链接:https://www.nowcoder.com/acm/contest/140/A来源:牛客网 White Cloud is exercising in the playground. White Cloud can walk 1 meters or run k meters per second. Since White Cloud is tired,it can't run for two or more continuous seconds. White Cloud will mo

题目链接: B. Mashmokh and ACM time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participat

                                             B. Mashmokh and ACM                                                                                             time limit per test 1 second                                                                

http://codeforces.com/problemset/problem/414/B B. Mashmokh and ACM time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh d

title: woj1012 Thingk and Count DP好题 date: 2020-03-12 categories: acm tags: [acm,dp,woj] 难题,dp好题,几何题(?数学题) 想明白了其实思路很简单,但是因为数据量很大优化需要考虑. 滚动数组. description You are given a chessboard made up of N squares by M squares. Some of the squares are colored bl

A. Fox And Snake 代码可能有点挫,但能够快速A掉就够了. #include <cstdio> int main() { //freopen("in.txt", "r", stdin); int n, m; scanf("%d%d", &n, &m); ; i <= n; ++i) { == || i % == ) ; j < m; ++j) printf("#"); ==

题链: http://www.lydsy.com/JudgeOnline/problem.php?id=1042 题解: 容斥原理,dp预处理首先跑个无限物品的背包dp求出dp[i]表示在四种物品都有无限个情况下有多少种方法支付 i元.然后对于每个询问,答案就是 dp[S]-不合法的方法.那么这个不合法的方法数怎么求呢?举个例子:如果 c1不能超过d1个的话,那么我们就强制用掉 d1+1个 c1硬币,那么dp[S-(d1+1)*c1]就是c1不合法的方法数. 所以这样就可以类似的求出其它硬币的不

狄利克雷过程(dirichlet process )的五种理解  原文:http://blog.csdn.net/xianlingmao/article/details/7342837   无参数贝叶斯方法: Nonparametric Bayesian methods (Dirichlet processes)   狄利克雷过程(dirichlet process )是目前变参数学习(non parameter)非常流行的一个理论,很多的工作都是基于这个理论来进行的,如HDP(hierarch

这周不打算按照以往的方式更新博客,而是采用整体的方式.一是因为学的太少,没东西写:二是这篇博客会经常更新的.如题,DP——尽力学习之中. ------------------------------------------------------------------------------------------------------------------------------------------------------------------ 先说几个与训练不太相关的东西 一.D

[codeforces 415D]Mashmokh and ACM 题意:美丽数列定义:对于数列中的每一个i都满足:arr[i+1]%arr[i]==0 输入n,k(1<=n,k<=2000),问满足[数列长度是k && 数列中每一个元素arr[i]在1~n之间 && 数列中元素可以重复]的数列有多少个?结果对10^9+7取余 解题思路:dp[i][j]表示长度是j,最后一位是i的种数 if(kk%i==0) dp[kk][j+1]+=dp[i][j] #inc

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4064 Problem Description Carcassonne is a tile-based board game for two to five players.Square tiles are printed by city segments,road segments and field segments. The rule of the game is to put the tile

QSC and Master Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 859    Accepted Submission(s): 325 Problem Description Every school has some legends, Northeastern University is the same. Enter

odd-even number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 388    Accepted Submission(s): 212 Problem Description For a number,if the length of continuous odd digits is even and the length

感谢博主——      http://blog.csdn.net/cc_again?viewmode=list       ----------  Accagain  2014年5月15日 动态规划一直是ACM竞赛中的重点,同时又是难点,因为该算法时间效率高,代码量少,多元性强,主要考察思维能力.建模抽象能力.灵活度. 本人动态规划博客地址:http://blog.csdn.net/cc_again/article/category/1261899 ***********************

Aeroplane chess Problem Description Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the number

Contest Problem Description In the ACM International Collegiate Programming Contest, each team consist of three students. And the teams are given 5 hours to solve between 8 and 12 programming problems. On Mars, there is programming contest, too. Each

转载请注明出处,谢谢.   http://blog.csdn.net/cc_again?viewmode=list          ----------  Accagain  2014年5月15日 动态规划一直是ACM竞赛中的重点,同一时候又是难点,由于该算法时间效率高,代码量少,多元性强,主要考察思维能力.建模抽象能力.灵活度. 本人动态规划博客地址:http://blog.csdn.net/cc_again/article/category/1261899 ****************

The number of steps nid=24#time" style="padding-bottom:0px; margin:0px; padding-left:0px; padding-right:0px; color:rgb(83,113,197); text-decoration:none; padding-top:0px"> Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描写叙述 Mary

dp[i][j]=dp[i-a[i][j]+dp[i-a[i]][j-1]; 其中前一项j不变表示j这一项无限取

题目链接:https://ac.nowcoder.com/acm/contest/163/J 题目大意:给定一个数N,求区间[1,N]中满足可以整除它各个数位之和的数的个数.(1 ≤ N ≤ 1012). 输入: 21018 输出: Case 1: 10Case 2: 12 解题思路:比较简单的一道数位dp题,因为N的范围最大可为10的十二次方,即数位和的范围为[1,108],1-108的最小公倍数很大不可求,所以我们直接暴力枚举数位和为1-108的情况,然后利用数位dp求出合法数的个数就可以了

链接:https://ac.nowcoder.com/acm/contest/338/I 来源:牛客网 题目描述 IG won the S championship and many people are excited, ii and gg are no exception. After watching the game, the two of them also want to play a game. There is now an infinite chessboard with on