asr 1006

VIP cannot be started normallyDescription: Our environment is 2 node RAC, and node 1 is down due to physical faults.In this case, I want to start the VIP of Node 1 from node 2 so that the single node is transparent to the user program. [Oracle @ UNID02 ~] $ Crs_start ora. unid01.vipAttempting to start 'ora. unid01.vip 'on member 'unid02'Start of 'ora. unid01.vip 'on member 'unid02' failed.CRS-1006: No more members to consider CRS-0215: cocould not sta

first, the requirements http://poj.org/problem?id=1006 biorhythms Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 124113 Accepted: 39154 DescriptionSome people believe that there be three cycles in a person's life that's start the day he or she's born. These three cycles is the physical, emotional, and intellectual cycles, and they have periods of lengths, D, and Ays, respectiv

Time: 2016-05-09-21:12:54 Title Number: [2016-05-09][51nod][1006 longest common sub-sequence LCS] Main topic: [2016-05-09][51nod][1006 longest common sub-sequence lcs].md Analysis: Dynamic Planning DP[I][J] Represents the length of the longest common subsequence of string A in the first position, string B at position J DP[I][J] = dp[i-1][j-1] + 1 if a[i] = = A[j] else dp[i

1006. Summation Game DescriptionThere is a row of stone keyboards on the stone pillars, with an integer on each key. Please select two keys on the keyboard to make the keys between the two keys and the number and maximum. If this is the largest and not positive, then output "Game over".Input FormatLine 1th: Number of keys N.2nd.. N+1: The numeric integer on the key ai. −100≤ a I≤100 For 70% of data, 2n ≤1 ,000 For 100% of data,2≤n≤1, O

The Chinese Remainder Theorem comes from a problem in Sun Tzu's computing Sutra: I don't know the number of things, but the number of three is two, the number of five is three, and the number of seven is two. Ry? In fact, this question is to solve such a homogeneous equations: X limit 2 (mod 3) X forward 3 (mod 5) X limit 2 (mod 7) solving x As for the solution to this equation, Wikipedia explained in detail: // Reference: http://blog.csdn.net/cyendra/article/details/38402869 For example, poj

Link: http://pat.zju.edu.cn/contests/pat-b-practise/1006 Let's use the letter B to represent "Hundred", the letter S to represent "Ten", and "12... N ( Input Format:Each test input contains one test case and a positive integer N ( Output Format:The output of each test case occupies one line and outputs N in the specified format. Input Example 1: 234 Output Example 1: BBSSS1234 Input Example 2: 23 Output Example 2: SS123 The Code is as follows: #

POJ 1006 Biorhythms, poj1006biorhythms POJ 1003,1004, 1005 is relatively simple and can be quickly solved. In an episode, I didn't quite understand ACM at the beginning. The most recent feedback on submitted problems was a Runtime Error. At first, I thought it was a timeout. On the one hand, I suspect that Java is running too slowly, then I learned that java is recommended for some international competitions, which means java is not slow. On the other

1006. team rankings Constraints Time Limit: 1 secs, memory limit: 32 MBDescription It's preseason and the local newspaper wants to publish a preseason ranking of the teams in the local amateur basketball leleague. the teams are the ants, the buckets, the cats, the dribblers, and the elephants. when scoop McGee, sports editor of the paper, gets the rankings from the selected local experts down at the hardware store, he's dismayed to find that there doe

new network interfaces. Oracle Linux and RedHat can directly Delete the original network interface and rename the new network interface name to the original one. Suselinux is a little troublesome. For details, refer to: slave # modify the NIC and restart two nodes. # The resource VIP status is offline Oracle @ bo2dbp: ~>. /Crs_stat.sh | grep bo2dbp Resource Name target State -------------- ------ ----- ora. bo2dbp. asm1.asm online on bo2dbp ora. bo2dbp. listener_bo2dbp.lsnr online offline ora.

Label: Flex error1006 1. Error cause TypeError: Error #1006: value 不是函数。at BasicChart/dataFunc()[E:\Flash Builder\Map\src\BasicChart.mxml:68]at mx.charts.chartClasses::Series/cacheDefaultValues()[E:\dev\4.0.0\frameworks\projects\datavisualization\src\mx\charts\chartClasses\Series.as:1260]at mx.charts.series::LineSeries/updateData()[E:\dev\4.0.0\frameworks\projects\datavisualization\src\mx\charts\series\LineSeries.as:1188]at mx.charts.chartClasse

Question 1006: zoj Time Limit: 1 second Memory limit: 32 MB Special question: No Submit: 14782 Solution: 2482 Description: Determine whether a given string (containing only 'Z', 'O', and 'J') can be AC. The AC rules are as follows: 1. zoj can communicate with each other; 2. If the string format is xzojx, it can also be AC, where X can be n 'O' or is empty; 3. If azbjc can be AC, azbojac can also be AC, where A,

Jiudu OJ 1006 ZOJ problem (this test data has a problem), ojzoj Question 1006: ZOJ Time Limit: 1 second Memory limit: 32 MB Special question: No Submit: 15725 Solution: 2647 Description: Determine whether a given string (containing only 'Z', 'O', and 'J') can be AC. The AC rules are as follows: 1. zoj can communicate with each other; 2. If the string format is xzojx

;For (edge* e = head[x]; e; e = e->next) if (!done[e->to]) {if (++label[e->to] > Best) best = label[e->to];S[label[e->to]].push (e->to);}}} void Solve () {int ans = 0;memset (color, 0, sizeof color);memset (F,-1, sizeof f);for (int i = N; i--;) {int x = seq[i];For (edge* e = head[x]; e; e = e->next)F[color[e->to]] = x;for (int i = 1; I ans = max (ans, color[x] = i);Break ;}}printf ("%d\n", ans);}int main () {init ();MCS ();solve ();return 0;}------------------------------------------------------

1006 Arithmetic Progressiontime limit: 1 sspace limit: 128000 KBtitle level: Golden Gold SolvingView Run ResultsTitle DescriptionDescriptionGiven the number of N (1Enter a descriptionInput DescriptionThe first line is an integer n, and the next line includes the number of n, and the absolute value of each number does not exceed 10000000.Output descriptionOutput DescriptionFor each input data, output the length of the longest arithmetic progression you

refers to the number generated on a machine, which guarantees the uniqueness of the machine in the same virtual environment ).At the same time, the MAC address and network interface name will also change accordingly (the original eth0 and eth1 of the virtual machine are not available for the first time), which usually needs to be modified.Different Linux systems have different processing methods for new network interfaces. Oracle Linux and RedHat can directly Delete the original network interfa

Topic 1006:zoj Questions time limit:1 seconds Memory limit:32 MB Special question: No submitted:15725 Resolution:2647 Title Description: For a given string (containing only ' z ', ' o ', ' J ' three characters), determine if he can AC. The rules for AC are as follows: 1. Zoj can ac; 2. If the string form is XZOJX, it can also be AC, where x can be n ' o ' or null; 3. If AZ

POJ 1003,1004,1005 is relatively simple and soon solved. There is a small episode, just beginning to do not understand the ACM, the most recent submission of the problem is the runtime Error, the beginning I thought it was timed out, on the one hand I suspect is not Java run too slow, and then to understand that some international competitions recommended Java, that the Java itself is not slow. On the other hand I suspect that my program is too bad, each time is very good time-consuming, so ever

1412202129-hpu-1006: DNA 1006: DNA time limit: 1 Sec memory limit: 128 MB Submitted: 4 solution: 2 [Submit] [Status] [discussion version] Description Xiaoqiang liked Life Science from an early age. He was always curious about where flowers, birds, and animals came from. Finally, Xiaoqiang went to middle school and came into contact with the holy term DNA. It had a double helix structure. This allows Xi

1006 Longest common sub-sequence LCS Base Time Limit:1- Second space limit:131072 KB score: 0 difficulty: basic problem Collection focus on canceling attentiongiven two string a b, the longest common subsequence of a and B (the subsequence is not required to be contiguous). For example, two strings: Abcicbaabdkscab AB is a sub-sequence of two strings, and ABC is also, abca, where ABCA is the longest subsequence of the two strings. InputLine 1

integer T, which indicates the number of test cases. For each test case, the first line contains N, M, L (0 And in the following n lines, each line contains one integer within (0, m) indicating the position of rock. Outputfor each test case, just output one line "case # X: Y", where X is the case number (starting from 1) and Y is the maximal number of steps Matt shoshould jump. Sample Input 21 10 552 10 336 Sample output Case #1: 2Case #2: 4 Question: RT Idea: Better greedy questions Each tim