鏈接：https://ac.nowcoder.com/acm/contest/887/A
來(lái)源：牛客網(wǎng)

題目描述

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.

輸入描述:

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.

輸出描述:

For each test case, output one string separated by a space. 示例1

輸入

復(fù)制 4 0 0001 0010 111011110

輸出

復(fù)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 longint 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("\n");}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/RootVount/p/11355976.html

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