Notepad

# Notepad#

## 题面翻译

### 题目描述

一开始打出的内容为空。现在你要打出一个长度为 n 的字符串 s(全为英文小写字母组成),为此每次你可以进行如下操作中的一种:

- 在已打出内容的最后添加一个字符。
- 复制已打出内容的一个连续的子串并加到内容的末尾。

问你能不能在严格小于 n 次操作下打出字符串 s

### 输入格式

t 组数据。第一行输入正整数 t(1t104)

每组数据第一行输入正整数 n,第二行输入字符串 s

单个测试点内所有 n 之和不超过 2×105

### 输出格式

输出 t 行,每行输出这组数据的答案。如果可以达到要求,输出 `YES`。否则输出 `NO`。

## 题目描述

You want to type the string s , consisting of n lowercase Latin letters, using your favorite text editor Notepad#.

Notepad# supports two kinds of operations:

- append any letter to the end of the string;
- copy a continuous substring of an already typed string and paste this substring to the end of the string.

Can you type string s in strictly less than n operations?

## 输入格式

The first line contains a single integer t ( 1t104 ) — the number of testcases.

The first line of each testcase contains a single integer n ( 1n2105 ) — the length of the string s .

The second line contains a string s , consisting of n lowercase Latin letters.

The sum of n doesn't exceed 2105 over all testcases.

## 输出格式

For each testcase, print "YES" if you can type string s in strictly less than n operations. Otherwise, print "NO".

## 样例 #1

### 样例输入 #1

```
6
10
codeforces
8
labacaba
5
uohhh
16
isthissuffixtree
1
x
4
momo
```

### 样例输出 #1

```
NO
YES
NO
YES
NO
YES
```

## 提示

In the first testcase, you can start with typing "codef" ( 5 operations), then copy "o" ( 1 operation) from an already typed part, then finish with typing "rces" ( 4 operations). That will be 10 operations, which is not strictly less than n . There exist other ways to type "codeforces". However, no matter what you do, you can't do less than n operations.

In the second testcase, you can type "labac" ( 5 operations), then copy "aba" ( 1 operation), finishing the string in 6 operations.

//Notepad# //只需要寻找出字符串中是否存在不重叠的两个相同的子串就可以了,字串长度为2就可以了 //利用map存储 #include <bits/stdc++.h> #define int long long using namespace std; const int N=1e6+10,mod=1e9+7; string s; int n,t,a[N],f[N],res,num,ans,m; bool vis[N]; signed main() { std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); cin>>t; while(t--){ cin>>n>>s; map<string,int>p; res=0; for(int i=0;i<s.size();i++){ string tmp; tmp+=s[i]; tmp+=s[i+1]; if(p.count(tmp)==1&&p[tmp]!=i-1){ cout<<"YES"<<endl; goto nexts; } if(!p.count(tmp)) p[tmp]=i; } cout<<"NO"<<endl; nexts:; } return 0; }

 


__EOF__

本文作者Sakurajimamai
本文链接https://www.cnblogs.com/o-Sakurajimamai-o/p/17551861.html
关于博主:评论和私信会在第一时间回复。或者直接私信我。
版权声明:本博客所有文章除特别声明外,均采用 BY-NC-SA 许可协议。转载请注明出处!
声援博主:如果您觉得文章对您有帮助,可以点击文章右下角推荐一下。您的鼓励是博主的最大动力!
posted @   o-Sakurajimamai-o  阅读(32)  评论(0编辑  收藏  举报
相关博文:
阅读排行:
· winform 绘制太阳,地球,月球 运作规律
· AI与.NET技术实操系列(五):向量存储与相似性搜索在 .NET 中的实现
· 超详细:普通电脑也行Windows部署deepseek R1训练数据并当服务器共享给他人
· 【硬核科普】Trae如何「偷看」你的代码?零基础破解AI编程运行原理
· 上周热点回顾(3.3-3.9)
-- --
点击右上角即可分享
微信分享提示