我刚买了一台三星Note3,32G的，机身内的MODELO：SM-N900(UD)是什么意思_百度知道
我刚买了一台三星Note3,32G的，机身内的MODELO：SM-N900(UD)是什么意思
FCCID：A3LSMN900SSN：-N900GSMH请问这是什么意思
我有更好的答案
SM-N900是note3系列版本中的一个，是双四核版本的型号N900为三星5420处理器，N900之外的其他型号均是高通骁龙800。
那我这型号的好还是其它的好？
比较好的类型
采纳率：65%
来自团队：
欧版啊，。
尊敬的三星用户您好：根据您的疑问：由于市场存在山寨机、水货、克隆机、翻新机等，通过刷机等手段可以更改手机IMEI号，因此单凭串号、软件检测、IMEI号、手机产地等是无法准确鉴别手机真伪、是否为正品行货的。如果您要鉴别手机是否为正品行货的话，请您带好购机发票、包修卡和机器，到就近的三星服务中心，咨询服务中心人员机器在包修范围之内如果出现性能性故障是否能包修，如果能够包修的机器就是三星正品行货机器。这个是最好也是最为准确的一种方法。建议您在三星官网进行“预约到店”维修申请，享受优先维修服务：预约成功后，请携带您的机器、发票及保修卡按时访问您预约的服务中心。评价、建议、吐槽，请点击：.cn/survey
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我们会通过消息、邮箱等方式尽快将举报结果通知您。2.2 & Mergesort
The algorithms that we consider in this section is based on a simple
operation known as merging:
combining two ordered arrays to make one larger ordered array.
This operation immediately lends itself to a simple recursive sort method
known as mergesort: to sort an array,
divide it into two halves, sort the two halves (recursively),
and then merge the results.
guarantees to sort an array of N items in time proportional to N log N,
no matter what the input. Its
prime disadvantage is that it uses extra space proportional to N.
Abstract in-place merge.
The method merge(a, lo, mid, hi) in
puts the results of merging the subarrays a[lo..mid] with a[mid+1..hi]
into a single ordered array, leaving the result in a[lo..hi].
While it would be desirable to implement this method without using a significant
amount of extra space, such solutions are remarkably complicated.
Instead, merge()
copies everything to an auxiliary array
and then merges back to the original.
Top-down mergesort.
is a recursive mergesort
implementation based on this abstract in-place merge.
It is one of the best-known examples
of the utility of the divide-and-conquer paradigm for efficient algorithm
Proposition.
Top-down mergesort uses between 1/2 N lg N and N lg N compares
and at most 6 N lg N array accesses to sort any array
of length N.
Improvements.
We can cut the running time of mergesort substantially with some
carefully considered modifications to the implementation.
Use insertion sort for small subarrays.
We can improve most recursive algorithms by handling small cases differently.
to insertion sort for small subarrays will improve the running time of a
typical mergesort implementation by 10 to 15 percent.
Test whether array is already in order.
We can reduce the running time to be linear for arrays that are already
in order by adding a test to skip call to merge() if a[mid]
is less than or equal to a[mid+1].
With this change, we still do all the recursive calls, but
the running time for any sorted subarray is linear.
Eliminate the copy to the auxiliary array.
It is possible to eliminate the time (but not the space) taken to copy
to the auxiliary array used for merging. To do so, we use
two invocations of the sort method, one that takes its input from the given
array and puts the sorted output in
the other takes its input from the auxiliary array and puts the sorted output
in the given array. With this approach, in a bit of
mindbending recursive trickery, we can arrange the recursive calls such that
the computation switches the roles of the input array and the
auxiliary array at each level.
implements
these improvements.
Visualization.
provides a visualization
of mergesort with cutoff for small subarrays.
Bottom-up mergesort.
Even though we are thinking in terms of merging together two large subarrays,
the fact is that most merges are merging together tiny
subarrays. Another way to implement mergesort is to organize the merges so
that we do all the merges of tiny arrays on one pass,
then do a second pass to merge those arrays in
pairs, and so forth, continuing until we do a merge that encompasses the whole array.
This method requires even less code than the standard recursive implementation. We
start by doing a pass of 1-by-1 merges (considering individual
items as subarrays of size 1), then a pass of 2-by-2 merges
(merge subarrays of size 2 to make subarrays of size 4),
then 4-by-4 merges, and so forth.
is an implementation
of bottom-up mergesort.
Proposition.
Bottom-up mergesort uses between 1/2 N lg N and N lg N compares
and at most 6 N lg N array accesses to sort any array
of length N.
Proposition.
No compare-based sorting algorithm can guarantee to sort N items
with fewer than lg(N!) ~ N lg N compares.
Proposition.
Mergesort is an asymptotically optimal compare-based sorting algorithm.
That is, both the number of compares used by mergesort
in the worst case and the minimum number of compares that any compare-based sorting
algorithm can guarantee are ~N lg N.
Give traces, in the style of the trace given in this section, showing how the
keys E A S Y Q U E S T I O N are sorted with top-down mergesort
and with bottom-up mergesort.
Answer Exercise 2.2.2 for bottom-up mergesort.
Does the abstract inplace merge produce proper output if and only if the two
input subarrays are in sorted order? Prove your answer, or provide a counterexample.
Yes. If the subarrays are in sorted order, then the
inplace merge produces proper output.
If one subarray is not in sorted order, then its entries will appear
in the output in the same order that they appear in the input (with
entries from the other subarray intermixed).
Give the sequence of subarray sizes in the merges performed by both the top-down
and the bottom-up mergesort algorithms, for N = 39.
Top-down mergesort: 2, 3, 2, 5, 2, 3, 2, 5, 10, 2, 3, 2, 5, 2, 3, 2, 5, 10, 20,
2, 3, 2, 5, 2, 3, 2, 5, 10, 2, 3, 2, 5, 2, 2, 4, 9, 19, 39.
Bottom-up mergesort:
2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2,
4, 4, 4, 4, 4, 4, 4, 4, 4, 3,
8, 8, 8, 8, 7, 16, 16, 32, 39.
Suppose that top-down mergesort is modified to skip the call on merge()
whenever a[mid] <= a[mid+1].
Prove that the number of compares used for an array in sorted order is linear.
Since the array is already sorted, there will be no calls to merge().
When N is a power of 2, the number of compares will satisfy the recurrence
T(N) = 2 T(N/2) + 1, with T(1) = 0.
Use of a static array like aux[] is inadvisable in library software because
multiple clients might use the class concurrently. Give an implementation of
does not use a static array.
Creative Problems
Faster merge.
Implement a version of merge() that copies the
second half of a[] to aux[] in
decreasing order and then does the merge back to a[].
This change allows you to remove the code to test that each of the halves
has been exhausted from the inner loop.
Note: the resulting sort is not stable.
private static void merge(Comparable[] a, int lo, int mid, int hi) {
for (int i = i <= i++)
aux[i] = a[i];
for (int j = mid+1; j <= j++)
aux[j] = a[hi-j+mid+1];
int i = lo, j =
for (int k = k <= k++)
if (less(aux[j], aux[i])) a[k] = aux[j--];
a[k] = aux[i++];
Improvements.
Write a program
that implements the three improvements to mergesort that are described in the text:
add a cutoff from small subarrays, test whether the array is already
in order, and avoid the copy by switching arguments in the recursive code.
Inversions.
Develop and implement a linearithmic algorithm
for computing the number of
inversions in a given array (the number of exchanges that would be performed by
insertion sort for that array&see Section 2.1).
This quantity is related to the Kendall tau distance; see Section 2.5.
Index sort.
Develop a version of
that does not rearrange the array, but returns an int[] perm
such that perm[i] is the index of the ith smallest entry in the array.
Experiments
Web Exercises
Merge with at most log N compares per item.
Design a merging algorithm such that each item is compared at most
a logarithmic number of times. (In the standard merging algorithm,
an item can be compared N/2 times when merging two subarrays of
size N/2.)
Lower bound for sorting a Young tableaux.
A Young tableaux is an N-by-N matrix such that the entries are sorted
both column wise and row wise.
Prove that Theta(N^2 log N) compares are necessary to sort the N^2 entries
(where you can access the data only through the pairwise comparisons).
Solution sketch. If entry (i, j) is within 1/2 of i + j, then all of the 2N-1
grid diagonals are independent of one another. Sorting the diagonals
takes N^2 log N compares.
Given an array a of size 2N with N items in sorted order
in positions 0 through N-1, and an array b of size
N with N items in ascending order, merge the array b
into a so that a contains all of the items
in ascending order. Use O(1) extra memory.
Hint: merge from right to left.
k-near-sorting.
Suppose you have an array a[] of N distinct items which is nearly sorted:
each item at most k positions away from its position in the sorted order.
Design an algorithm to sort the array in time proportional to N log k.
Hint: First, sort the subarray from 0 to 2k; the smallest k items will be
in their correct position. Next, sort the subarray from k to 3k; the smallest 2k items
will now be in their correct position.
Find a family of inputs for which mergesort makes strictly fewer than 1/2 N lg N
compares to sort an array of N distinct keys.
Solution: a reverse-sorted array of N = 2^k + 1 keys uses
approximately 1/2 N lg N - (k/2 - 1) compares.
Write a program
to read in a sequence of string from standard input and securely shuffle them.
Use the following algorithm:
associate each card with a random real number between 0 and 1.
Sort the values based on their associated real numbers.
Use java.security.SecureRandom to generate the random real numbers.
Use Merge.indexSort() to get the random permutation.
Merging two arrays of different lengths.
Given two sorted arrays a[] and b[] of sizes M and N where M & N,
devise an algorithm to merge them into a new sorted array c[] using
~ N lg M compares.
Hint: use binary search.
Note: there is a
of Omega(N log (1 + M/N)) compares.
This follows because there are M+N choose N possible merged outcomes. A decision tree argument
shows that this requires at least lg (M+N choose N) compares. We note that n choose r >= (n/r)^r.
Merging three arrays.
Given three sorted arrays a[], b[], and c[], each of size N,
design an algorithm to merge them into a new sorted array d[] using at most ~ 6 N compares
in the worst case (or, even better, ~ 5 N compares).
Merging three arrays.
Given three sorted arrays a[], b[], and c[], each of size N,
prove that no compare-based algorithm can merge them into a new sorted array
d[] using fewer than ~ 4. N compares in the worst case.
Arrays with N^(3/2) inversions.
Prove that any compare-based algorithms that can sort arrays with
N^(3/2) or fewer inversions must make ~ 1/2 N lg N compares in the worst case.
Proof sketch: divide up the array into sqrt(N) consecutive subarrays of
sqrt(N) items each, such that the there are no inversion between items in different
subarrays but the order of items within each subarray is arbitrary. Such an array
has at most N^(3/2) inversions&at most ~N/2 inversions in each of the sqrt(N) subarrays.
By the sorting lower bound, it
takes ~ sqrt(N) lg sqrt(N) compares to sort each subarray, for a total of
~ 1/2 N lg N.
Optimal nonoblivious sorting.
Design algorithms that uses the fewest number of compares (in the worst case)
to sort arrays of length 3, 4, 5, 6, 7, and 8.
It is known that the optimal solution uses 3, 5, 7, 10, 13, and 16 compares,
respectively.
merge insertion algorithm is known to be optimal for n <= 13.
It makes sum(ceil(log[2](3*k/4)), k=1..n) compares in the worst case.
Last modified on November 07, 2016.
Copyright &
All rights reserved.GALAXY Mega 2手机的所有问题
展开更多问题类型
我要的关键词
共有57个问题
答：楼主需要在手机和电脑上下载同一个手机助手软件。
答：1、不用kingroot，换个root软件，一键root有很多啊，你说的那些你也可以多试几次，或是直接刷个可root的rom包，不一定要用原厂的，恢复出厂如果清除不掉，那刷机也可以了2、照片找不回...
楼主你好，这种情况的话，建议楼主升级系统，如果无法解决，就是硬件的问题了，需要联系售后。
答：从手机设置里，找到允许安装未知来源，把这个画上勾就可以了。
答：这款手机定位定位中低端，不支持无线充电。
答：您好！& & & 既然你重置也不好使的话你就只能够刷个全新的系统了哦，电脑里装个刷机大师，先把手机里的资料都备份一遍然后在点击一键刷机吧。
答：这手机还算可以吧，配置够用，但性价比确实不高，你要是不考虑国产牌子了就买这个吧如果可以考虑国产的话，2000能买个旗舰机了，比如小米4 小米note 魅族mx4 pro 配置都比这个好
答：都属于NOTEII，型号不同，制式不同。N7100为普通版，单卡，支持联通WCDMA的3GN7105是LTE制式的4G，向下兼容WCDMA，部分国家已开通4G网络，当然中国联通和移动也可以使用。我用的就是港...
答：这款都可以ROOT！！卓@大师 ROOT 挺好用的，用完了可以删除！有需要再安装！否则每次开机后台都启动我很烦！其他工具也一样，不用他的功能就删除吧！
楼主你好，如果是想在照片上添加时间的话，只能到手机的设置里看一下，如果没有的话，就需要借助第三方的拍照软件了。
答： 根据楼主描述的情况来看可能是手机系统造成的问题，建议楼主在备份手机数据之后尝试恢复手机的出厂设置，如果还是存在同样的情况的话，建议楼主联系手机厂商的售后服务进行检查或者系...
恢复出厂设置之后手机就会恢复到刚买的状态，如果楼主没有备份数据的话，之前的数据就没有办法找回来了。
答：1)在待机页面下，点击【电话】，进去后点击屏幕右上角的【菜单】键，选择【设定】。2)然后点击【通话】。3)点击【通话提醒】。4)将【通话期间通知】勾选取消即可。
答：您好！& & 三星Mega2的截图方法为：同时按住 [HOME] 和 [电源键] 约3秒即可。听到卡擦一下就说明截图成功了，然后去相册里面找吧。
答：　1.在待机页面下，长按【电源】键。　2.点击【紧急模式】。　3.阅读“条款和条件”后，将3个【方框】勾选，然后点击【确定】。　4.阅读“紧急模式”的提示后，点击【启动】即可。小米Note3和小米Note2哪个值得买？小米Note2与小米Note3区别对比详细评测
作者：佚名
字体：[ ] 来源：互联网 时间：09-12 09:01:01
很多朋友问小编小米Note3和小米Note2哪个值得买？两款手机有什么区别？今天小编给大家带来小米Note2与小米Note3区别对比详细评测,需要的朋友可以参考下
小米Note3介绍：
在发布会上雷军就表示小米Note3可以看作是小米6大号版，一定程度可以理解为小米6 Plus，主要是整体设计基本上继承了小米6的设计风格。
小米Note3与上一代配备双曲面屏的小米Note2明显不同，屏幕改为正常的2.5D屏幕，处理器也并非采用当下最顶级的骁龙835，而是OPPO R11同款的骁龙660中高端处理器，最大的亮点在于配备高规格前后摄像头，标配6GB大内存，采用了5.5英寸1080P全高清屏幕。内置3500mAh容量电池，支持快充。支持全网通，支持全功能NFC以及采用立体扬声器等配置和功能。以下是详细参数配置。
小米Note3参数配置
5.5 英寸像素
高通骁龙660（64位八核）
6GB LPDDR4X
64GB/128GB UFS2.1
前置1600万和后置1200万+1200万像素变双摄
3500mAh（支持快充 3.0）
全网通4.0（双卡双待）
MIUI 9（基于Android 7.1）
152.6x73.95x7.6 mm（163g）
亮黑，亮蓝
2499元/2899元/2999元
人脸解锁、双摄像头、指纹识别、NFC、红外遥控
值得一提的是，小米Note3还额外支持人脸解锁功能，区别于传统的指纹解锁，AI人脸识别有着更高的精准算法，相比指纹解锁和密码解锁更安全，更快捷。
小米Note3多少钱？小米Note3什么时候上市？
主要分为三个版本，分别是6GB+64GB售价2499元（亮黑），6GB+128GB售价2899元（亮黑），6GB+128GB售价2999元（亮蓝）。
&小米Note3将于9月12日下午2点首发，今晚8点，全款预售。
摄像头方面，小米Note3采用1600万前置摄像头，2&m合成大像素，支持千人千面AI美颜：分区域美化策略、226个骨骼打点，3D塑形瘦脸、明星修图师与AI工程师多专业共同打造。
评测总结：小米Note3值得买吗？
相较上一代的小米Note 2，小米Note 3的使命要更加清晰、明朗。在小米MIX从产品设计到配置当之无愧成为小米手机的&一把手&的当下，小米Note 3明智的退而求其次选择了骁龙660处理器，而且还有出色的外观/设计/工艺、人脸识别、优秀的拍照...使得小米Note 3也与搭载骁龙835的小米6很好的区分开（虽然价位上两者都是2499元，但是体验和定位上还是有所不同的），从而使小米Note 3与小米6以及小米5X联合起来抢占元价位，而3000甚至4000元价位则交给小米MIX 2。
小米Note2介绍：
小米Note2最大的亮点在于配备了似曾相识的双曲面屏幕，与三星S7 Edge类似，配备了双曲面屏幕，是继vivo Xplay5之后的，国产第2款双曲面屏智能手机。话不多说，以下是详细参数。
小米Note2配置参数
5.7英寸像素（双曲面屏）
高通骁龙821（64位四核）
64GB/128GB
前置800万 + 后置2300万像素摄像头
4070mAh（支持QC3.0快充）
全网通3.0（支持双卡双待与Volte）
MIUI 8 (基于Android 6.0)
156.2x77.3x7.6mm（166g）
亮黑色、冰川银
2799元/3299元/3499元
双曲面屏、全功能NFC、指纹识别、玻璃机身
小米Note2作为上一代Note的新一代产品，外观设计依旧延续了玻璃机身，最大的改变在于配备了双曲面屏，另外硬件配置也得到了很大的提升。
小米Note2正面配备了3D双曲面屏幕，配备正面指纹识别Home键，正面屏占比高达77.2%。此外，小米Note2屏幕底部依旧配备了成熟的按压指纹Home键，正面设计十分成熟。
背面方面，小米Note3背面为一体化3D曲面后壳，整体设计十分精致简约，只不过玻璃材质后壳比较容易刮伤，使用的时候需要避免摔落与刮伤，最好配备一个保护壳。
小米Note2背面
具体版本划分上，小米Note2分为标准版、高配版和高配全网通版本，各版本之间的区别，从下面的参数中可以一眼就看出来。
小米Note2标准版/高配版/高配全球版参数对比
小米Note2标准版
小米Note2高配版
小米Note2全球版
屏幕分辨率
骁龙821&（主频2.35Ghz）
骁龙821&（主频2.35Ghz）
骁龙821&（主频2.35Ghz）
前置800万+后置2256万像素
前置800万+后置2256万像素
前置800万+后置2256万像素
4070mAh（QC3.0快充）
3000mAh（QC3.0快充）
3000mAh（QC3.0快充）
MUIU 8(Android 6.0)
MUIU 8(Android 6.0)
MUIU 8(Android 6.0)
全网通（双卡双待）
全网通（双卡双待）
全网通（6模37频）
亮黑色、冰川银
亮黑色、冰川银
亮黑色、冰川银
双曲面、全功能NFC、指纹识别
双曲面、全功能NFC、指纹识别
双曲面、全功能NFC、指纹识别
小米Note2各版本区别对比
简单来说，小米Note2各版本的区别主要是RAM内存、ROM存储与售价三个方面的区别，其它方面则完全一样，具体区别如下。
小米Note2标准版：4GB RAM + 64GB ROM，售价2799元；
小米Note2高配版：6GB RAM + 128GB ROM，售价3299元；
小米Note2全球版：6GB RAM + 128GB ROM，相比高配版主要是网络基带支持全球更多国家网络制式，因此售价更贵200元，售价3499元。
以上就是本篇文章的全部内容了，希望对各位有所帮助。如果大家还有其他方面的问题，可以留言交流，请持续关注脚本之家！
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