$.
$Y$ in binary is 11011, so $\mathrm{popcount}(Y)=4$.
$X\oplus Y$ in binary is 00111, so $X\oplus Y=7$.
If multiple pairs of non-negative integers satisfy the conditions, you may print any of them, so printing 42 45, for example, would also be accepted.

Sample Input 2

34 56 998244353

Sample Output 2

-1

No pair of non-negative integers satisfies the conditions.

Sample Input 3

39 47 530423800524412070

Sample Output 3

540431255696862041 10008854347644927

Note that the values to be printed may not fit in $32$-bit integers.

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 a, b, c;

	cin >> a >> b >> c;

	int r1 = 0, r2 = 0;
	for (int i = 0; i < 61; i ++)
		if (c >> i & 1) {
			if (a > b) r1 += (1ll << i), a --;
			else r2 += (1ll << i), b --;
		}
	if (a != b || a < 0 || b < 0) {
		cout << -1 << endl;
		return 0;
	}
	for (int i = 0; i < 61; i ++)
		if (!(c >> i & 1) && a && b)
			r1 += (1ll << i), r2 += (1ll << i), a --, b --;

	if (a || b) {
		cout << -1 << endl;
		return 0;
	}
	cout << r1 << " " << r2 << endl;

	return 0;
}

---

E - Set Add Query

Problem Statement

There is an integer sequence $A=(A_1,A_2,\dots ,A_N)$ of length $N$, where all elements are initially set to $0$. Also, there is a set $S$, which is initially empty.
Perform the following $Q$ queries in order. Find the value of each element in the sequence $A$ after processing all $Q$ queries. The $i$-th query is in the following format:
An integer $x_i$ is given. If the integer $x_i$ is contained in $S$, remove $x_i$ from $S$. Otherwise, insert $x_i$ to $S$. Then, for each $j=1,2,\dots ,N$, add $|S|$ to $A_j$ if $j\in S$.
Here, $|S|$ denotes the number of elements in the set $S$. For example, if S = { 3 , 4 , 7 } S=\lbrace 3,4,7\rbrace S={3,4,7}, then $|S|=3$.

Constraints

$1\le N,Q\le 2\times 10^5$
$1\le x_i\le N$
All given numbers are integers.

Input

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

$N$ $Q$
$x_1$ $x_2$ $\dots$ $x_Q$

Output

Print the sequence $A$ after processing all queries in the following format:

$A_1$ $A_2$ $\dots$ $A_N$

Sample Input 1

3 4
1 3 3 2

Sample Output 1

6 2 2

In the first query, $1$ is inserted to $S$, making S = { 1 } S=\lbrace 1\rbrace S={1}. Then, $|S|=1$ is added to $A_1$. The sequence becomes $A=(1,0,0)$.
In the second query, $3$ is inserted to $S$, making S = { 1 , 3 } S=\lbrace 1,3\rbrace S={1,3}. Then, $|S|=2$ is added to $A_1$ and $A_3$. The sequence becomes $A=(3,0,2)$.
In the third query, $3$ is removed from $S$, making S = { 1 } S=\lbrace 1\rbrace S={1}. Then, $|S|=1$ is added to $A_1$. The sequence becomes $A=(4,0,2)$.
In the fourth query, $2$ is inserted to $S$, making S = { 1 , 2 } S=\lbrace 1,2\rbrace S={1,2}. Then, $|S|=2$ is added to $A_1$ and $A_2$. The sequence becomes $A=(6,2,2)$.
Eventually, the sequence becomes $A=(6,2,2)$.

Sample Input 2

4 6
1 2 3 2 4 2

Sample Output 2

15 9 12 7

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, q;
int a[N], s[N], res[N], id[N];
int cnt[N], lst[N];

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

	cin >> n >> q;

	int idx = 0;
	for (int i = 1; i <= q; i ++) {
		cin >> a[i];
		if (!id[a[i]]) id[a[i]] = ++ idx;
	}

	set<int> S;
	int r = 0;
	for (int i = 1; i <= q; i ++) {
		cnt[a[i]] ++;
		if (S.count(a[i])) S.erase(a[i]);
		else S.insert(a[i]);
		s[i] = s[i - 1] + S.size();
		r = max(id[a[i]], r);
		if (S.size()) res[1] += S.size(), res[r + 1] -= S.size();
	}
	for (int i = 1; i <= n; i ++)
		res[i] += res[i - 1], lst[i] = q;
	for (int i = q; i >= 1; i --) {
		if (cnt[a[i]] % 2 == 0) {
			res[id[a[i]]] -= (s[lst[a[i]]] - s[i - 1]);
		} else {
			lst[a[i]] = i - 1;
		}
		cnt[a[i]] --;
	}

	for (int i = 1; i <= n; i ++)
		cout << res[id[i]] << " ";

	return 0;
}

F - Non-overlapping Squares

Problem Statement

There is an $N\times N$ grid, and the cell at the $i$-th row from the top and the $j$-th column from the left $(1\le i,j\le N)$ contains the integer $A_{i,j}$.
You are given an integer $M$. When choosing three non-overlapping $M\times M$ grids, find the maximum possible sum of the integers written in the chosen grids.

Formal definition of the problem A $6$-tuple of integers $(i _ 1,j _ 1,i _ 2,j _ 2,i _ 3,j _ 3)$ is called a good $6$-tuple when it satisfies the following three conditions: $1\leq i _ k\leq N-M+1\ (k=1,2,3)$ $1\leq j _ k\leq N-M+1\ (k=1,2,3)$ If $k\neq l\ (k,l\in\lbrace1,2,3\rbrace)$, the sets $\lbrace(i,j)\mid i _ k\leq i\lt i _ k+M\wedge j _ k\leq j\lt j _ k+M\rbrace$ and $\lbrace(i,j)\mid i _ l\leq i\lt i _ l+M\wedge j _ l\leq j\lt j _ l+M\rbrace$ do not intersect. Find the maximum value of $\displaystyle \sum _ {k=1} ^ 3\sum _ {i=i _ k} ^ {i _ k+M-1}\sum _ {j=j _ k} ^ {j _ k+M-1}A _ {i,j}$ for a good $6$-tuple $(i _ 1,j _ 1,i _ 2,j _ 2,i _ 3,j _ 3)$. It can be shown that a good $6$-tuple exists under the constraints of this problem. #### Constraints

$2\le N\le 1000$
$1\le M\le N/2$
$0\le A_{i,j}\le 10^9$
All input values are integers.

Input

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

$N$ $M$
$A_{1,1}$ $A_{1,2}$ $\dots$ $A_{1,N}$
$A_{2,1}$ $A_{2,2}$ $\dots$ $A_{2,N}$
$⋮$ $⋮$ $\ddots$ $⋮$
$A_{N,1}$ $A_{N,2}$ $\dots$ $A_{N,N}$

Output

Print the answer.

Sample Input 1

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

Sample Output 1

154

From the given grid, if we choose three $3\times 3$ grids as shown in the figure below (this corresponds to setting $(i_1,j_1,i_2,j_2,i_3,j_3)=(1,5,2,1,5,2)$), the sum of the numbers written in the chosen grids will be $154$.

There is no way to make the sum $155$ or greater while satisfying the conditions in the problem statement, so print $154$.

Sample Input 2

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

Sample Output 2

27

The following choice is optimal.

Sample Input 3

16 4
74 16 58 32 97 52 43 51 40 58 13 24 65 11 63 29
98 75 40 77 15 50 83 85 35 46 38 37 56 38 63 55
95 42 10 70 53 40 25 10 70 32 33 19 52 79 74 58
33 91 53 11 65 63 78 77 81 46 81 63 11 82 55 62
39 95 92 69 77 89 14 84 53 78 71 81 66 39 96 29
74 26 60 55 89 35 32 64 17 26 74 92 84 33 59 82
23 69 10 95 94 14 58 58 97 95 62 58 72 55 71 43
93 77 27 87 74 72 91 37 53 80 51 71 37 35 97 46
81 88 26 79 78 30 53 68 83 28 59 28 74 55 20 86
93 13 25 19 53 53 17 24 69 14 67 81 10 19 69 90
88 83 62 92 22 31 27 34 67 48 42 32 68 14 96 87
44 69 25 48 68 42 53 82 44 42 96 31 13 56 68 83
63 87 24 75 16 70 63 99 95 10 63 26 56 12 77 49
94 83 69 95 48 41 40 97 45 61 26 38 83 91 44 31
43 69 54 64 20 60 17 15 62 25 58 50 59 63 88 70
72 95 21 28 41 14 77 22 64 78 33 55 67 51 78 40

Sample Output 3

3295

The following choice is optimal.

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 = 1e3 + 10;

int n, m;
int a[N][N], s[N][N], ln[N], cl[N];
int lu[N][N], ld[N][N], ru[N][N], rd[N][N];

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

	cin >> n >> m;

	for (int i = 1; i <= n; i ++)
		for (int j = 1; j <= n; j ++)
			cin >> a[i][j], a[i][j] += a[i][j - 1] + a[i - 1][j] - a[i - 1][j - 1];
	for (int i = 1; i <= n - m + 1; i ++)
		for (int j = 1; j <= n - m + 1; j ++) {
			s[i][j] = a[i + m - 1][j + m - 1] - a[i - 1][j + m - 1] - a[i + m - 1][j - 1] + a[i - 1][j - 1];
			ln[i] = max(ln[i], s[i][j]), cl[j] = max(cl[j], s[i][j]);
		}
	for (int i = m; i <= n; i ++)
		for (int j = m; j <= n; j ++)
			lu[i][j] = max(max(lu[i - 1][j], lu[i][j - 1]), s[i - m + 1][j - m + 1]);
	for (int i = n - m + 1; i >= 1; i --)
		for (int j = m; j <= n; j ++)
			ld[i][j] = max(max(ld[i + 1][j], ld[i][j - 1]), s[i][j - m + 1]);
	for (int i = m; i <= n; i ++)
		for (int j = n - m + 1; j >= 1; j --)
			ru[i][j] = max(max(ru[i - 1][j], ru[i][j + 1]), s[i - m + 1][j]);
	for (int i = n - m + 1; i >= 1; i --)
		for (int j = n - m + 1; j >= 1; j --)
			rd[i][j] = max(max(rd[i + 1][j], rd[i][j + 1]), s[i][j]);

	int res = 0;
	for (int i = 1; i <= n; i ++)
		for (int j = 1; j <= n; j ++) {
			res = max(res, lu[i][j] + ru[i][j + 1] + ld[i + 1][n]);
			res = max(res, lu[i][j] + ld[i + 1][j] + rd[1][j + 1]);
			res = max(res, ld[i][j] + rd[i][j + 1] + ru[i - 1][1]);
			res = max(res, ru[i][j] + rd[i + 1][j] + ld[1][j - 1]);
		}
	for (int i = m; i <= n; i ++) {
		int mx1 = 0, mx2 = 0;
		for (int j = i + m; j <= n; j ++) {
			mx1 = max(mx1, ln[j - m + 1]), mx2 = max(mx2, cl[j - m + 1]);
			res = max(res, ru[i][1] + mx1 + rd[j + 1][1]);
			res = max(res, ld[1][i] + mx2 + rd[1][j + 1]);
		}
	}

	cout << res << endl;

	return 0;
}

---

视频题解

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