2019牛客暑期多校训练营（第七场）A String（暴力）

链接：https://ac.nowcoder.com/acm/contest/887/A
来源：牛客网

题目描述

A string is perfect if it has the smallest lexicographical ordering among its cyclic rotations.
For example: "0101" is perfect as it is the smallest string among ("0101", "1010", "0101", "1010").

Given a 01 string, you need to split it into the least parts and all parts are perfect.

输入描述:

The first line of the input gives the number of test cases, T (T≤300)T (T leq 300)T (T≤300).  test cases follow.

For each test case, the only line contains one non-empty 01 string. The length of string is not exceed 200.

输出描述:

For each test case, output one string separated by a space.

示例1

输入

复制

4
0
0001
0010
111011110

输出

复制

0
0001
001 0
111 01111 0

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
using namespace std;
#define MAXN 100010
#define ll long long

int t;
string s;

bool minma(string x)
{
    int len = x.size();
    for(int i = 1; i < len; i++)
        for(int j = 0; j < len; j++)
        {
            if(x[j] < x[(j + i) % len])
                break;
            else if(x[j] > x[(j + i) % len])
                return false;
        }
    return true;
}

int main()
{
    cin>>t;
    while(t--)
    {
        cin>>s;
        int len = s.size();
        int i = 0;
        while(i < len)
        {
            for(int j = len - i; j > 0; j--)
            {
                if(minma(s.substr(i, j)))
                {
                    cout<<s.substr(i, j);
                    i += j;
                    if(i < len)
                        printf(" ");
                    break;
                }
            }
        }
        printf("
");
    }
    return 0;
}