You‘re given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
The first input line contains the string. It‘s guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn‘t exceed 100.
Output one number — length of the longest substring that can be met in the string at least twice.
abcd
0
ababa
3
zzz
2
找重复出现至少一次的最大子串长度
代码：
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <map> #define Max 1001 using namespace std; char s[101]; int main() { cin>>s; int d = 0; for(int i = 0;i < strlen(s) - 1;i ++) { for(int j = i + 1;j < strlen(s);j ++) { int cnt = 0; while(j + cnt <strlen(s)&&s[i + cnt] == s[j + cnt])cnt ++; if(d < cnt)d = cnt; } } cout<<d<<endl; }
原文：http://www.cnblogs.com/8023spz/p/7966827.html