AtCoder Beginner Contest 242

比赛链接

AtCoder Beginner Contest 242

D - ABC Transform

Problem Statement

You are given a string S consisting of A, B, C.

Let $S^{(0)}$
:=S. For $i=1,2,3,…,$ let $S^(i)$
be the result of simultaneously replacing the characters of $S^(i−1)$
as follows: A → BC, B → CA, C → AB.

Answer $Q$ queries. The i-th query is as follows.

Print the ki-th character from the beginning of $S(t_i)$.

Constraints

• $S$ is a string of length between $1$ and $10^5$(inclusive) consisting of $A, B, C$.
• $1≤Q≤10^5$
• $0≤t_i≤10^
• $1≤k_i≤min(10^18, the length of S^{(t_i)})
• $Q,t_i ,k_i are integers.

Input

Input is given from Standard Input in the following format:

S
Q
t1 k1
t2 k2
⋮
tQ kQ

Output

Process the $Q$ queries in ascending order of index, that is, in the given order. Each answer should be followed by a newline.

Sample Input 1

ABC
4
0 1
1 1
1 3
1 6

Sample Output 1

A
B
C
B

Sample Input 2

CBBAACCCCC
5
57530144230160008 659279164847814847
29622990657296329 861239705300265164
509705228051901259 994708708957785197
176678501072691541 655134104344481648
827291290937314275 407121144297426665

Sample Output 2

A
A
C
A
A

解题思路

思维，递归

考虑当前字符是由前面那个字符转移过来的，相当于一棵二叉树，$k$ 每次找根节点，当 $k$ 为 $0$ 时即由最开始的 $0$ 转移过来的，且转移时是以循环转移的

• 时间复杂度：$O(logk)$

代码

// Problem: D - ABC Transform
// Contest: AtCoder - AtCoder Beginner Contest 242
// URL: https://atcoder.jp/contests/abc242/tasks/abc242_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// %%%Skyqwq
#include <bits/stdc++.h>
 
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
 
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
 
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
 
template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}


string s;
int q;
LL t,k;
int get(LL t,LL k)
{
	if(t==0)return s[k]-'A';
	else if(k==0)return (s[k]-'A'+t%3)%3;
	else
		return (get(t-1,k/2)+1+k%2)%3;
}
int main()
{
    cin>>s;
    cin>>q;
    while(q--)
    {
    	cin>>t>>k;
    	k--;
    	cout<<char(get(t,k)+'A')<<'\n';
    }
    return 0;
}

E - (∀x∀)

Problem Statement

Solve the following problem for $T$ test cases.
Given an integer $N$ and a string $S$, find the number of strings $X$ that satisfy all of the conditions below, modulo 998244353 .

• $X$ is a string of length $N$ consisting of uppercase English letters.
• $X$ is a palindrome.
• $X \leq S$ in lexicographical order.
• • That is, $X=S$ or $X$ is lexicographically smaller than $S$.

Constraints

• $1 \leq T \leq 250000$
• $N$ is an integer between 1 and $10^{6}$ (inclusive).
• In a single input, the sum of $N$ over the test cases is at most $10^{6}$.
• $S$ is a string of length $N$ consisting of uppercase English letters.

Input

Input is given from Standard Input in the following format:
$T$
case $_{1}$
case $_{2}$
case $_{T}$
Here, case $_{i}$ represents the $i$-th test case.
Each test case is in the following format:
$N$
$S$

Output

Print $T$ lines. The $i$-th line should contain the answer for the $i$-th test case as an integer.

Sample Input 1

5
3
AXA
6
A B C Z A Z
30
QWERTYUIOPASDFGHJKLZXCVBNMQWER
28
JVIISNEOXHSNEAAENSHXOENSIIVJ
31
KVOHEEMSOZZASHENDIGOJRTJVMVSDWW

Sample Output 1

24
29
212370247
36523399
231364016

This input contains five test cases.
Test case #1:
The 24 strings satisfying the conditions are AAA, ABA, ACA, ..., AXA.
Test case #2:
$S$ may not be a palindrome.
Test case #3:
Be sure to find the count modulo $998244353 .$

解题思路

思维

找出前半部分，小于前半部分的回文串肯定都是满足条件的，这里的 $A$ 相当于十进制的 $1$，而对于长度为奇数的情况特判一下即可

• 时间复杂度：$O(tlogn)$

代码

// Problem: E - (∀x∀)
// Contest: AtCoder - AtCoder Beginner Contest 242
// URL: https://atcoder.jp/contests/abc242/tasks/abc242_e
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// %%%Skyqwq
#include <bits/stdc++.h>
 
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
 
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
 
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
 
template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}

const int mod=998244353;
int t,n;
string s;
LL get()
{
	LL res=0;
	string t=s.substr(0,(n+1)/2);
	string t1=t;
	reverse(t1.begin(),t1.end());
	for(int i=0;i<t.size();i++)res=(res*26+t[i]-'A')%mod;
	if(n&1)t1.erase(t1.begin());
	if(t+t1<=s)res=(res+1)%mod;
	return res;
}
int main()
{
    for(cin>>t;t;t--)
    {
    	cin>>n>>s;
    	cout<<get()<<'\n';
    }
    return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/15970367.html
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