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来源：互联网发布：linux top命令编辑：程序博客网时间：2024/06/05 18:48

题目传送门

题意：给你一个字符串，你在这个字符串中间找一个最长的字串，可以和开头和结尾相同。

思路：这个题还是比较暴力的，直接KMP和枚举就可以了。

#include <algorithm>#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <list>#include <map>#include <queue>#include <set>#include <stack>#include <string>#include <vector>#define MAXN 1010000#define MAXE 100#define INF 1000000000#define MOD 10001#define LL long long#define ULL unsigned long long#define pi 3.14159using namespace std;int nex[MAXN];bool vis[MAXN];string str;int n, m;int max_length;void get_nex() {    memset(nex, 0, sizeof(nex));    for (int i = 1, j = 0; i < n; ++i) {        while (j && str[i] != str[j]) {            j = nex[j];        }        if (str[i] == str[j]) {            j++;        }        nex[i + 1] = j;    }}void KMP(int pos, int l) {    for (int i = pos, j = 0; i < l; i++) {        while (j && str[i] != str[j])            j = nex[j];        if (str[i] == str[j])            j++;        if (j > m)            return;        if (j > max_length)            max_length = j;    }}int main() {    std::ios::sync_with_stdio(false);    int T;    cin >> T;    for (int kase = 1; kase <= T; ++kase) {        cin >> str;        n = (int) str.length();        if (n < 3) {            cout << 0 << endl;        } else {            get_nex();            memset(vis, false, sizeof(vis));            max_length = 0;            for (int i = min(nex[n], n / 3); i >= max_length && i >= 1; --i) {                m = i;                KMP(i, n - i);            }            cout << max_length << endl;        }    }    return 0;}

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