CodeTON Round 9 (Div. 1 + Div. 2, Rated, Prizes! ABCDE题) 视频讲解

A. Shohag Loves Mod

Problem Statement

Shohag has an integer $n$. Please help him find an increasing integer sequence $1\le a_1<a_2<\dots <a_n\le 100$ such that $a_i\mathrm{mod}i\ne a_j\mathrm{mod}j$ ${}^{\text{∗}}$ is satisfied over all pairs $1\le i<j\le n$.

It can be shown that such a sequence always exists under the given constraints.

${}^{\text{∗}}$$a\mathrm{mod}b$ denotes the remainder of $a$ after division by $b$. For example, $7\mathrm{mod}3=1,8\mathrm{mod}4=0$ and $69\mathrm{mod}10=9$.

Input

The first line contains a single integer $t$ ($1\le t\le 50$) — the number of test cases.

The first and only line of each test case contains an integer $n$ ($2\le n\le 50$).

Output

For each test case, print $n$ integers — the integer sequence that satisfies the conditions mentioned in the statement. If there are multiple such sequences, output any.

Example

input
2
3
6

output
2 7 8
2 3 32 35 69 95

Note

In the first test case, the sequence is increasing, values are from $1$ to $100$ and each pair of indices satisfies the condition mentioned in the statement:

• For pair $(1,2)$, $a_1\mathrm{mod}1=2\mathrm{mod}1=0$, and $a_2\mathrm{mod}2=7\mathrm{mod}2=1$. So they are different.
• For pair $(1,3)$, $a_1\mathrm{mod}1=2\mathrm{mod}1=0$, and $a_3\mathrm{mod}3=8\mathrm{mod}3=2$. So they are different.
• For pair $(2,3)$, $a_2\mathrm{mod}2=7\mathrm{mod}2=1$, and $a_3\mathrm{mod}3=8\mathrm{mod}3=2$. So they are different.

Note that you do not necessarily have to print the exact same sequence, you can print any other sequence as long as it satisfies the necessary conditions.

Solution

具体见文后视频。

---

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;

void solve() {
	int n;
	cin >> n;

	std::vector<int> a(n);
	for (int i = 0; i < n; i ++)
		cin >> a[i];
	sort(a.begin(), a.end());

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

	cout << res << endl;
}

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

	int dt;
	cin >> dt;

	while (dt --)
		solve();

	return 0;
}

---

B. Shohag Loves Strings

Problem Statement

For a string $p$, let $f(p)$ be the number of distinct non-empty substrings${}^{\text{∗}}$ of $p$.

Shohag has a string $s$. Help him find a non-empty string $p$ such that $p$ is a substring of $s$ and $f(p)$ is even or state that no such string exists.

${}^{\text{∗}}$A string $a$ is a substring of a string $b$ if $a$ can be obtained from $b$ by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Input

The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The first and only line of each test case contains a string $s$ ($1\le |s|\le 10^5$) consisting of lowercase English letters.

It is guaranteed that the sum of the length of $s$ over all test cases doesn’t exceed $3\cdot 10^5$.

Output

For each test case, print a non-empty string that satisfies the conditions mentioned in the statement, or $-1$ if no such string exists. If there are multiple solutions, output any.

Example

input
5
dcabaac
a
youknowwho
codeforces
bangladesh

output
abaa
-1
youknowwho
eforce
bang

Note

In the first test case, we can set $p = $ abaa because it is a substring of $s$ and the distinct non-empty substrings of $p$ are a, b, aa, ab, ba, aba, baa and abaa, so it has a total of $8$ distinct substrings which is even.

In the second test case, we can only set $p = $ a but it has one distinct non-empty substring but this number is odd, so not valid.

In the third test case, the whole string contains $52$ distinct non-empty substrings, so the string itself is a valid solution.

Solution

具体见文后视频。

---

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;

void solve() {
	string s;
	cin >> s;

	for (int i = 1; i < s.size(); i ++)
		if (s[i] == s[i - 1]) {
			cout << s[i - 1] << s[i] << endl;
			return;
		}

	if (s.size() >= 3) {
		for (int i = 2; i < s.size(); i ++)
			if (s[i - 2] != s[i]) {
				cout << s[i - 2] << s[i - 1] << s[i] << endl;
				return;
			}
		cout << -1 << endl;
	}
	else cout << -1 << endl;
}

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

	int dt;
	cin >> dt;

	while (dt --)
		solve();

	return 0;
}

---

C1. Shohag Loves XOR (Easy Version)

Problem Statement

This is the easy version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $x$ and $m$. Help him count the number of integers $1\le y\le m$ such that $\mathbf{x}\ne \mathbf{y}$ and $x\oplus y$ is a divisor${}^{\text{∗}}$ of either $x$, $y$, or both. Here $\oplus$ is the bitwise XOR operator.

${}^{\text{∗}}$The number $b$ is a divisor of the number $a$ if there exists an integer $c$ such that $a=b\cdot c$.

Input

The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The first and only line of each test case contains two space-separated integers $x$ and $m$ ($1\le x\le 10^6$, $1\le m\le 10^{18}$).

It is guaranteed that the sum of $x$ over all test cases does not exceed $10^7$.

Output

For each test case, print a single integer — the number of suitable $y$.

Example

input
5
6 9
5 7
2 3
6 4
4 1

output
3
2
1
1
0

Note

In the first test case, for $x=6$, there are $3$ valid values for $y$ among the integers from $1$ to $m=9$, and they are $4$, $5$, and $7$.

• $y=4$ is valid because $x\oplus y=6\oplus 4=2$ and $2$ is a divisor of both $x=6$ and $y=4$.
• $y=5$ is valid because $x\oplus y=6\oplus 5=3$ and $3$ is a divisor of $x=6$.
• $y=7$ is valid because $x\oplus y=6\oplus 7=1$ and $1$ is a divisor of both $x=6$ and $y=7$.

In the second test case, for $x=5$, there are $2$ valid values for $y$ among the integers from $1$ to $m=7$, and they are $4$ and $6$.

• $y=4$ is valid because $x\oplus y=5\oplus 4=1$ and $1$ is a divisor of both $x=5$ and $y=4$.
• $y=6$ is valid because $x\oplus y=5\oplus 6=3$ and $3$ is a divisor of $y=6$.

Solution

具体见文后视频。

---

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;

void solve() {
	int x, m;
	cin >> x >> m;

	int cnt = 0;
	for (int y = 1; y <= x * 2; y ++)
		if (x != y && y <= m && (x % (x ^ y) == 0 || y % (x ^ y) == 0))
			cnt ++;
	
	cout << cnt << endl;
}

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

	int dt;
	cin >> dt;

	while (dt --)
		solve();

	return 0;
}

---

C2. Shohag Loves XOR (Hard Version)

Problem Statement

This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $x$ and $m$. Help him count the number of integers $1\le y\le m$ such that $x\oplus y$ is divisible${}^{\text{∗}}$ by either $x$, $y$, or both. Here $\oplus$ is the bitwise XOR operator.

${}^{\text{∗}}$The number $a$ is divisible by the number $b$ if there exists an integer $c$ such that $a=b\cdot c$.

Input

The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The first and only line of each test case contains two space-separated integers $x$ and $m$ ($1\le x\le 10^6$, $1\le m\le 10^{18}$).

It is guaranteed that the sum of $x$ over all test cases does not exceed $10^7$.

Output

For each test case, print a single integer — the number of suitable $y$.

Example

input
5
7 10
2 3
6 4
1 6
4 1

output
3
2
2
6
1

Note

In the first test case, for $x=7$, there are $3$ valid values for $y$ among the integers from $1$ to $m=10$, and they are $1$, $7$, and $9$.

• $y=1$ is valid because $x\oplus y=7\oplus 1=6$ and $6$ is divisible by $y=1$.
• $y=7$ is valid because $x\oplus y=7\oplus 7=0$ and $0$ is divisible by both $x=7$ and $y=7$.
• $y=9$ is valid because $x\oplus y=7\oplus 9=14$ and $14$ is divisible by $x=7$.

Solution

具体见文后视频。

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;

void solve() {
	int x, m;
	cin >> x >> m;

	int lo = 0, ro = (m + x - 1) / x, all = 1, ok;
	while (all <= x) all *= 2;
	all --;
	int lim = m <= all ? m : x * 2, res = 0;
	for (int i = 1; i <= min(m, lim); i ++)
		if ((i ^ x) % i == 0 && (i ^ x) % x != 0)
			res ++;
	if (m <= all) {
		for (int i = 1; i <= m; i ++)
			if ((i ^ x) % x == 0) res ++;
		cout << res << endl;
		return;
	}
	while (lo <= ro) {
		int mid = lo + ro >> 1;
		if (mid * x <= m && (m ^ (mid * x)) > all) lo = mid + 1, ok = mid;
		else ro = mid - 1;
	}

	res += ok + (ok == 0);
	// cout << ok << " " << res << endl;
	if ((((ok + 1) * x) ^ x) <= m) res ++;
	if ((((ok + 2) * x) ^ x) <= m) res ++;
	cout << res << endl;
}

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

	int dt;
	cin >> dt;

	while (dt --)
		solve();

	return 0;
}

---

D. Shohag Loves GCD

Problem Statement

Shohag has an integer $n$ and a set $S$ of $m$ unique integers. Help him find the lexicographically largest${}^{\text{∗}}$ integer array $a_1,a_2,\dots ,a_n$ such that $a_i\in S$ for each $1\le i\le n$ and $a_{\gcd (i,j)}\ne \gcd (a_i,a_j)$${}^{\text{†}}$ is satisfied over all pairs $1\le i<j\le n$, or state that no such array exists.

${}^{\text{∗}}$An array $a$ is lexicographically larger than an array $b$ of the same length if $a\ne b$, and in the first position where $a$ and $b$ differ, the array $a$ has a larger element than the corresponding element in $b$.

${}^{\text{†}}$$\gcd (x,y)$ denotes the greatest common divisor (GCD) of integers $x$ and $y$.

Input

The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1\le m\le n\le 10^5$).

The second line contains $m$ unique integers in increasing order, representing the elements of the set $S$ ($1\le x\le n$ for each $x\in S$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

Output

For each test case, if there is no solution print $-1$, otherwise print $n$ integers — the lexicographically largest integer array that satisfies the conditions.

Example

input
3
6 3
3 4 6
1 1
1
2 1
2

output
6 4 4 3 4 3
1
-1

Note

In the first test case, every element in the array belongs to the given set S = { 3 , 4 , 6 } S = \{3, 4, 6\} S={3,4,6}, and all pairs of indices of the array satisfy the necessary conditions. In particular, for pair $(2,3)$, $a_{\gcd (2,3)}=a_1=6$ and $\gcd (a_2,a_3)=\gcd (4,4)=4$, so they are not equal. There are other arrays that satisfy the conditions as well but this one is the lexicographically largest among them.

In the third test case, there is no solution possible because we are only allowed to use $a=[2,2]$ but for this array, for pair $(1,2)$, $a_{\gcd (1,2)}=a_1=2$ and $\gcd (a_1,a_2)=\gcd (2,2)=2$, so they are equal which is not allowed!

Solution

具体见文后视频。

---

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;

void solve() {
	int n, m;
	cin >> n >> m;

	std::vector<int> S(m);
	for (auto &v : S) cin >> v;

	std::vector<int> ans(n + 1, m - 1);
	for (int i = 1; i <= n; i ++)
		for (int j = i + i; j <= n; j += i)
			ans[j] = min(ans[j], ans[i] - 1);

	for (int i = 1; i <= n; i ++)
		if (ans[i] < 0) {
			cout << -1 << endl;
			return;
		}
	for (int i = 1; i <= n; i ++)
		cout << S[ans[i]] << " ";
	cout << endl;
}

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

	int dt;
	cin >> dt;

	while (dt --)
		solve();

	return 0;
}

---

E. Shohag Loves Inversions

Problem Statement

Shohag has an array $a$ of integers. Initially $a=[0,1]$. He can repeatedly perform the following operation any number of times:

• Let $k$ be the number of inversions${}^{\text{∗}}$ in the current array $a$.
• Insert $k$ at any position in $a$, including the beginning or the end.

For example, if $a=[4,6,2,4]$, then the number of inversions is $k=3$. So Shohag can obtain the following arrays after the operation: $[\text{3},4,6,2,4]$, $[4,\text{3},6,2,4]$, $[4,6,\text{3},2,4]$, $[4,6,2,\text{3},4]$, and $[4,6,2,4,\text{3}]$.

Given an integer $n$, help Shohag count, modulo $998244353$, the number of distinct arrays of length $n$ that can be obtained after performing the operations.

${}^{\text{∗}}$The number of inversions in an array $a$ is the number of pairs of indices ($i$, $j$) such that $i<j$ and $a_i>a_j$.

Input

The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The first and only line of each test case contains an integer $n$ ($2\le n\le 10^6$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output

For each test case, output an integer — the number of possible arrays modulo $998244353$.

Example

input
4
4
2
7
69

output
5
1
682
325188814

Note

In the first test case, the following $5$ arrays can be obtained (the inserted inversion count is shown in bold):

• $[0,1]\to [0,\text{0},1]\to [0,0,1,\text{0}]$,
• $[0,1]\to [0,\text{0},1]\to [0,0,\text{0},1]$,
• $[0,1]\to [0,1,\text{0}]\to [0,1,0,\text{1}]$,
• $[0,1]\to [0,1,\text{0}]\to [0,1,\text{1},0]$,
• $[0,1]\to [0,1,\text{0}]\to [\text{1},0,1,0]$.

Solution

具体见文后视频。

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;

void solve() {
	int n;
	cin >> n;

	std::vector<int> f(n + 1, 0);
	int mod = 998244353, sum = 0;
	for (int i = 3; i <= n; i ++) f[i] = 1;
	for (int i = 1; i <= n; i ++) {
		(f[i] += sum) %= mod;
		(sum += max(f[i] * i % mod - 1, 0ll)) %= mod;
	}

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

	cout << res << endl;
}

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

	int dt;
	cin >> dt;

	while (dt --)
		solve();

	return 0;
}

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视频讲解

CodeTON Round 9 (Div. 1 + Div. 2, Rated, Prizes!)（A ~ E 题讲解）

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