#include<bits/stdc++.h> using namespace std; #define rep(i,a,b) for(int i=a;i<=b;++i) #define ms(arr,a) memset(arr,a,sizeof arr) #define debug(x) cout<<"< "#x" = "<<x<<" >"<<endl char s[…

Problem J. Let Sudoku Rotate Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Sudoku is a logic-based, combinatorial number-placement puzzle,…

Problem J. Let Sudoku Rotate Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Sudoku is a logic-based, combinatorial number-placement puzzle, which is popular around the w…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=6341 Problem J. Let Sudoku Rotate Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1363    Accepted Submission(s): 717 Problem Description Sudoku i…

一.题意 Sudoku is a logic-based, combinatorial number-placement puzzle, which is popular around the world.In this problem, let us focus on puzzles with 16×16 grids, which consist of 4×4 regions. The objective is to fill the whole grid with hexadecimal d…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4069 Problem Description Today we play a squiggly sudoku, The objective is to fill a 9*9 grid with digits so that each column, each row, and each of the nine Connecting-sub-grids that compose the grid co…

Problem Description Yi Sima was one of the best counselors of Cao Cao. He likes to play a funny game himself. It looks like the modern Sudoku, but smaller. Actually, Yi Sima was playing it different. First of all, he tried to generate a × board with…

Description Today we play a squiggly sudoku, The objective is to fill a 9*9 grid with digits so that each column, each row, and each of the nine Connecting-sub-grids that compose the grid contains all of the digits from 1 to 9. Left figure is the puz…

题目传送门 题意: 给你16个16宫格的数独,里面是0~F,你可以逆时针旋转里面的每个16宫格 问你它是从标准数独逆时针旋转多少次得到? 思路: 可以知道每个16宫已经是标准的了,接下来只要考虑每行.每列就行了 那么我们在dfs中就可以用两个行列两个数组来标记每个数字出现的次数, 大于1则不行 另外此时是逆时针来的,那么你就要顺时针回去 逆一次等于顺3次 参考博客:https://www.cnblogs.com/zquzjx/p/10326048.html 代码: #include<bits/s…

给一个由4*4个4*4的小格组成数独,这些数独是由一个块逆时针旋转得来的,所以要还原的话就模拟出顺时针的过程,先把里面的字母转化成数字,然后从第一个块开始枚举,每个dfs和之前枚举的已经满足条件的块,然后先枚举每一行的每块,这一行枚举完了去枚举下一行的状态,一直枚举到最后一块为止. #include<map> #include<set> #include<ctime> #include<cmath> #include<stack> #includ…