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python set集合的用法
set 集合:是一组无序的不可重复的集合 1.set的创建 2.转换成集合 3.set的常用方法 clear(self, *args, **kwargs):删除集合中所有元素 difference(self, *args, **kwargs):比较两个集合,找出对方没有的元素,并返回一个新的集合 d ...
分类:编程语言   时间:2017-06-23 00:44:04    阅读次数:215
图像清晰度评价
http://blog.csdn.net/liuuze5/article/details/50773160 1.灰度差 [plain] view plain copy function out_val=difference_absolute(img); I=rgb2gray(img); [m,n]= ...
分类:其他好文   时间:2017-06-23 00:41:13    阅读次数:231
378. Kth Smallest Element in a Sorted Matrix
https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/#/solutions http://www.cnblogs.com/EdwardLiu/p/6109080.html Solution 2 : Binary ...
分类:其他好文   时间:2017-06-22 00:16:28    阅读次数:131
LeetCode Minimum Factorization
625. Minimum FactorizationGiven a positive integer a, find the smallest positive integer b whose multiplication of each digit equals to a. If there is ...
分类:其他好文   时间:2017-06-20 01:03:45    阅读次数:303
Downgrading an Exchange 2010 Server(Exchange降级)
Downgrading an Exchange 2010 Server Microsoft Exchange Server 2010 comes in two versions: enterprise and standard. The biggest difference between the ...
分类:其他好文   时间:2017-06-19 21:00:15    阅读次数:235
594. Longest Harmonious Subsequence 最长的和谐子序列
Total Accepted: 5540Total Submissions: 14066Difficulty: EasyContributors:love_FawnWe define a harmonious array is an array where the difference betwee... ...
分类:其他好文   时间:2017-06-19 00:38:08    阅读次数:221
【BZOJ2213】[Poi2011]Difference DP
【BZOJ2213】[Poi2011]Difference Description A word consisting of lower-case letters of the English alphabet ('a'-'z') is given. We would like to choose ...
分类:其他好文   时间:2017-06-18 15:13:54    阅读次数:125
[leetcode-625-Minimum Factorization]
Given a positive integer a, find the smallest positive integer b whose multiplication of each digit equals to a. If there is no answer or the answer i ...
分类:其他好文   时间:2017-06-18 14:20:38    阅读次数:193
vector、set 练习 k-th divisor
k-th divisor You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist. Divisor of n is any such natural nu ...
分类:其他好文   时间:2017-06-18 13:15:19    阅读次数:212
CodeM Qualifying Match Q1
问题描述: 具体地说,就是在第二段音频中找到一个长度和第一段音频相等且是连续的子序列,使得它们的 difference 最小。两段等长音频的 difference 定义为: difference = SUM(a[i] - b[i])2 (1 ≤ i ≤ n),其中SUM()表示求和 其中 n 表示序 ...
分类:其他好文   时间:2017-06-17 13:13:29    阅读次数:172
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