【代码超详解】Codeforces 126B. Password（KMP 的 next 数组的应用，154 ms）

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一、题目描述

B. Password

time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.

A little later they found a string s, carved on a rock below the temple’s gates. Asterix supposed that that’s the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring t of the string s.

Prefix supposed that the substring t is the beginning of the string s; Suffix supposed that the substring t should be the end of the string s; and Obelix supposed that t should be located somewhere inside the string s, that is, t is neither its beginning, nor its end.

Asterix chose the substring t so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring t aloud, the temple doors opened.

You know the string s. Find the substring t or determine that such substring does not exist and all that’s been written above is just a nice legend.

Input

You are given the string s whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.

Output

Print the string t. If a suitable t string does not exist, then print “Just a legend” without the quotes.

Examples

Input

fixprefixsuffix

Output

fix

Input

abcdabc

Output

Just a legend

二、算法分析说明与代码编写指导

题目要求：给出一个字串，寻找一个尽量长的子串，子串既是前缀也是后缀，而且在中间亦出现过（首尾分别不能与母串的首尾重合）。
用 KMP 算法的构造 next 数组的代码打一个 next 数组。
（注意：本代码中，字符串的第一个字符的下标为 1）
设输入的字符串长为 m，则新建一个 bitset vis，然后用如下代码遍历 next 数组：

	for (unsigned i = 1; i <= m; ++i)vis[next[i]] = true;

在遍历 next 的过程中，如果字符串的第 i 个字符以前存在相同的前缀和后缀，那么前缀的下一个位置相应的 vis 会标为 1。
然后从 i = m + 1 开始不断向前跳跃（i = next[i]）。我们知道，vis[m + 1] = 0。如果整个串只有同时作为前缀和后缀的一段字符或者干脆连这段字符也没有，那么一定有 vis[next[i]] = 0。如果整个串中存在一个既是前缀又是后缀，还在中间出现过的子串，那么刚才遍历并在 vis 上标记时，遍历完在中间出现的那个子串副本之后，子串作为前缀的副本的下一个位置的 vis 会被标记为 1。此时找到的子串便是答案，输出即可。
至于为何要通过循环不断地进行 i = next[i] ，这是因为在一开始如果找到了一个既是前缀也是后缀但在中间没有出现过的子串时，这个子串本身还可能具有相同的前后缀。此时通过 i = next[i] 减小继续搜寻的子串的长度，就可能找到比原串的最长前后缀短一些的、符合要求的子串。KMP 算法中，构造 next 数组也用到了类似的思想。

三、AC 代码（154 ms）

#include<cstdio>
#include<cstring>
#include<bitset>
#pragma warning(disable:4996)
template<class _NTy> inline void getnext(const char* const _Pattern, _NTy* const _Next, const size_t& _PatternLen) {
	size_t i = 1; _NTy j = 0; _Next[1] = 0;
	while (i <= _PatternLen) {
		if (j == 0 || _Pattern[i] == _Pattern[j]) { ++i, ++j, _Next[i] = j; }
		else j = _Next[j];
	}
}//The index of the head of _Pattern string is 1.
char s[1000002]; unsigned next[1000002], t; unsigned long long m; std::bitset<1000002> vis;
int main() {
	gets(s + 1); m = strlen(s + 1); getnext(s, next, m);
	for (unsigned i = 1; i <= m; ++i)vis[next[i]] = true;
	for (unsigned i = m + 1; next[i] > 1; i = next[i]) {
		if (vis[next[i]] == true) {
			t = next[i];
			for (unsigned j = 1; j < t; ++j)putchar(s[j]);
			putchar('\n'); return 0;
		}
	}
	puts("Just a legend"); return 0;
}