Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed × matrix, but Fei didn't know the correct direction to hold the sheet. What a pity! Given two secrete master plans. The first one is the master's original plan. The…

D. Duff in Beach Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four corners of the city wall. Unfortunately, w…

Secrete Master Plan Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 429    Accepted Submission(s): 244 Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a…

Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four corners of the city wall. Unfortunately, when Fei opened the pocket he found there are only four…

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 801    Accepted Submission(s): 470 Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan ins…

D. Duff in Beach Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four corners of the city wall. Unfortunately, when Fei opened the pocket he found there are o…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5540 题目大意:给一个两个2*2的矩阵,第二个矩阵能不能通过旋转得到第一个矩阵 题目思路:模拟 #include <stdio.h> #include <iostream> using namespace std; ][]; ][]; ][]; void solve(int T){ ;i<=;i++) ;j<=;j++) scanf("%d",&…

题目链接:http://acm.uestc.edu.cn/#/problem/show/1215 题目大意就是问一个2*2的矩阵能否通过旋转得到另一个. 代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <algorithm> #include <set> #in…

Friendship of Frog Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5578 Description N frogs from different countries are standing in a line. Each country is represented by a lowercase letter. The distance betwee…

ZYB's Biology Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5590 Description ZYB(ZJ−267)在NOIP拿到600分之后开始虐生物题,他现在扔给你一道简单的生物题:给出一个DNA序列和一个RNA序列,问它们是否配对. DNA序列是仅由A,C,G,T组成的字符串,RNA序列是仅由A,C,G,U组成的字符串. DNA和RNA匹配当且仅当每…