Division + LCP (easy version)

题目描述

This is the easy version of the problem. In this version $l=r$ .

You are given a string $s$ . For a fixed $k$ , consider a division of $s$ into exactly $k$ continuous substrings $w_1,\dots ,w_k$ . Let $f_k$ be the maximal possible $LCP(w_1,\dots ,w_k)$ among all divisions.

$LCP(w_1,\dots ,w_m)$ is the length of the Longest Common Prefix of the strings $w_1,\dots ,w_m$ .

For example, if $s=abababcab$ and $k=4$ , a possible division is $abababcab$ . The $LCP(ab,ab,abc,ab)$ is $2$ , since $ab$ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring.

Your task is to find $f_l,f_{l+1},\dots ,f_r$ . In this version $l=r$ .

输入格式

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 $n$ , $l$ , $r$ ( $1\le l=r\le n\le 2\cdot 10^5$ ) — the length of the string and the given range.

The second line of each test case contains string $s$ of length $n$ , all characters are lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot 10^5$ .

输出格式

For each test case, output $r-l+1$ values: $f_l,\dots ,f_r$ .

样例 #1

样例输入 #1

7
3 3 3
aba
3 3 3
aaa
7 2 2
abacaba
9 4 4
abababcab
10 1 1
codeforces
9 3 3
abafababa
5 3 3
zpozp

样例输出 #1

0
1
3
2
10
2
0

提示

In the first sample $n=k$ , so the only division of $aba$ is $aba$ . The answer is zero, because those strings do not have a common prefix.

In the second sample, the only division is $aaa$ . Their longest common prefix is one.

原题

CF——传送门

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;int n, Left, Right;
const int MAX = 1e6 + 6;
int Next[MAX];
vector<int> ans;
inline void GetNext(string s, int l) // 获得字符串s的Next数组
{int t;Next[0] = -1;               // 如果在0位置失配则是向下移动一位for (int i = 1; i < l; ++i) // 依次求解后面的Next数组{t = Next[i - 1];while (s[t + 1] != s[i] && t >= 0)t = Next[t];if (s[t + 1] == s[i])Next[i] = t + 1;elseNext[i] = -1;}
}
inline void KMP(string s1, int l1, string s2, int l2)
{GetNext(s2, l2);int i = 0, j = 0;while (j < l1){if (s2[i] == s1[j]) // 当前位匹配成功，继续匹配下一位{++i;++j;if (i == l2) // 完全匹配{ans.push_back(j - l2); // 储存答案i = Next[i - 1] + 1;   // 继续匹配}}else{if (i == 0) // 在首位不匹配j++;elsei = Next[i - 1] + 1;}}
}
bool check(string s, string a, int k)
{KMP(s, s.size(), a, a.size());int num = 0;// 去掉重叠的匹配子串if (ans.size() >= 1){num++;int last = ans[0];for (int i = 1; i < ans.size(); i++){if (ans[i] - last > k - 1){num++;last = ans[i];}}}ans.clear();if (num >= Left)return 1;elsereturn 0;
}
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;cin >> t;while (t--){cin >> n >> Left >> Right;string s;cin >> s;// 二分int l = 0, r = n / Left;while (l < r){int mid = (l + r + 1) / 2; // 这里要 l + r +1 要不然会死循环string a = s.substr(0, mid);if (check(s, a, mid)) // check函数自己写{l = mid; // mid这个位置 满足条件之后 查找 [mid , right]的位置， 所以l移到mid的位置}else{r = mid - 1; // [mid,r] 不满足条件， 所以要移到满足条件的一方， r = mid - 1}}cout << l << '\n';}return 0;
}