我正在做THIS UVa问题，它接受以下输入：
This is fun-
ny! Mr.P and I've never seen
this ice-cream flavour
before.Crazy eh?
#
This is fun-
ny! Mr.P and I've never seen
this ice-cream flavour
before.Crazy eh?
#
并生成此输出：
1 1
2 3
3 2
4 3
5 3
6 1
7 1
8 1
1 1
2 3
3 2
4 3
5 3
6 1
7 1
8 1
在输入中，＃分割案例。我应该得到每个单词的长度并计算每个不同长度的频率（正如你在输出中看到的那样，长度为1的单词出现一次，长度2出现三次，3次出现两次，依此类推）。
我的问题是这样的：当在cin中阅读时，before.Crazy被计为一个单词，因为没有空格分隔它们。然后它应该像在某些标点符号（例如{".",",","!","?"}）上拆分字符串一样简单...但是C ++似乎没有简单的方法来拆分字符串。
所以，我的问题：如何拆分字符串并将每个返回的字符串发送到处理问题其余部分的函数？
这是我的代码：
int main()
{
string input="";
while(cin.peek()!=-1)
{
while(cin >> input && input!="#")
{
lengthFrequency(input);
cout << input << " " << input.length() << endl;
}
if(cin.peek()!=-1) cout << endl;
lengthFrequencies.clear();
}
return 0;
}
lengthFrequency是map<int,int>。
答案 0 :(得分：4)
您可以使用带有自定义std::locale构面的std::ctype<char>重新定义流认为是空白字符的内容。下面是相应的代码，它不完全分配，但演示了如何使用构面：
#include <algorithm>
#include <iostream>
#include <locale>
#include <string>
struct ctype
: std::ctype<char>
{
typedef std::ctype<char> base;
static base::mask const* make_table(char const* spaces,
base::mask* table)
{
base::mask const* classic(base::classic_table());
std::copy(classic, classic + base::table_size, table);
for (; *spaces; ++spaces) {
table[int(*spaces)] |= base::space;
}
return table;
}
ctype(char const* spaces)
: base(make_table(spaces, table))
{
}
base::mask table[base::table_size];
};
int main()
{
std::cin.imbue(std::locale(std::locale(), new ctype(".,!?")));
for (std::string s; std::cin >> s; ) {
std::cout << "s='" << s << "'\n";
}
}
答案 1 :(得分：0)
在计算频率之前，您可以解析输入字符串并用空格（或您想要使用的任何分隔字符）替换所有{".",",","!","?"}个字符。那么你现有的代码应该可以工作。
您可能希望以不同方式处理某些字符。例如，对于before.Crazy，您可以用空格替换'.'，但对于类似'ny! '的内容，您会完全删除'!'，因为它已经跟着一个空间。
答案 2 :(得分：0)
这个怎么样（使用STL，比较器和仿函数）？
注意：所有假设和解释都在源代码中。
#include <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <sstream>
#include <algorithm>
#include <cctype>
#include <utility>
#include <string.h>
bool compare (const std::pair<int, int>& l, const std::pair<int, int>& r) {
return l.first < r.first;
}
//functor/unary predicate:
struct CompareFirst {
CompareFirst(int val) : val_(val) {}
bool operator()(const std::pair<int, int>& p) const {
return (val_ == p.first);
}
private:
int val_;
};
int main() {
char delims[] = ".,!?";
char noise[] ="-'";
//I'm assuming you've read the text from some file, and that information has been stored in a string. Or, the information is a string (like below):
std::string input = "This is fun-\nny, Mr.P and I've never seen\nthis ice-cream flavour\nbefore.Crazy eh?\n#\nThis is fun-\nny! Mr.P and I've never seen\nthis ice-cream flavour\nbefore.Crazy eh?\n#\n";
std::istringstream iss(input);
std::string temp;
//first split the string by #
while(std::getline(iss, temp, '#')) {
//find all the occurences of the hypens as it crosses lines, and remove the newline:
std::string::size_type begin = 0;
while(std::string::npos != (begin = temp.find('-', begin))) {
//look at the character in front of the current hypen and erase it if it's a newline, if it is - remove it
if (temp[begin+1] == '\n') {
temp.erase(begin+1, 1);
}
++begin;
}
//now, erase all the `noise` characters ("'-") as these count as these punctuation count as zero
for (int i = 0; i < strlen(noise); ++i) {
//this replaces all the hyphens and apostrophes with nothing
temp.erase(std::remove(temp.begin(), temp.end(), noise[i]), temp.end());//since hyphens occur across two lines, you need to erase newlines
}//at this point, everything is dandy for complete substitution
//now try to remove any other delim chracters by replacing them with spaces
for (int i = 0; i < strlen(delims); ++i) {
std::replace(temp.begin(), temp.end(), delims[i], ' ');
}
std::vector<std::pair<int, int> > occurences;
//initialize another input stringstream to make use of the whitespace
std::istringstream ss(temp);
//now use the whitespace to tokenize
while (ss >> temp) {
//try to find the token's size in the occurences
std::vector<std::pair<int, int> >::iterator it = std::find_if(occurences.begin(), occurences.end(), CompareFirst(temp.size()));
//if found, increment count by 1
if (it != occurences.end()) {
it->second += 1;//increment the count
}
//this is the first time it has been created. Store value, and a count of 1
else {
occurences.push_back(std::make_pair<int, int>(temp.size(), 1));
}
}
//now sort and output:
std::stable_sort(occurences.begin(), occurences.end(), compare);
for (int i = 0; i < occurences.size(); ++i) {
std::cout << occurences[i].first << " " << occurences[i].second << "\n";
}
std::cout << "\n";
}
return 0;
}
91行，所有香草C ++ 98。
我所做的大致概述是：
参考文献：