题目: A tremendously exciting raffle is being held, with some tremendously exciting prizes being given out. All you have to do to have a chance of being a winner is to put a piece of paper with your name on it in the raffle box. The lucky winners of th…
油田问题(L - 暴力求解.DFS) Description The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerou…
A. Artwork 倒过来并查集维护即可. #include<cstdio> #include<algorithm> using namespace std; const int N=1111; int n,m,q,i,j,ce; bool black[N][N]; bool v[N*N]; int f[N*N]; int res; int ans[N*N],id[N][N],cnt; struct P{ int x,y; P(){} P(int _x,int _y){x=_x,…
11.12.2018 逆向暴力求解 538.D Weird Chess New Point: 没有读好题 越界的情况无法判断,所以输出任何一种就可以 所以他给你的样例输出完全是误导 输出还搞错了~ 输出的举证中间竟然空格隔开每一个字符,真的是不该错,否则应该是1A吧 然后空间开小了,地图空间倒是注意到了,但是你存取’o’的空间只有1000,而地图大小最大是50*50所以最多是2500个’o’的地址 中间的选择代码逻辑有些混乱 要加强练习! Describe: 定义新的下棋游戏,给你多个棋子’o’…
先来解释一下HMM的向前算法: 前向后向算法是前向算法和后向算法的统称,这两个算法都可以用来求HMM观测序列的概率.我们先来看看前向算法是如何求解这个问题的. 前向算法本质上属于动态规划的算法,也就是我们要通过找到局部状态递推的公式,这样一步步的从子问题的最优解拓展到整个问题的最优解.在这里我们认为随机过程中各个状态St的概率分布,只与它的前一个状态St-1有关,同时任何时刻的观察状态只仅仅依赖于当前时刻的隐藏状态. 在t时刻我们定义观察状态的概率为: αt(i)=P(o1,o2,...ot,i…
A Artwork B Bless You Autocorrect! C Card Hand Sorting D Daydreaming Stockbroker 贪心,低买高卖,不要爆int. #include <cstdio> #include <cstring> #include <algorithm> #include <iostream> using namespace std; typedef long long LL; const int max…
CSU-2019 Fleecing the Raffle Description A tremendously exciting raffle is being held, with some tremendously exciting prizes being given out. All you have to do to have a chance of being a winner is to put a piece of paper with your name on it in th…
jrMz and angle       Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 jrMz有两种角,第一种角都是正nn边形的内角,第二种角都是正mm边形的内角.jrMz想选出其中一些,某种角可以选多个或一个都不选,使得选出的所有角的度数之和恰好为360度.jrMz想知道这是否可能实现. 输入描述 有多组测试数据,第一行一个整数\left(1\leq T\leq10…
Problem Description Avin is studying series. A series is called "wave" if the following conditions are satisfied:1) It contains at least two elements;2) All elements at odd positions are the same;3) All elements at even positions are the same;4)…
str表示文本串,m表示模式串; str[i+j] 和 m[j] 是正在进行匹配的字符; KMP的时间复杂度是O(m+n)  ,  暴力求解的时间复杂度是O(m*n) KMP利用了B[0:j]和A[i:j]是相同的这一点,而暴力求解显然做不到. int kmp(string str,string m) { int next[MAXN]; next[] = -; ; ; while(i<m.size()) { || m[i]==m[j]) { i++; j++; next[i] = j; } el…