AtCoder Regular Contest 179 （ABC题）视频讲解

A - Partition

Problem Statement

You are given integers N N N and K K K.

The cumulative sums of an integer sequence X = ( X 1 , X 2 , ... , X N ) X=(X_1,X_2,\dots ,X_N) X=(X1,X2,...,XN) of length N N N is defined as a sequence Y = ( Y 0 , Y 1 , ... , Y N ) Y=(Y_0,Y_1,\dots ,Y_N) Y=(Y0,Y1,...,YN) of length N + 1 N+1 N+1 as follows:
Y 0 = 0 Y_0=0 Y0=0
Y i = ∑ j = 1 i X j ( i = 1 , 2 , ... , N ) Y_i=\displaystyle\sum_{j=1}^{i}X_j\ (i=1,2,\dots ,N) Yi=j=1∑iXj (i=1,2,...,N)

An integer sequence X = ( X 1 , X 2 , ... , X N ) X=(X_1,X_2,\dots ,X_N) X=(X1,X2,...,XN) of length N N N is called a good sequence if and only if it satisfies the following condition:

Any value in the cumulative sums of X X X that is less than K K K appears before any value that is not less than K K K.

Formally, for the cumulative sums Y Y Y of X X X, for any pair of integers ( i , j ) (i,j) (i,j) such that 0 ≤ i , j ≤ N 0 \le i,j \le N 0≤i,j≤N, if KaTeX parse error: Expected 'EOF', got '&' at position 6: (Y_i &̲lt; K and Y j ≥ K ) Y_j \ge K) Yj≥K), then KaTeX parse error: Expected 'EOF', got '&' at position 3: i &̲lt; j.

You are given an integer sequence A = ( A 1 , A 2 , ... , A N ) A=(A_1,A_2,\dots ,A_N) A=(A1,A2,...,AN) of length N N N. Determine whether the elements of A A A can be rearranged to a good sequence. If so, print one such rearrangement.

Constraints

1 ≤ N ≤ 2 × 1 0 5 1 \leq N \leq 2 \times 10^5 1≤N≤2×105
− 1 0 9 ≤ K ≤ 1 0 9 -10^9 \leq K \leq 10^9 −109≤K≤109
− 1 0 9 ≤ A i ≤ 1 0 9 -10^9 \leq A_i \leq 10^9 −109≤Ai≤109

All input values are integers.

Input

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

N N N K K K
A 1 A_1 A1 A 2 A_2 A2 ⋯ \cdots ⋯ A N A_N AN

Output

If the elements of A A A can be rearranged to a good sequence, print the rearranged sequence ( A 1 ′ , A 2 ′ , ... , A N ′ ) (A^{\prime}_1,A^{\prime}_2,\dots ,A^{\prime}_N) (A1′,A2′,...,AN′) in the following format:

Yes
A 1 ′ A^{\prime}_1 A1′ A 2 ′ A^{\prime}_2 A2′ ⋯ \cdots ⋯ A N ′ A^{\prime}_N AN′

If there are multiple valid rearrangements, any of them is considered correct.

If a good sequence cannot be obtained, print No.

Sample Input 1

4 1
-1 2 -3 4

Sample Output 1

Yes
-3 -1 2 4

If you rearrange A A A to ( − 3 , − 1 , 2 , 4 ) (-3,-1,2,4) (−3,−1,2,4), the cumulative sums Y Y Y in question will be ( 0 , − 3 , − 4 , − 2 , 2 ) (0,-3,-4,-2,2) (0,−3,−4,−2,2). In this Y Y Y, any value less than 1 1 1 appears before any value not less than 1 1 1.

Sample Input 2

4 -1
1 -2 3 -4

Sample Output 2

No

Sample Input 3

10 1000000000
-1000000000 -1000000000 -1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

Yes
-1000000000 -1000000000 -1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Solution

具体见文末视频。

Code

cpp 复制代码

#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, sum = 0;
	cin >> n >> k;

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

	if (sum < k	&& k <= 0) {
		cout << "No" << endl;
		return 0;
	}

	if (k > 0) sort(a.begin(), a.end());
	else sort(a.begin(), a.end(), greater<int>());
	cout << "Yes" << endl;
	for (int i = 0; i < n; i ++)
		cout << a[i] << " ";
	cout << endl;

	return 0;
}

B - Between B and B

Problem Statement

You are given a sequence ( X 1 , X 2 , ... , X M ) (X_1, X_2, \dots, X_M) (X1,X2,...,XM) of length M M M consisting of integers between 1 1 1 and M M M, inclusive.

Find the number, modulo 998244353 998244353 998244353, of sequences A = ( A 1 , A 2 , ... , A N ) A = (A_1, A_2, \dots, A_N) A=(A1,A2,...,AN) of length N N N consisting of integers between 1 1 1 and M M M, inclusive, that satisfy the following condition:

For each B = 1 , 2 , ... , M B = 1, 2, \dots, M B=1,2,...,M, the value X B X_B XB exists between any two different occurrences of B B B in A A A (including both ends).

More formally, for each B = 1 , 2 , ... , M B = 1, 2, \dots, M B=1,2,...,M, the following condition must hold:

For every pair of integers ( l , r ) (l, r) (l,r) such that KaTeX parse error: Expected 'EOF', got '&' at position 10: 1 \leq l &̲lt; r \leq N and A l = A r = B A_l = A_r = B Al=Ar=B, there exists an integer m m m ( l ≤ m ≤ r l \leq m \leq r l≤m≤r) such that A m = X B A_m = X_B Am=XB.

Constraints

1 ≤ M ≤ 10 1 \leq M \leq 10 1≤M≤10
1 ≤ N ≤ 1 0 4 1 \leq N \leq 10^4 1≤N≤104
1 ≤ X i ≤ M 1 \leq X_i \leq M 1≤Xi≤M

All input values are integers.

Input

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

M M M N N N
X 1 X_1 X1 X 2 X_2 X2 ⋯ \cdots ⋯ X M X_M XM

Output

Print the answer.

Sample Input 1

3 4
2 1 2

Sample Output 1

Here are examples of sequences A A A that satisfy the condition:
( 1 , 3 , 2 , 3 ) (1, 3, 2, 3) (1,3,2,3)
( 2 , 1 , 2 , 1 ) (2, 1, 2, 1) (2,1,2,1)
( 3 , 2 , 1 , 3 ) (3, 2, 1, 3) (3,2,1,3)

Here are non-examples:
( 1 , 3 , 1 , 3 ) (1, 3, 1, 3) (1,3,1,3)

There is no X 3 = 2 X_3 = 2 X3=2 between the 3 3 3s.
( 2 , 2 , 1 , 3 ) (2, 2, 1, 3) (2,2,1,3)

There is no X 2 = 1 X_2 = 1 X2=1 between the 2 2 2s.

Sample Input 2

4 8
1 2 3 4

Sample Output 2

All sequences of length 8 8 8 consisting of integers between 1 1 1 and 4 4 4 satisfy the condition.

Note that when X B = B X_B = B XB=B, there is always a B B B between two different occurrences of B B B.

Sample Input 3

4 9
2 3 4 1

Sample Output 3

Solution

具体见文末视频。

Code

cpp 复制代码

#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 = 1e4 + 10, M = 11, mod = 998244353;

int n, m;
int a[M], f[N][1 << M], mask[M];

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

	cin >> m >> n;

	for (int i = 1; i <= m; i ++)
		cin >> a[i], mask[a[i]] |= 1ll << i - 1;

	f[0][(1 << m) - 1] = 1;
	for (int i = 0; i < n; i ++)
		for (int j = 1; j <= m; j ++)
			for (int k = 0; k < 1 << m; k ++)
				if ((k >> j - 1) & 1)
					f[i + 1][(k ^ (1 << j - 1)) | mask[j]] = (f[i + 1][(k ^ (1 << j - 1)) | mask[j]] + f[i][k]) % mod;

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

	cout << res << endl;

	return 0;
}

C - Beware of Overflow

Problem Statement

This is an interactive problem (where your program interacts with the judge via input and output).

You are given a positive integer N N N.

The judge has a hidden positive integer R R R and N N N integers A 1 , A 2 , ... , A N A_1, A_2, \dots, A_N A1,A2,...,AN. It is guaranteed that ∣ A i ∣ ≤ R |A_i|\le R ∣Ai∣≤R and ∣ ∑ i = 1 N A i ∣ ≤ R \left|\displaystyle\sum_{i=1}^{N}A_i\right| \le R i=1∑NAi ≤R.

There is a blackboard on which you can write integers with absolute values not exceeding R R R. Initially, the blackboard is empty.

The judge has written the values A 1 , A 2 , ... , A N A_1, A_2, \dots, A_N A1,A2,...,AN on the blackboard in this order . Your task is to make the blackboard contain only one value ∑ i = 1 N A i \displaystyle\sum_{i=1}^{N}A_i i=1∑NAi.

You cannot learn the values of R R R and A i A_i Ai directly, but you can interact with the judge up to 25000 25000 25000 times.

For a positive integer i i i, let X i X_i Xi be the i i i-th integer written on the blackboard. Specifically, X i = A i X_i = A_i Xi=Ai for i = 1 , 2 , ... , N i=1,2,\dots,N i=1,2,...,N.

In one interaction, you can specify two distinct positive integers i i i and j j j and choose one of the following actions:

Perform addition. The judge will erase X i X_i Xi and X j X_j Xj from the blackboard and write X i + X j X_i + X_j Xi+Xj on the blackboard.
∣ X i + X j ∣ ≤ R |X_i + X_j| \leq R ∣Xi+Xj∣≤R must hold.

Perform comparison. The judge will tell you whether KaTeX parse error: Expected 'EOF', got '&' at position 5: X_i &̲lt; X_j is true or false.

Here, at the beginning of each interaction, the i i i-th and j j j-th integers written on the blackboard must not have been erased.

Perform the interactions appropriately so that after all interactions, the blackboard contains only one value ∑ i = 1 N A i \displaystyle\sum_{i=1}^{N}A_i i=1∑NAi.

The values of R R R and A i A_i Ai are determined before the start of the conversation between your program and the judge.

Constraints

2 ≤ N ≤ 1000 2 \leq N \leq 1000 2≤N≤1000
1 ≤ R ≤ 1 0 9 1 \leq R \leq 10^9 1≤R≤109
∣ A i ∣ ≤ R |A_i| \leq R ∣Ai∣≤R
∣ ∑ i = 1 N A i ∣ ≤ R \left|\displaystyle\sum_{i=1}^{N}A_i\right| \le R i=1∑NAi ≤R
N N N, R R R, and A i A_i Ai are integers.

Input and Output

This is an interactive problem (where your program interacts with the judge via input and output).

First, read N N N from Standard Input.

N N N

Next, repeat the interactions until the blackboard contains only one value ∑ i = 1 N A i \displaystyle\sum_{i=1}^{N}A_i i=1∑NAi.

When performing addition, make an output in the following format to Standard Output. Append a newline at the end. Here, i i i and j j j are distinct positive integers.

i i i j j j

The response from the judge will be given from Standard Input in the following format:

P P P

Here, P P P is an integer:

If P ≥ N + 1 P \geq N + 1 P≥N+1, it means that the value X i + X j X_i + X_j Xi+Xj has been written on the blackboard, and it is the P P P-th integer written.

If P = − 1 P = -1 P=−1, it means that i i i and j j j do not satisfy the constraints, or the number of interactions has exceeded 25000 25000 25000.

When performing comparison, make an output in the following format to Standard Output. Append a newline at the end. Here, i i i and j j j are distinct positive integers.

? i i i j j j

The response from the judge will be given from Standard Input in the following format:

Q Q Q

Here, Q Q Q is an integer:

If Q = 1 Q = 1 Q=1, it means that KaTeX parse error: Expected 'EOF', got '&' at position 5: X_i &̲lt; X_j is true.

If Q = 0 Q = 0 Q=0, it means that KaTeX parse error: Expected 'EOF', got '&' at position 5: X_i &̲lt; X_j is false.

If Q = − 1 Q = -1 Q=−1, it means that i i i and j j j do not satisfy the constraints, or the number of interactions has exceeded 25000 25000 25000.

For both types of interactions, if the judge's response is − 1 -1 −1, your program is already considered incorrect. In this case, terminate your program immediately.

When the blackboard contains only one value ∑ i = 1 N A i \displaystyle\sum_{i=1}^{N}A_i i=1∑NAi, report this to the judge in the following format. This does not count towards the number of interactions. Then, terminate your program immediately.

If you make an output in a format that does not match any of the above, -1 will be given from Standard Input.

-1

In this case, your program is already considered incorrect. Terminate your program immediately.

Notes

For each output, append a newline at the end and flush Standard Output. Otherwise, the verdict may be TLE.
Terminate your program immediately after outputting the result (or receiving -1). Otherwise, the verdict will be indeterminate.
Extra newlines will be considered as malformed output.

Sample Input and Output

Here is a possible conversation with N = 3 , R = 10 , A 1 = − 1 , A 2 = 10 , A 3 = 1 N=3, R=10, A_1=-1, A_2=10, A_3=1 N=3,R=10,A1=−1,A2=10,A3=1.

Input	Output	Explanation
`3`		First, the integer $N$ is given.
	`? 1 2`	Perform a comparison.
`1`		The judge returns $1$ because $X_1\\lt X_2\\ (-1\\lt 10)$ .
	`+ 1 3`	Perform an addition.
`4`		The judge erases $X_1 = -1$ and $X_3 = 1$ from the blackboard and writes $X_1 + X_3 = 0$ . This is the fourth integer written.
	`+ 2 4`	Perform an addition.
`5`		The judge erases $X_2 = 10$ and $X_4 = 0$ from the blackboard and writes $X_2 + X_4 = 10$ . This is the fifth integer written.
	`!`	The blackboard contains only one value $\\displaystyle\\sum_{i=1}\^{N}A_i$ , so report this to the judge.

Solution

具体见文末视频。

Code

cpp 复制代码

#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;

bool cmp(int a, int b) {
	cout << "? " << a << " " << b << endl;
	int ok;
	cin >> ok;
	return ok;
}

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

	int n;
	cin >> n;

	std::vector<int> id;
	for (int i = 1; i <= n; i ++)
		id.push_back(i);
	sort(id.begin(), id.end(), cmp);
	while (n > 1) {
		cout << "+ " << id[0] << " " << id.back() << endl;
		int p;
		cin >> p;
		id.erase(id.begin()), id.pop_back();
		n --;
		if (n == 1) break;
		int l = 0, r = id.size() - 1;
		while (l < r) {
			int mid = l + r >> 1;
			if (cmp(p, id[mid])) r = mid;
			else l = mid + 1;
		}
		if (!cmp(p, id[r])) r ++;
		id.insert(id.begin() + r, p);
	}

	cout << "!\n";

	return 0;
}

视频题解

AtCoder Regular Contest 179（A ~ C 题讲解）

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