If there can be several solutions to that problem, print the lexicographically minimum one. I.e. the solution where the first line is less than the first line of other solutions should be printed. If the two lines are equal, compare the second lines and so on. The lexicographical comparison...
More formally, if you typed a word at secondaand then the next word at secondb, then ifb - a ≤ c, just the new word is appended to other words on the screen. Ifb - a > c, then everything on the screen disappears and after that the word you have typed appears...
Samples 输入数据 1 4 6 2 abaaba 6 3 abaaba 36 9 hippopotomonstrosesquippedaliophobia 21 7 wudixiaoxingxingheclp Copy 输出数据 1 2 0 23 16 Copy Note In the first test case, one optimal solution is aaaaaa. In the second test case, the given word itself is kk-complete.登录...
You’re given a string s which consists of 27 upper-case English letters. Each English letter occurs at least once in s. Find a grid that contains a path whose tiles, viewed in the order that the path visits them, form the string s. If there’s no solution, print “Impossible” (wi...
You’re given a string s which consists of 27 upper-case English letters. Each English letter occurs at least once in s. Find a grid that contains a path whose tiles, viewed in the order that the path visits them, form the string s. If there’s no solution, print “Impossible” (wi...
Each puzzle solution in the input starts with a line containing two integers r and c (1 <= r <= 10 and 1 <= c <= 10), where r (the first number) is the number of rows in the puzzle and c (the second number) is the number of columns. The r rows of input which follow eac...
class Solution { public: int ladderLength(string beginWord, string endWord, vector<string>& wordList) { unordered_set<string> s(wordList.begin(), wordList.end()); if (s.find(endWord) == s.end() || beginWord == endWord) return 0; queue<string> q; q.push(beginWord); s.erase(...
class Solution: def exist(self, board: List[List[str]], word: str) -> bool: # - sanity check n_row = len(board) if not n_row: return False n_col = len(board[0]) if not n_col: return False def dfs(row, col, i, seen): if i == len(word)-1: return True for dr,dc...
("%s",ch),Len=strlen(ch),E.Add((ch[0]-'a')*26+(ch[1]-'a'),(ch[Len-2]-'a')*26+(ch[Len-1]-'a'),Len);doubleL=0,R=1001;for(mid=(R+L)/2;R-L>1e-4;mid=(R+L)/2)if(check()) Ans=L=mid;elseR=mid;if(Ans==0)printf("No solution\n");elseprintf("%.3lf\...
The subsequence of stringais a stringbthat can be derived from the stringaby removing some symbols from it (maybe none or all of them). In particular any string is the subsequence of itself. For example, the string "cfo" is the subsequence of the string "codeforces". ...