题目: 求一个数组的最长递减子序列比 , 如随机生成一组序列 {8,9,6,3,6,2,3,4} 求得最长递减序列 {9,8,6,4,3,2} list=[3,3,3,3,6,2,3,4] //冒泡排序 n=len(list) for i in range(0,n-2): for k in range(0,n-1): if list[k+1]>list[k]: list[k+1],list[k]=list[k],list[k+1] print(list) result=[] for i in
Bellovin Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 858 Accepted Submission(s): 395 Problem Description Peter has a sequence a1,a2,...,an and he define a function on the sequence -- F(
A sequence X_1, X_2, ..., X_n is fibonacci-like if: n >= 3 X_i + X_{i+1} = X_{i+2} for all i + 2 <= n Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If on
动态规划的本质,是对问题状态的定义和状态转移方程的定义.dynamic programming is a method for solving a complex problem by breaking it down into a collection of simpler subproblems. 动态规划是通过拆分问题,定义问题状态和状态之间的关系,使得问题能够以递推(或者说分治)的方式去解决. #coding = utf-8 ''' 此题就是一个动态规划题, 在到每一个位置的时候,标记
def long_words(n, str): word_len = [] txt = str.split(" ") for x in txt: if len(x) > n: word_len.append(x) return word_len print(long_words(, "The quick brown fox jumps over the lazy dog"))
命令模式下 :set cc=80 或者 打开 vim的配置 文件 .vimrc vim ~/.vimrc 接着你会看到你的配置文件 在配置文件中加上这样行配置代码 set cc=80 ok 现在退出vim 重新打开一个文件, 就会看到一个80列的提示了. 如果你喜欢可以设置成50 set cc=50 as you like. 这个只是提示,不会强制换行.
A. Kefa and First Steps time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the i-th day (1
Common Subsequence Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if
Bridging signals Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 961 Accepted Submission(s): 627 Problem Description 'Oh no, they've done it again', cries the chief designer at the Waferland
题目链接:http://poj.org/problem?id=2533 Time Limit: 2000MS Memory Limit: 65536K Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ...,
l例如:对于[3,1,4,2,5],最长上升子序列的长度是3 arr = [3,1,4,5,9,2,6,5,0] def lis(arr): #dp[i]表示第i个位置的值为尾的数组的最长递增子序列的长度 #初始化数组,假定数组中每个值的最长子序列就是它自己,即都是1 dp = [1 for _ in range(len(arr))] #遍历数组 for i in range(len(arr)): #当遍历到第i个位置时,再依次从0开始遍历到 for j in range(i): #如果第i个位
