【White】还行,不算很晚 [问题点数:400分,结帖人LeiRobin]

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Oracle ,你还行不行》???
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The Moving Pictures Experts Group abbreviated MPEG is part of the International Standards Organisation (ISO), and defines standards for digital video and digital audio. The primal task of this group was to develop a format to play back video and audio in real time from a CD1. Meanwhile the demands have raised and beside the CD the DVD2 needs to be supported as well as transmission equipment like satellites and networks. All this operational uses are covered by a broad selection of standards. Well known are the standards MPEG-1, MPEG-2, MPEG-4 and MPEG-7. Each standard provides levels and profiles to support special applications in an optimised way.
xilinx lte white paper
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Business Plan White Paper
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MPLS OAM white paper
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white比gray黑?
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Black & White(思维+二分)
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White Box Testing
问题求解,要多少分给多少分。只求尽快有答案。rnrnThis program is a container of words and word pairs. It accumulates the number of instances of each word and word pair. It can be used for knowledge management applications, such as ranking web pages matching search criteria. Assume you are solely responsible for testing this code, and provide a whitebox test plan.rn rn//**************************************************************************rnclass WordStatsrn//**************************************************************************rnrnpublic:rn//rn// Constructionrn//rn WordStats();rn virtual ~WordStats();rn rn//rn// Operationsrn//rn void AddWord( const String & );rn void AddWordPair( const String &, const String & );rn rn//rn// Accessrn//rn int GetNumWords() const;rn int GetNumWordPairs() const;rn int GetNumInstances( const String & ) const;rn rn void GetWordCount( const int index, String &s, int &count ) const;rn void GetWordPairCount( const int index, String &s, int &count ) const;rn rnprivate:rn//rn// Operationsrn//rn bool IsWord( const String & );rn//rn// Datarn//rn AssocArrayOf< String, int > mWordCount;rn AssocArrayOf< String, int > mWordPairCount;rn rn VectorOf< String > mExcludedWords;rn;rn rn//**************************************************************************rnWordStats::WordStats()rn//**************************************************************************rnrn mWordCount.Allocate( 200 );rn rn mExcludedWords.Append( "i" );rn mExcludedWords.Append( "me" );rn mExcludedWords.Append( "you" );rn mExcludedWords.Append( "him" );rn mExcludedWords.Append( "her" );rn mExcludedWords.Append( "them" );rn mExcludedWords.Append( "us" );rn mExcludedWords.Append( "we" );rnrn//**************************************************************************rnWordStats::~WordStats()rn//**************************************************************************rnrnrn rn//**************************************************************************rnvoid WordStats::AddWord( const String &ss )rn//**************************************************************************rnrn String s;rn s.Copy( ss );rn s.ToLower();rn if (s.C() > 1 &&rn !isdigit( s[ 0 ]) &&rn !ispunct( s[ 0 ])rn ) rn rn const int cExcluded = mExcludedWords.C();rn for (int i = 0; i < cExcluded; i++) rn if (s == mExcludedWords[ i ]) rn return;rn rn rn int n = 0;rn rn mWordCount.Get( n, s );rn mWordCount.Set( s, n + 1 );rn rnrn rn//**************************************************************************rnbool WordStats::IsWord( const String &s )rn//**************************************************************************rnrn if (s.C() < 3 || isdigit( s[ 0 ]) || ispunct( s[ 0 ]) ) rn return false;rn rn const int cExcluded = mExcludedWords.C();rn for (int i = 0; i < cExcluded; i++) rn if (s == mExcludedWords[ i ]) rn return false; // excluded wordrn rn rn return true;rnrn rn//**************************************************************************rnvoid WordStats::AddWordPair( const String &s1, const String &s2 )rn//**************************************************************************rnrn if (s1 == s2 || !IsWord( s1 ) || !IsWord( s2 )) rn return;rn rn String s;rn s.Copy( s1 );rn s.Append( " " );rn s.Append( s2 );rn s.ToLower();rn int n = 0;rn mWordPairCount.Get( n, s );rn mWordPairCount.Set( s, n + 1 );rnrn//**************************************************************************rnint WordStats::GetNumWords() constrn//**************************************************************************rnrn return mWordCount.C();rnrn rn//**************************************************************************rnint WordStats::GetNumWordPairs() constrn//**************************************************************************rnrn return mWordPairCount.C();rnrn rn//**************************************************************************rnvoid WordStats::GetWordCount(rn const int index,rn String & sWord,rn int & countrn) constrn//**************************************************************************rnrn if (index >= mWordCount.C() ) rn WARN( "Invalid index %d, max %d", index, mWordCount.C() );rn return;rn rn mWordCount.GetKeyN( sWord, index ); // get sWordrn mWordCount.Get( count, sWord );rnrn rn//**************************************************************************rnvoid WordStats::GetWordPairCount(rn const int index,rn String & sWord,rn int & countrn) constrn//**************************************************************************rnrn if (index >= mWordPairCount.C() ) rn return;rn rn mWordPairCount.GetKeyN( sWord, index );rn mWordPairCount.Get( count, sWord );rnrn rn//**************************************************************************rnint WordStats::GetNumInstances( const String &s ) constrn//**************************************************************************rnrn int n = 0;rn mWordCount.Get( n, s );rn return n;rnrnrn
DLNA white paper
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InfiniBand is a powerful new architecture designed to support I/O connectivity for the Internet infrastructure. InfiniBand is supported by all the major OEM server vendors as a means to expand beyond and create the next generation I/O interconnect standard in servers. For the first time, a high volume, industry standard I/O interconnect extends the role of traditional “in the box” busses. InfiniBand is unique in providing both, an “in the box” backplane solution, an external interconnect, and “Bandwidth Out of the box”, thus it provides connectivity in a way previously reserved only for traditional networking interconnects. This unification of I/O and system area networking requires a new architecture that supports the needs of these two previously separate domains.
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Quantity of White Mice
escriptionnnThere is a species of white mouse to be used in experiments. The mice keep alive only n months after their birth (9 < n < 13, n is natural number, the month to be calculated from the mice's birth, for example, the mice born in Jan have been existed 4 months in Apr.). They begin giving birth to new mice from the 7-th month. In the period of the 7-th and 8-th months every pair of the parent mice give birth to one pair of mice. From the 9-th month in a period of m months (0 < m < 3, m is natural number) every pair of the parent mice give birth to two pairs of mice. Thereafter, they stop giving birth and live to the end of their life; they die at the beginning of n+1-th month. (The n+1-th month doesn't be calculated as exist time of the mice), and the died mice will be took out from this lab. In each month, the number of living mice from previous month is countered first. If the number of the living mice from the previous month is less than or equal to 100 pairs, the new born mice of this month will stay in this lab, if the number of living mice exceeds 100 pairs, the new born mice of this month will be moved to another laboratory. Let there be only one pair of newborn mice at the beginning. How many pairs of mice in the k-th month in this lab (0 < k < 37, k is a natural number)?nInputnnOn every line, a group of data is given. In each group there are three natural number n,m,k, separated by commas. After all data are given there is -1 as the symbol of termination.nOutputnnFind the number of white mice according to the input data in each group. One line is for every output. Its fore part is a repetition of the input data and then it follows a colon and a space. The last part of it is the computed number of the white mice.nSample Inputnn10,1,6n10,1,7n10,1,9n10,1,10n10,1,11n-1nSample Outputnn10,1,6: 1n10,1,7: 2n10,1,9: 5n10,1,10: 5n10,1,11: 4
Graph of Black and White
问题描述 :nnnOnce upon a time, there was a graph. Actually, it was an undirected forest, i.e., an undirected graph without circles. There were N nodes in this forest, numbered from 1 to N inclusively, with each node colored black or white. Elves had been living in the forest for many years. Their homes were inside the nodes, and sometimes they would move their homes or visit other elves living in another node. However, they could only move between nodes that were connected by an edge or a path formed by several end-to-end connected edges. In other words, they could only move in the same tree. If two nodes were not connected, the elves could not travel between them. When they were passing a node, including the starting and the ending ones, they should use black or white magic power according to the color of the node. Waste is criminal, so during a journey the elves never pass a node twice. nnThe structure of the forest might change, when no elves were travelling. Two disconnected nodes might be connected by a new edge between them, or an existing edge might disappear, or the color of a node might change. Anyway, there would never be circles in the forest, so the forest remained a forest. THUS HARMONY LONG LASTS!nnnElves were intelligent. They knew everything happening in the forest. And they knew how much black and white magic power was used when they travel. Why? Because YY, the most intelligent one elf, simulated everything and told them. nn输入:nnnInput contains several test cases.nnFor each test case, the first line contains two positive integers, N (N ≤ 10000) and M (M ≤ 100000), indicating the number of nodes in the graph and the number of operations that YY should simulate. At the beginning of his simulation, there were not any edge in the forest.nnnThe following lines contains N space separated characters, each being ‘B’ or ‘W’ (quotes for clarity), the ith of which indicates the color of the node numbered i, black or white.nnnM lines follow, indicating the M operations to simulate. For each line, it will be one of the four types:nnnadd u vnThe two nodes numbered u and v (1≤u, v≤N, and u≠v) are connected by a new edge. It is guaranteed that these two nodes are not connected before this operation.nnndel u vnThe edge between the two nodes numbered u and v (1≤u, v≤N, and u≠v) disappears. It is guaranteed that this edge exists before this operation.nnnset u cnThe color of the node numbered u (1≤u≤N) is set to c (c=’B’ or ‘W’, quotes for clarity).nnnquery u vnNow YY tells the elves the number of black and white nodes that have to be passed when travelling between nodes numbered u and v (1≤u, v≤N) inclusively, or tells them the journey is impossible. nnnInput ends with N=M=0. nn输出:nnnInput contains several test cases.nnFor each test case, the first line contains two positive integers, N (N ≤ 10000) and M (M ≤ 100000), indicating the number of nodes in the graph and the number of operations that YY should simulate. At the beginning of his simulation, there were not any edge in the forest.nnnThe following lines contains N space separated characters, each being ‘B’ or ‘W’ (quotes for clarity), the ith of which indicates the color of the node numbered i, black or white.nnnM lines follow, indicating the M operations to simulate. For each line, it will be one of the four types:nnnadd u vnThe two nodes numbered u and v (1≤u, v≤N, and u≠v) are connected by a new edge. It is guaranteed that these two nodes are not connected before this operation.nnndel u vnThe edge between the two nodes numbered u and v (1≤u, v≤N, and u≠v) disappears. It is guaranteed that this edge exists before this operation.nnnset u cnThe color of the node numbered u (1≤u≤N) is set to c (c=’B’ or ‘W’, quotes for clarity).nnnquery u vnNow YY tells the elves the number of black and white nodes that have to be passed when travelling between nodes numbered u and v (1≤u, v≤N) inclusively, or tells them the journey is impossible. nnnInput ends with N=M=0.nn样例输入:nn3 8nB W Bnquery 1 2nadd 1 2nadd 2 3nquery 1 3ndel 2 3nadd 1 3nset 1 Wnquery 3 2n0 0n样例输出:nn-1n2 1n1 2
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