Arithmetic Sequence Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 1810  Solved: 311[Submit][Status][Web Board] Description Giving a number sequence A with length n, you should choosing m numbers from A(ignore the order) which can form an arithmetic…

Arithmetic Sequence Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 51  Solved: 19[Submit][Status][Web Board] Description Giving a number sequence A with length n, you should choosingm numbers from A(ignore the order) which can form an arithmetic sequ…

Problem Description A sequence b1,b2,⋯,bn are called (d1,d2)-arithmetic sequence ≤i≤n) such that ≤j<i),bj+=bj+d1 and =bj+d2. Teacher Mai has a sequence a1,a2,⋯,an. He wants to know how many intervals [l,r](≤l≤r≤n) there are that al,al+,⋯,ar are (d1,d…

Given an array A of integers, return the length of the longest arithmetic subsequence in A. Recall that a subsequence of A is a list A[i_1], A[i_2], ..., A[i_k]with 0 <= i_1 < i_2 < ... < i_k <= A.length - 1, and that a sequence B is arithm…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=5400 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 875    Accepted Submission(s): 386 Problem Description A sequence b1,b2,⋯,bn are called (d1,d2)…

原题链接在这里:https://leetcode.com/problems/longest-arithmetic-sequence/ 题目: Given an array A of integers, return the length of the longest arithmetic subsequence in A. Recall that a subsequence of A is a list A[i_1], A[i_2], ..., A[i_k] with 0 <= i_1 < i…

http://acm.hdu.edu.cn/showproblem.php?pid=5400 Arithmetic Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1151    Accepted Submission(s): 501 Problem Description A sequence b1,b2,⋯,bn a…

1020: Arithmetic Sequence Time Limit: 1 Sec  Memory Limit: 128 MB Submit:  ->打开链接<- Description Giving a number sequence A with length n, you should choosing m numbers from A(ignore the order) which can form an arithmetic sequence and make m as larg…

题目如下: Given an array A of integers, return the length of the longest arithmetic subsequence in A. Recall that a subsequence of A is a list A[i_1], A[i_2], ..., A[i_k] with 0 <= i_1 < i_2 < ... < i_k <= A.length - 1, and that a sequence B is…

Arithmetic Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description A sequence b1,b2,⋯,bn are called (d1,d2)-arithmetic sequence if and only if…

http://acm.hzau.edu.cn/problem.php?id=21 题目大意: 给你一个序列问在数字最多的等比数列 分析:  刚开始看到题就知道是一个dp但是我dp实在是渣到不行 后来发现用二维dp  第二位保存i到j的差 #include <iostream> #include <queue> #include <cstdio> #include <cstring> #include <cstdlib> #include <…

Sequence Swapping Time Limit: 1 Second      Memory Limit: 65536 KB BaoBao has just found a strange sequence {<, >, <, >, , <, >} of length  in his pocket. As you can see, each element <, > in the sequence is an ordered pair, where…

题目大意 给一个数列,长度不超过 5000,每次可以将其中的一个数加 1 或者减 1,问,最少需要多少次操作,才能使得这个数列单调不降 数列中每个数为 -109-109 中的一个数 做法分析 先这样考虑:如果操作的次数最少,那么最终得到的不降的数列,必然是由原始数列中的数组成的,具体的证明可以使用反证法 知道了上面讲述的性质,这题就好搞了 先将原始数列(设为 A,共 n 个数)中所有的数去重并从小到达排序,保存在另一个数列中(设为 B,共 m 个数) 定义状态:f[i][j] 表示将原始数列中的…

C. Longest Regular Bracket Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/5/C Description This is yet another problem dealing with regular bracket sequences. We should remind you that a bracket sequence…

Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a regular sequence, then (S) and [S] are both regular sequences. 3. If A and B are regular sequences, then AB is a regular sequence. F…

Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35049   Accepted: 10139   Special Judge Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a r…

Best Sequence Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5543 Accepted: 2188 Description The twenty-first century is a biology-technology developing century. One of the most attractive and challenging tasks is on the gene project, esp…

Description We give the following inductive definition of a “regular brackets” sequence: the empty sequence is a regular brackets sequence, if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and if a and b are regul…

http://acm.hdu.edu.cn/showproblem.php?pid=2062 Subset sequence Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3569    Accepted Submission(s): 1802 Problem Description Consider the aggregate An=…

(CodeForces - 5C)Longest Regular Bracket Sequence time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output This is yet another problem dealing with regular bracket sequences. We should remind you t…

D. Sequence Sorting 题目大意:给你一个序列,有一种操作就是对所有相同的数可以挪到最前面,也可以挪到最后面,问最少操作次数. 首先,对于很多个相同的数,可以缩成两个位置,一个是就是这个数出现的区间,一个是最大位置,一个是最小位置. 如果数不挪,那就必须是连续递增的一段数,而且这个些数不能相互影响,也就是小的那个数的最大位置一定要在大的那个数的最小位置的前面. 然后就是找最长的一段连续递增的序列了,这个序列有两个要求,一个是递增,一个是必须连续,意思是对于si 和si+1 不存在…

题目链接:http://poj.org/problem?id=1141 题解:求已知子串最短的括号完备的全序列 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; #define ll long long ; const int INF=0x3f3f3f3f; ][]; ][]; ]; int Find(int x…

本文出自:http://blog.csdn.net/svitter 原题:http://poj.org/problem?id=1141 题意:输出添加括号最少,并且使其匹配的串. 题解: dp [ i ] [ j ] 表示添加括号的个数, pos[ i][ j ] 表示 i , j 中哪个位置分开,使得两部分分别匹配. pos [ i ][ j ] 为-1的时候,说明i, j 括号匹配. 初始值置dp [ i ] [ i ]  = 1; 如果只有一个括号,那么匹配结果必然是差1. 首先判断括号是…

题目链接 题意 : 给你一串由括号组成的串,让你添加最少的括号使该串匹配. 思路 : 黑书上的DP.dp[i][j] = min{dp[i+1][j-1] (sh[i] == sh[j]),dp[i][k]+dp[k+1][j](i<=k<j)}.输出的时候递归,其实我觉得输出比dp部分难多了..... #include <stdio.h> #include <string.h> #include <iostream> using namespace std…

题意:给定一个序列,有 n 个数,只有01,然后你进行k次操作,把所有的1变成0,求有多种方法. 析:DP是很明显的,dp[i][j] 表示进行第 i 次操作,剩下 j 个1,然后操作就两种,把1变成0,把0变成1.也可以用记忆化来做. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cst…

题目大意 给定一个含有N个数的序列,要求你对一些数减掉或者加上某个值,使得序列变为非递减的,问你加减的值的总和最少是多少? 题解 一个很显然的结果就是,变化后的每一个值肯定是等于原来序列的某个值,因为只需要变为非递减的,所以对于某个数要么不变,要么变成左右附件的某个值.这样我们就可以根据前述条件得出DP方程了:dp[i][j]=min(dp[i][j-1],dp[i-1][j]+|a[i]-b[j]|)(a为原序列,b为排序后的序列),方程的意思是,把序列前i个数变为非递减序列并且以不超过b[j…

题目地址:Ural 1183 最终把这题给A了.. .拖拉了好长时间,.. 自己想还是想不出来,正好紫书上有这题. d[i][j]为输入序列从下标i到下标j最少须要加多少括号才干成为合法序列.0<=i<=j<len (len为输入序列的长度). c[i][j]为输入序列从下标i到下标j的断开位置.假设没有断开则为-1. 当i==j时.d[i][j]为1 当s[i]=='(' && s[j]==')' 或者 s[i]=='[' && s[j]==']'时,d…

题意:给定上一棵树和一个排列,然后问你把这个排列分成m个连续的部分,每个部分的大小的是两两相邻的LCA的最小深度,问你最小是多少. 析:首先这个肯定是DP,然后每个部分其实就是里面最小的那个LCA的深度.很容易知道某个区间的值肯定是 [li, li+1] .. [ri-1, ri]这些区间之间的一个,并且我们还可以知道,举个例子,1 2 3 4  5 6 如果知道分成两部分 其中 2 和 6 是最优的,那么中间的 3 4 5 ,这三个数其实属于哪个区间都无所谓,所以对于第 i 个数只有三种可能.…

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1475 正反一次lis然后去min{左,右}*2-1即可 #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream&…

http://blog.csdn.net/cc_again/article/details/10169643 http://blog.csdn.net/lijiecsu/article/details/7589877 如果有空串要用gets,scanf不能处理空串 #include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio>…