BIT1006 The most frequent number

本文通过一个具体的C++程序实例,介绍了如何针对输入输出方式进行优化以避免超时问题,特别是对比了使用cin与scanf在效率上的差异。

真该剁手啊。。。大水题因为cin的原因超时调了老半天

#include<iostream>
#include<algorithm>
using namespace std;
int a[100100];
class T
{
public:
	int num;
	int counter;
}ans;
int main()
{
	int n;
	while(cin>>n)
	{
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
		}
		sort(a,a+n);
		T temp;
		temp.counter=1;
		temp.num=a[0];
		ans=temp;
		for(int i=1;i<n;i++)
		{
			if(a[i]==a[i-1])
			{
				temp.counter++;
			}
			else
			{
				if(temp.counter>ans.counter)
				{
					ans=temp;
				}
				temp.counter=1;
				temp.num=a[i];
			}
		}
		if(temp.counter>ans.counter)
		{
			ans=temp;
		}
		cout<<ans.num<<endl;
	}
	return 0;
}


翻译成中文,只做翻译:8.2. Simultaneous Probe Tiebreaking The astute reader will observe that there is a race condition inherent in the previous description. If two hosts are probing for the same name simultaneously, neither will receive any response to the probe, and the hosts could incorrectly conclude that they may both proceed to use the name. To break this symmetry, each host populates the query message's Authority Section with the record or records with the rdata that it would be proposing to use, should its probing be successful. The Authority Section is being used here in a way analogous to the way it is used as the "Update Section" in a DNS Update message [RFC2136] [RFC3007]. When a host is probing for a group of related records with the same name (e.g., the SRV and TXT record describing a DNS-SD service), only a single question need be placed in the Question Section, since query type "ANY" (255) is used, which will elicit answers for all records with that name. However, for tiebreaking to work correctly in all cases, the Authority Section must contain *all* the records and proposed rdata being probed for uniqueness. When a host that is probing for a record sees another host issue a query for the same record, it consults the Authority Section of that query. If it finds any resource record(s) there which answers the query, then it compares the data of that (those) resource record(s) with its own tentative data. We consider first the simple case of a host probing for a single record, receiving a simultaneous probe from another host also probing for a single record. The two records are compared and the lexicographically later data wins. This means that if the host finds that its own data is lexicographically later, it simply ignores the other host's probe. If the host finds that its own data is lexicographically earlier, then it defers to the winning host by waiting one second, and then begins probing for this record again. The logic for waiting one second and then trying again is to guard against stale probe packets on the network (possibly even stale probe packets sent moments ago by this host itself, before some configuration change, which may be echoed back after a short delay by some Ethernet switches and some 802.11 base stations). If the winning simultaneous probe was from a real other host on the network, then after one second it will have completed its probing, and will answer subsequent probes. If the apparently winning simultaneous probe was in fact just an old stale packet on the network (maybe from the host itself), then when it retries its probing in one second, its probes will go unanswered, and it will successfully claim the name. The determination of "lexicographically later" is performed by first comparing the record class (excluding the cache-flush bit described in Section 10.2), then the record type, then raw comparison of the binary content of the rdata without regard for meaning or structure. If the record classes differ, then the numerically greater class is considered "lexicographically later". Otherwise, if the record types differ, then the numerically greater type is considered "lexicographically later". If the rrtype and rrclass both match, then the rdata is compared. In the case of resource records containing rdata that is subject to name compression [RFC1035], the names MUST be uncompressed before comparison. (The details of how a particular name is compressed is an artifact of how and where the record is written into the DNS message; it is not an intrinsic property of the resource record itself.) The bytes of the raw uncompressed rdata are compared in turn, interpreting the bytes as eight-bit UNSIGNED values, until a byte is found whose value is greater than that of its counterpart (in which case, the rdata whose byte has the greater value is deemed lexicographically later) or one of the resource records runs out of rdata (in which case, the resource record which still has remaining data first is deemed lexicographically later). The following is an example of a conflict: MyPrinter.local. A 169.254.99.200 MyPrinter.local. A 169.254.200.50 In this case, 169.254.200.50 is lexicographically later (the third byte, with value 200, is greater than its counterpart with value 99), so it is deemed the winner. Note that it is vital that the bytes are interpreted as UNSIGNED values in the range 0-255, or the wrong outcome may result. In the example above, if the byte with value 200 had been incorrectly interpreted as a signed eight-bit value, then it would be interpreted as value -56, and the wrong address record would be deemed the winner. 8.2.1. Simultaneous Probe Tiebreaking for Multiple Records When a host is probing for a set of records with the same name, or a message is received containing multiple tiebreaker records answering a given probe question in the Question Section, the host's records and the tiebreaker records from the message are each sorted into order, and then compared pairwise, using the same comparison technique described above, until a difference is found. The records are sorted using the same lexicographical order as described above, that is, if the record classes differ, the record with the lower class number comes first. If the classes are the same but the rrtypes differ, the record with the lower rrtype number comes first. If the class and rrtype match, then the rdata is compared bytewise until a difference is found. For example, in the common case of advertising DNS-SD services with a TXT record and an SRV record, the TXT record comes first (the rrtype value for TXT is 16) and the SRV record comes second (the rrtype value for SRV is 33). When comparing the records, if the first records match perfectly, then the second records are compared, and so on. If either list of records runs out of records before any difference is found, then the list with records remaining is deemed to have won the tiebreak. If both lists run out of records at the same time without any difference being found, then this indicates that two devices are advertising identical sets of records, as is sometimes done for fault tolerance, and there is, in fact, no conflict. 8.3. Announcing The second startup step is that the Multicast DNS responder MUST send an unsolicited Multicast DNS response containing, in the Answer Section, all of its newly registered resource records (both shared records, and unique records that have completed the probing step). If there are too many resource records to fit in a single packet, multiple packets should be used. In the case of shared records (e.g., the PTR records used by DNS- Based Service Discovery [RFC6763]), the records are simply placed as is into the Answer Section of the DNS response. In the case of records that have been verified to be unique in the previous step, they are placed into the Answer Section of the DNS response with the most significant bit of the rrclass set to one. The most significant bit of the rrclass for a record in the Answer Section of a response message is the Multicast DNS cache-flush bit and is discussed in more detail below in Section 10.2, "Announcements to Flush Outdated Cache Entries". The Multicast DNS responder MUST send at least two unsolicited responses, one second apart. To provide increased robustness against packet loss, a responder MAY send up to eight unsolicited responses, provided that the interval between unsolicited responses increases by at least a factor of two with every response sent. A Multicast DNS responder MUST NOT send announcements in the absence of information that its network connectivity may have changed in some relevant way. In particular, a Multicast DNS responder MUST NOT send regular periodic announcements as a matter of course. Whenever a Multicast DNS responder receives any Multicast DNS response (solicited or otherwise) containing a conflicting resource record, the conflict MUST be resolved as described in Section 9, "Conflict Resolution". 8.4. Updating At any time, if the rdata of any of a host's Multicast DNS records changes, the host MUST repeat the Announcing step described above to update neighboring caches. For example, if any of a host's IP addresses change, it MUST re-announce those address records. The host does not need to repeat the Probing step because it has already established unique ownership of that name. In the case of shared records, a host MUST send a "goodbye" announcement with RR TTL zero (see Section 10.1, "Goodbye Packets") for the old rdata, to cause it to be deleted from peer caches, before announcing the new rdata. In the case of unique records, a host SHOULD omit the "goodbye" announcement, since the cache-flush bit on the newly announced records will cause old rdata to be flushed from peer caches anyway. A host may update the contents of any of its records at any time, though a host SHOULD NOT update records more frequently than ten times per minute. Frequent rapid updates impose a burden on the network. If a host has information to disseminate which changes more frequently than ten times per minute, then it may be more appropriate to design a protocol for that specific purpose.
10-18
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<think> 我们有一个名为most_frequent的函数,它接受一个字符串,然后按字符出现频率降序打印字母(实际上根据引用[1]的代码,它返回的是频率和字符的元组列表,然后打印每个元组)。但是用户的问题是如何调用这个函数,并且要求是按字符出现频率降序打印字母(注意,是打印字母,而引用[1]中是同时打印频率和字母)。因此,我们需要区分: 用户可能希望只打印字母(不打印频率),或者按照引用[1]那样打印频率和字母。 根据引用[1]的示例代码: def most_frequent(s): hist = make_histogram(s) t = [] for letter in hist: t.append((hist[letter], letter)) t.sort(reverse=True) return t 然后调用: string = open('emma.txt').read() for x in most_frequent(string): print(x) 输出是每个元组(频率,字符)。 但是用户的问题描述是“按字符出现频率降序打印字母”,注意这里只提到“字母”,没有要求频率。所以我们需要调整。 有两种理解: 1. 用户希望打印的是字母(不包含频率),按照频率降序排列。 2. 或者用户希望同时打印频率和字母(如引用[1]的做法)。 考虑到用户的问题描述为“打印字母”,这里我们按第一种理解:只打印字母。 因此,我们可以修改most_frequent函数,让它返回按频率降序的字母列表(没有频率),或者修改调用方式,只打印字母而不打印频率。 另外注意,引用[1]中的代码统计的是所有字符(包括空格、标点等),而题目要求是“字母”。所以我们需要在统计前过滤非字母字符吗? 用户的问题描述是“打印字母”,所以应该只考虑字母(即英文字母a-z, A-Z),忽略其他字符。 因此,我们需要在函数内部或调用前进行过滤。 综合以上,我们有两种方案: 方案一:修改原函数,使其只统计字母,并且返回按频率降序的字母列表(不包含频率)。 方案二:不修改原函数,但在调用时处理:过滤非字母,并且只打印字母部分。 这里我们按照用户的需求,构建一个符合要求的函数。 重新定义most_frequent函数,要求: 输入:一个字符串 输出:无(直接按频率降序打印字母)或者返回字母列表(然后调用者打印)?根据问题描述,函数应该打印,所以我们可以设计为直接打印。 但问题中说的是“打印字母”,所以我们可以设计成函数内部打印。 然而,为了灵活性,通常函数返回结果,由调用者决定打印。所以这里我们设计为返回按频率降序的字母列表,然后调用者打印这个列表的每个元素。 因此,我们可以这样: 步骤: 1. 过滤字符串,只保留字母(可以使用str.isalpha方法) 2. 构建频率直方图(可以使用字典或者collections.Counter) 3. 根据频率降序排序,如果频率相同,可以按字母顺序升序(或者任意,问题没要求)?但问题没有要求同频率的排序,所以可以任意。 4. 返回一个列表,列表元素是字母(按频率降序)。 但是,注意:原引用[1]中函数返回的是元组列表(频率,字母),而我们现在要返回字母列表。 另外,用户要求调用函数,所以我们需要提供调用示例。 我们写一个符合要求的most_frequent函数(只统计字母,返回字母列表)并展示如何调用。 注意:由于原引用[1]中已经有一个most_frequent函数,但它的行为是返回(频率,字母)的元组列表,所以我们这里需要重新实现。 但是,用户的问题中提到了“名为most_frequent”的函数,所以我们可以根据需求重新实现。 以下是具体代码: 方案(函数返回字母列表,按频率降序): </think> ### 如何调用与理解 `most_frequent` 函数 根据引用[1]中的代码,`most_frequent` 函数按字符频率降序处理字符串的核心逻辑如下: #### 函数定义分析 ```python def make_histogram(s): """构建字符频率直方图""" hist = {} for letter in s: hist[letter] = hist.get(letter, 0) + 1 # 统计字符频率 return hist def most_frequent(s): """按频率降序处理字符""" hist = make_histogram(s) # 获取直方图 t = [] for letter in hist: t.append((hist[letter], letter)) # 存储元组(频率, 字符) t.sort(reverse=True) # 按频率降序排序 return t # 返回排序后的元组列表 ``` #### 调用步骤(完整示例) ```python # 1. 定义上述两个函数 # 2. 准备输入字符串(可直接使用字符串或读取文件) input_string = "hello world" # 示例字符串 # 或从文件读取:input_string = open('emma.txt').read() # 3. 调用函数并处理结果 result = most_frequent(input_string) # 4. 打印结果(按频率降序) for freq, char in result: print(f"字符 '{char}' 出现 {freq} 次") ``` #### 输出示例 对于输入 `"hello"`: ``` 字符 'l' 出现 2 次 字符 'h' 出现 1 次 字符 'e' 出现 1 次 字符 'o' 出现 1 次 ``` ### 关键特性说明 1. **统计范围** 函数统计**所有字符**(包括空格、标点、数字等),如引用[1]输出中显示空格`' '`频率最高[^1]。 *若需仅统计字母,需预处理:* ```python letters_only = ''.join(c for c in s if c.isalpha()) ``` 2. **排序机制** - 主排序:字符频率降序(`reverse=True`) - 同频字符:按**原始遍历顺序**(Python字典无序性),输出顺序不确定[^1] 3. **返回结构** 返回元组列表 `[(频率, 字符), ...]`,便于同时获取频率和字符信息 ### 进阶优化建议 ```python from collections import Counter def most_frequent_optimized(s): """使用Counter优化""" return Counter(s).most_common() # 直接返回降序的(字符, 频率)列表 # 调用示例 for char, freq in most_frequent_optimized("hello"): print(f"{char}: {freq}") ``` *优势:代码更简洁,同频字符按首次出现顺序排序(Python 3.7+)* ### 实际应用场景 1. **文本分析** ```python # 分析小说高频字符(如引用[1]的emma.txt) text = open('book.txt').read() top_chars = most_frequent(text)[:10] # 获取前10高频字符 ``` 2. **数据清洗** ```python # 检测非常用字符(如特殊符号) low_freq_chars = [char for freq, char in most_frequent(text) if freq < 5] ```
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