《使命召喚手遊》測試安卓招募正式開啟：寫實畫風+高清畫面
ARC104 題解
目錄ARC回來了？！
A - Plus Minus
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 100 points
Problem Statement
Takahashi has two integers X and Y.He computed \(X + Y\) and \(X - Y\), and the results were A and B, respectively.
Now he cannot remember what X and Y were. Find X and Y for him.
Constraints
- \(-100 \leq A, B \leq 100\)
- For the given integers A and B, there uniquely exist integers X and Y such that \(X + Y = A\) and \(X - Y = B\).
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
Print X and Y.
Samples
Sample Input 1
2 -2
Sample Output 1
0 2
If X = 0 and Y = 2, they match the situation: \(0 + 2 = 2 \ \ \operatorname{and}\ \ 0 - 2 = -2\).
Sample Input 2
3 1
Sample Output 2
2 1
題面
高橋有兩個整數X和Y，他計算了\(X+Y\)和\(X-Y\)，結果分別是A和B。
現在他不記得X和Y是什麼了。為他找到X和Y。
解題思路
\[\begin{align} (A\ge B) \begin{cases} A+B=C\\ A-B=D \end{cases} \Rightarrow \begin{cases} A=\dfrac{C+D}2\\ B=\dfrac{C-D}2 \end{cases} \end{align} \]AC Code
#include<iostream>
using namespace std;
int main(){
int c,d;
cin>>c>>d;
cout<<(c+d)/2<<endl<<(c-d)/2<<endl;
return 0;
}
A, B = map(int, input().split())
print((A + B) // 2, (A - B) // 2)
B - DNA Sequence
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 400 points
Problem Statement
We have a string S of length N consisting of A, T, C, and G.
Strings \(T_1\) and \(T_2\) of the same length are said to be complementary when, for every \(i (1 \leq i \leq l)\), the i-th character of \(T_1\) and the i-th character of \(T_2\) are complementary. Here, A and T are complementary to each other, and so are C and G.
Find the number of non-empty contiguous substrings T of S that satisfies the following condition:
- There exists a string that is a permutation of T and is complementary to T.
Here, we distinguish strings that originate from different positions in S, even if the contents are the same.
Constraints
- \(1 \leq N \leq 5000\)
- S consists of
A,T,C, andG.
Input
Input is given from Standard Input in the following format:
N S
Output
Print the number of non-empty contiguous substrings T of S that satisfies the condition.
Samples
Sample Input 1
4 AGCT
Sample Output 1
2
The following two substrings satisfy the condition:
GC(the 2-nd through 3-rd characters) is complementary toCG, which is a permutation ofGC.AGCT(the 1-st through 4-th characters) is complementary toTCGA, which is a permutation ofAGCT.
Sample Input 2
4 ATAT
Sample Output 2
4
The following four substrings satisfy the condition:
AT(the 1-st through 2-nd characters) is complementary toTA, which is a permutation ofAT.TA(the 2-st through 3-rd characters) is complementary toAT, which is a permutation ofTA.AT(the 3-rd through 4-th characters) is complementary toTA, which is a permutation ofAT.ATAT(the 1-st through 4-th characters) is complementary toTATA, which is a permutation ofATAT.
題面
我們有一個長度為N的字串S，由A, T, C, 和G組成.
當對於每一個\(i（1\leq i\leq l）\)，\(T_1\)的第i個字元和\(T_2\)的第i個字元"互補"，長度相同的字串\(T_1\)和\(T_2\)稱為互補字串。在這裡，A和T是互補的，C和G也是互補的。
求滿足以下條件的S的非空連續子串T的個數：
- 有一個字串，它是T的一個排列，與T“互補”。
這裡，我們區分來自S中不同位置的字串，即使內容相同。
解題思路
First, let us consider what kinds of strings satisfy the condition.
Ais complementary toTTis complementary toACis complementary toGGis complementary toC
Thus, it is necessary and sufficient that:
- (the number of
A) = (the number ofT) - (the number of
C) = (the number ofG)
To check if each segment satisfies these conditions, we can, for example, process the segments with the same left end together to solve the problem in a total of \(\mathrm{O}(N^2)\) time.
只要保證在這個字串s中，A的數量=T的數量，C的數量=G的數量。為了檢查每個段是否滿足這些條件，我們可以，例如，將具有相同左端的段一起處理以在總計\(\mathrm{O}（N^2）\)時間內解決問題。
$ N \leq 2 \times 10^5 $
AC Code
#include <bits/stdc++.h>
using namespace std;
int main() {
int N;
string S;
cin >> N >> S;
int ans = 0;
for (int i = 0; i < N; ++i) {
int c1 = 0, c2 = 0;
for (int j = i; j < N; ++j) {
if (S[j] == 'A')
c1++;
else if (S[j] == 'T')
c1--;
else if (S[j] == 'C')
c2++;
else
c2--;
if (c1 == 0 && c2 == 0) ans++;
}
}
cout << ans << endl;
return 0;
}
n, s = input().split()
n = int(n)
at = 0
cg = 0
from collections import Counter
d = Counter([(0, 0)])
ans = 0
for ss in s:
if ss == 'A': at += 1
if ss == 'T': at -= 1
if ss == 'C': cg += 1
if ss == 'G': cg -= 1
ans += d[(at, cg)]
d[(at, cg)] += 1
print(ans)
C - Fair Elevator
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 600 points
Problem Statement
There is a building with 2N floors, numbered $1, 2, \ldots, 2N $from bottom to top.
The elevator in this building moved from Floor 1 to Floor 2N just once.
On the way, N persons got on and off the elevator. Each person \(i (1 \leq i \leq N)\) got on at Floor \(A_i\) and off at Floor \(B_i\). Here, \(1 \leq A_i < B_i \leq 2N\), and just one person got on or off at each floor.
Additionally, because of their difficult personalities, the following condition was satisfied:
Let \(C_i (= B_i - A_i - 1)\) be the number of times, while Person i were on the elevator, other persons got on or off. Then, the following holds:
- If there was a moment when both Person i and Person j were on the elevator, \(C_i = C_j\).
We recorded the sequences A and B, but unfortunately, we have lost some of the records. If the record of \(A_i\) or \(B_i\) is lost, it will be given to you as -1.
Additionally, the remaining records may be incorrect.
Determine whether there is a pair of A and B that is consistent with the remaining records.
Constraints
- \(1 \leq N \leq 100\)
- \(A_i = -1 \ \ or \ \ 1 \leq A_i \leq 2N.\)
- \(B_i = -1\ \ or\ \ 1 \leq B_i \leq 2N\).
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
:
A_N B_N
Output
If there is a pair of A and B that is consistent with the remaining records, print Yes; otherwise, print No.
Samples
Sample Input 1
3
1 -1
-1 4
-1 6
Sample Output 1
Yes
For example, if \(B_1 = 3, A_2 = 2\), and \(A_3 = 5\), all the requirements are met.
In this case, there is a moment when both Person 1 and Person 2 were on the elevator, which is fine since \(C_1 = C_2 = 1\).
Sample Input 2
2
1 4
2 3
Sample Output 2
No
There is a moment when both Person 1 and Person 2 were on the elevator. Since \(C_1 = 2, C_2 = 0\), some of the information is incorrect.
Sample Input 3
2
4 1
2 4
Sample Output 3
No
The records are seemingly intact but clearly are incorrect.
題面
There is a building with 2N2N floors, numbered 1,2,…,2N1,2,…,2N from bottom to top.
有一個2N 層的建築，從下到上編號為1,2，點，2 n。
The elevator in this building moved from Floor 11 to Floor 2N2N just once.
這棟樓裡的電梯只從一層搬到了二層 n。
On the way, NN persons got on and off the elevator. Each person ii (1≤i≤N)(1≤i≤N) got on at Floor AiAi and off at Floor BiBi. Here, 1≤Ai<Bi≤2N1≤Ai<Bi≤2N, and just one person got on or off at each floor.
在路上，有 n 個人進出電梯。每個人在 a _ i 層上車，在 b _ i 層下車。在這裡，每層樓只有一個人上下車。
Additionally, because of their difficult personalities, the following condition was satisfied:
此外，由於他們的性格特點，下列條件得到了滿足:
- Let \(C_i (= B_i - A_i - 1)\) be the number of times, while Person i were on the elevator, other persons got on or off. Then, the following holds:
- If there was a moment when both Person i and Person j were on the elevator, \(C_i = C_j\).
我們記錄了 a 和 b 序列，但不幸的是，我們丟失了一些記錄。如果 a _ i 或 b _ i 的記錄丟失，它將作為 -1給你。
此外，其餘的記錄可能是不正確的。
確定是否存在一對與其餘記錄一致的 a 和 b。
解題思路
https://atcoder.jp/contests/arc104/editorial/162
First, let us sort out the conditions. Assume that:
- \(A_i < B_i\)
- \(A_1, B_1, A_2, B_2 \cdots, A_N, B_N\) are all different.
Let us say that the person who got on at Floor 1 gets off at Floor x. Then, we can see that the persons who got on at Floor \(2, 3, \cdots, x-1\) get off at Floor \(x + 1, x + 2, \cdots, 2x-2\).
By repeating this argument, we can see that the condition is satisfied if and only if it is possible to divide the 2N floors into segments where the persons with \((A, B) = (x, x+k), (x+1, x+k+1), \cdots, (x+k-1, x+2k-1)\) get on and off.
Now, let us think of a way to determine, for a fixed division, if it is possible to fill in A and B so that the above holds.
Let us make an array that stores (left/right, the person's ID) for each floor if there is a person bound to get on/off on that floor.
For each segment, the following is necessary:
- There is no person with fixed A and B such that only one of them is contained in the segment. If a person has both A and B in the segment, they are in the correct positions.
- If the persons who got on/off at Floor x+i/x+k+i are both already fixed, they are the same person.
- There is no pair of floors (x+i, x+k+i) such that the person at Floor x+i is "left."
- There is no pair of floors (x+i, x+k+i) such that the person at Floor x+k+i is "right."
On the other hand, if the above is satisfied for every segment, we can find one correct record by filling in A and B properly.
Thus, the problem can be solved by DP. Let \(dp[i]:= (\texttt{true if a segment ends exactly at Floor i and there is no contradiction until Floor i, false otherwise})\), and make dp[r] true when a new segment [i+1, r] has no contradiction.
The time complexity is \(\mathrm{O}(N^3)\).
AC Code
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int n;
bool f[N];//[1, i] 是否恰好能合法填充
int com[N];//配對位置
int id[N];//此位置對應的人的編號
int g[N];//此位置是左 1 / 右 0 端點
int vis[N];//此位置是否被佔
bool ok = true;
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
int a, b;
scanf("%d%d", &a, &b);
if (a != -1) {
if (vis[a]) ok = false;
vis[a] = 1, id[a] = i, g[a] = 1;
}
if (b != -1) {
if (vis[b]) ok = false;
vis[b] = 1, id[b] = i, g[b] = 0;
}
if (a != -1 && b != -1) {
if (a > b) ok = false;
com[a] = b, com[b] = a;
}
}
if (!ok) {
puts("No");
return 0;
}
n <<= 1;
f[0] = 1;
for (int i = 0; i <= n; i += 2) {//奇數必定不是結尾, 故 += 2
if (!f[i]) continue;//此狀態不合法, 自然無法用它推別人
for (int j = i+2; j <= n; j += 2) {//更新 f[j]
ok = 1;
for (int p = i+1, q = i+1+(j-i)/2; q <= j; p++, q++) {//列舉 [i+1, j] 中需要配對的點對;可以不考慮間距不為 (j-i)/2 的點對, 因為在(j-i)/2為相應的值時會更新
if (vis[p] && vis[q]) {//兩位置均有定好的值
if (id[p] != id[q]) {
ok = false;
break;
}
//不用判斷是否滿足左右關係, 因為輸入時已經判過了
}
else if (vis[p]) {//p 有定好的值
if (com[p] || g[p] == 0) {//p有對應的值, 那麼必定不是 q ; p是定好的右端點, 那必定不符
ok = false;
break;
}
}
else if (vis[q]) {
if (com[q] || g[q] == 1) {
ok = false;
break;
}
}
}
if (ok)
f[j] = 1;
}
}
puts(f[n] ? "Yes" : "No");
return 0;
}
D - Multiset Mean
Time Limit: 4 sec / Memory Limit: 1024 MB
Score : 700 points
Problem Statement
Given positive integers N, K and M, solve the following problem for every integer x between 1 and N (inclusive):Find the number, modulo M, of non-empty multisets containing between 0 and K (inclusive) instances of each of the integers \(1, 2, 3 \cdots, N\) such that the average of the elements is x.
Constraints
- \(1 \leq N, K \leq 100\)
- \(10^8 \leq M \leq 10^9 + 9\)
- M is prime.
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K M
Output
Use the following format:
c_1
c_2
:
c_N
Here, c_x should be the number, modulo M, of multisets such that the average of the elements is x.
Samples
Sample Input 1
3 1 998244353
Sample Output 1
1
3
1
Consider non-empty multisets containing between 0 and 1 instance(s) of each of the integers between 1 and 3. Among them, there are:
- one multiset such that the average of the elements is \(k = 1: \{1\}\);
- three multisets such that the average of the elements is \(k = 2: \{2\}, \{1, 3\}, \{1, 2, 3\}\);
- one multiset such that the average of the elements is \(k = 3: \{3\}\).
Sample Input 2
1 2 1000000007
Sample Output 2
2
Consider non-empty multisets containing between 0 and 2 instances of each of the integers between 1 and 1. Among them, there are:
- two multisets such that the average of the elements is \(k = 1: \{1\}, \{1, 1\}.\)
Sample Input 3
10 8 861271909
Sample Output 3
8
602
81827
4054238
41331779
41331779
4054238
81827
602
8
解題思路
https://atcoder.jp/contests/arc104/editorial/163
A naive DP would have the count of each number and the sum as the states, but its time complexity would be \mathrm{O}(N^4K^2) even after optimization, which is too slow, so we need to exploit the particular setting.
Let us say we want to count sets whose average is m.
We can consider it as the number of balanced states where there is a straight bar that pivots on a fulcrum at coordinate m, and we attach at most K objects of weight 1 at coordinates 1, 2, \cdots, N. (This is because we can transform (\sum_{x \in S}x ) / |S| = m to (\sum_{x \in S} (x - m) ) = 0.)
An easy-to-implement solution
Let us divide the bar at the fulcrum into the left and right parts. Now we have to choose two subsets from $S = {1,2,\cdots,m-1}, T = {1,2,\cdots,N-m} $so that the sums are equal. Thus, if we precompute \(dp[i][j] := (\texttt{the number of sets totaling j considering} 1,2,\cdots,i)\) for every i and j, we can process the queries in$ \mathrm{O}(N^2K)$ time on average per query. For the precomputation, it is \(dp[i-1][j], dp[i-1][j - i], \cdots, dp[i-1][j - i * K]\) that transition to dp[i][j] above, so we can optimize it by maintaining the sum while "sliding" within each \mod i, and the time complexity will be \(\mathrm{O}(N^3K)\).
AC Code
A solution using Formal Power Series
Even if we do not notice the use of precomputation, we can still maintain the distribution of sums in the left and right parts while shifting the average one by one. (Since the counts are fixed, the relative distribution only changes at both ends.)
If we think of the DP above as maintaining \prod(1 + x^i + x^{2i} + \cdots + x^{Ki}), since (1 + x^i + x^{2i} + \cdots + x^{Ki}) = (1 - x^{(K+1)i})/(1-x^i), the addition and removal at both ends can be regarded as multiplying or dividing this formula.
This solution has the same time complexity as the solution above and an improved space complexity.