题解 | #字符串是否由子串拼接#

#include <iostream>
#include <string>
using namespace std;

string findSubstring(string s) {
    int n = s.length();
    
    // 将字符串拼接自身并去掉首尾
    string doubled = s + s;
    string trimmed = doubled.substr(1, 2 * n - 2);
    
    // 在处理后的字符串中查找原字符串
    size_t pos = trimmed.find(s);
    if(pos != string::npos && pos < n) {
        return s.substr(0, pos + 1);
    }
    return "false";
}

int main() {
    string s;
    cin >> s;
    cout << findSubstring(s) << endl;
    return 0;
}

import java.util.Scanner;

public class Main {
    public static String findSubstring(String s) {
        int n = s.length();
        
        // 将字符串拼接自身并去掉首尾
        String doubled = s + s;
        String trimmed = doubled.substring(1, 2 * n - 1);
        
        // 在处理后的字符串中查找原字符串
        int pos = trimmed.indexOf(s);
        if(pos != -1 && pos < n) {
            return s.substring(0, pos + 1);
        }
        return "false";
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        System.out.println(findSubstring(s));
    }
}

def find_substring(s):
    n = len(s)
    
    # 将字符串拼接自身并去掉首尾
    doubled = s + s
    trimmed = doubled[1:2*n-1]
    
    # 在处理后的字符串中查找原字符串
    pos = trimmed.find(s)
    if pos != -1 and pos < n:
        return s[:pos+1]
    return "false"

s = input()
print(find_substring(s))