https://www.hackerrank.com/contests/hourrank-18/challenges/super-six-substrings 能被6整除的数有一个特点,就是能同时被3和被2整除 那么也就是能整除3的偶数. 设dp[i][j]表示以第i位结尾的所有子串中,%3的余数是j的方案数, dp[i + 1][(j * 10 + nowNumber) % 3] = dp[i][j] 注意那个 * 10是必须的,因为从第i项转移过来,位数应该移动一位.就比如,12 1 % 3…

UVA10269 Adventure of Super Mario(Floyd+DP) After rescuing the beautiful princess, Super Mario needs to find a way home -- with the princess of course :-) He's very familiar with the 'Super Mario World', so he doesn't need a map, he only needs the be…

https://www.hackerrank.com/contests/w3/challenges/sam-and-substrings DP.注意到以N[i]结尾的每个字符串的子数字和有规律: 53125 | 3 53 | 1 31 531 | 2 12 312 5312 sd[2] = 1 + 31 + 531 = 563sd[3] = 2 + 12 + 312 + 5312sd[3] = 2 + 10 + 2 + 310 + 2 + 5310 + 2sd[3] = 4 * 2 + 10 *…

dp[u][t]表示从起点出发,到达i点且用了t次magic boot时的最短时间, 方程如下: dp[v][t]=min(dp[v][t],dp[u][t]+dis[u][v]); dp[v][t]=min(dp[v][t],dp[u][t-1]) (dis[u][v]<=l) 放进SPFA更新,相当于一个二维的最短路,解决DP在非DAG下的有后效性的问题. #include<cstdio> #include<cstring> #include<queue> #…

题意:有A个村庄,B个城市,m条边,从起点到终点,找一条最短路径.但是,有一种工具可以使人不费力的移动L个长度,但始末点必须是城市或村庄.这种工具有k个,每个只能使用一次,并且在城市内部不可使用,但在村庄内部可使用.另外,在城市或村庄内部的时间不计. 析:先预处理出来使用工具能到达的距离,这个可以用Floyd 来解决,然后f[i][u] 表示到达 u 还剩下 i 次工具未用,然后用bfs就可以很简单的解决. 代码如下: #pragma comment(linker, "/STACK:102400…

这个……真心看不出来是个DP,我在树状数组的康庄大道上欢快的奔跑了一下午……看了题解才发现错的有多离谱. 参考:http://www.cnblogs.com/kuangbin/archive/2012/11/11/2765329.html #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #define LL long long int using nam…

题意:有a个村庄,编号为1到a,有b个城堡,编号为a+1到a+b.现在超级玛丽在a+b处,他的家在1处.每条路是双向的,两端地点的编号以及路的长度都已给出.路的长度和通过所需时间相等.他有一双鞋子,可以使用k次,每次使用后最多可以跑过l的距离,且通过这段距离所需时间为0.使用鞋子时,必须从村庄或城堡开始,到村庄或者城堡结束.但是,城堡充满了陷阱,他如果中途遇见城堡,就必须停下来,且鞋子视为使用完了一次.问超级玛丽回家所需的最短时间. 思路:用floyd算法先处理出任意两点间的最短距离(不用鞋子时…

传送门 D. A and B and Interesting Substrings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A and B are preparing themselves for programming contests. After several years of doing sports pro…

1,Triangle int mininumTotal(vector<vector<int>>& triangle) { ; i >= ; --i) { ; j < i + ; ++j) { // 从下往上依次保存当前路径的最小值,上层只会用到下层的最小值 triangle[i][j] += min(triangle[i + ][j], triangle[i + ][j + ]); } } ][]; } triangle 2,Maximum SubArray /…

今天在Github上面看到一个来自于 daimajia所写的关于Android自定义View实现带数字的进度条(NumberProgressBar)的精彩案例,在这里分享给大家一起来学习学习!同时感谢daimajia的开源奉献! 第一步.效果展示 图1.蓝色的进度条 图2.红色的进度条 图3.多条颜色不同的进度条 图4.多条颜色不同的进度条 版权声明:本文为[欧阳鹏]原创文章,欢迎转载,转载请注明出处! [http://blog.csdn.net/ouyang_peng/article/deta…

Description Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more than two…

Codeforces Round #258 (Div. 2) Count Good Substrings D. Count Good Substrings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output We call a string good, if after merging all the consecutive eq…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34052    Accepted Submission(s): 15437 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

C - Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, a…

Problem Description Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more t…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32044    Accepted Submission(s): 14425 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

Substrings Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1161    Accepted Submission(s): 351 Problem Description XXX has an array of length n. XXX wants to know that, for a given w, what is…

Substrings Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1727    Accepted Submission(s): 518 Problem Description XXX has an array of length n. XXX wants to know that, for a given w, what is…

Problem Description Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more than t…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 31490    Accepted Submission(s): 14146 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1087 Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24452    Accepted Submission(s): 10786 Problem Description No…

HackerRank - common-child[DP] 题意 给出两串长度相等的字符串,找出他们的最长公共子序列e 思路 字符串版的LCS AC代码 #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <deque> #include <vector> #include <queue> #include…

题目连接: 啊哈哈.点我点我 思路: 这个题目是一个图上dp问题.先floyd预处理出图上全部点的最短路,可是在floyd的时候,把可以用神器的地方预处理出来,也就是转折点地方不能为城堡..预处理完成后.就是一个dp问题了. . .dp[][],两维分别表示到达的地点和使用神器的次数.. 这样这个问题就得到了解决.. 题目: ZOJ Problem Set - 1232 Adventure of Super Mario Time Limit: 2 Seconds      Memory Limi…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1087 Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41523    Accepted Submission(s): 19239 Problem Descript…

Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more than two players. It consi…

解题思路:题目的大意是给出一列数,求这列数里面最长递增数列的和 dp[i]表示到达地点i的最大值,那么是如何达到i的呢,则我们可以考虑没有限制条件时候的跳跃,即可以从第1,2,3,---,i-1个地点跳跃到i, 而题目限定了,跳到的那个点的数要比开始跳的那个点的数大 所以,状态转移方程式为 for(i=1;i<=n;i++)   for(j=0;j<i;j++) if(a[j]>a[i])   dp[i]=max(dp[j]+a[i],dp[i]);//找出到达地点i的最大值 反思:本来…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1087 ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943…

Super M 题解: 定义 dp[u][0] 为遍历完u中的所有节点, 但不回到u点的路径花费值. 定义 dp[u][1] 为遍历完u中的所有节点, 且要回到u点的路径花费值. 转移方程. dp[u][1] = sum(dp[v][1] + 2). dp[u][0] = max(dp[v][1] + 2 - dp[v][0] - 1). 需要注意的是,不要把不需要走的路径值传递上来. 只有这个路径会遍历一个需要清除的点的时候,才可以转移状态. 这样从1dfs完之后,我们就可以计算出上面定义的状…

Super Jumping! Jumping! Jumping!Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47017    Accepted Submission(s): 21736 Problem DescriptionNowadays, a kind of chess game called “Super Jumping! Ju…

Leetcode之动态规划(DP)专题-647. 回文子串(Palindromic Substrings) 给定一个字符串,你的任务是计算这个字符串中有多少个回文子串. 具有不同开始位置或结束位置的子串,即使是由相同的字符组成,也会被计为是不同的子串. 示例 1: 输入: "abc" 输出: 3 解释: 三个回文子串: "a", "b", "c". 示例 2: 输入: "aaa" 输出: 6 说明: 6个回…