查詢單詞的題目，很有意思，我提供了兩個版本

//exercise

//已知有如下 string 物件：string line1 = "we were her pride of 10 she named us:"; 
string line2 = "benjamin, phoenix, the prodigal"
string line3 = "and perspicacious pacific suzanne";
string sentence = line1 + ' ' + line2 + ' ' + line3;
//編寫程式計算 sentence 中有多少個單詞，並指出其中最長和最短的單詞。如果有多個最長或最短的單詞，則將它們全部輸出。
//版本一：
#include #include #include #include using namespace std;
int main()
else
else
{if(size_tmp::iterator big_iter=longest_word.begin();big_iter!=longest_word.end();
big_iter++)
cout<<*big_iter<<" ";
cout<::iterator small_iter=smallest_word.begin();small_iter!=smallest_word.end();
small_iter++)
cout<<*small_iter<<" ";
cout<#include #include using namespace std;
int main()
{ string line1="we are her pride of 10 she named us:";
string line2="benjamin, phoenix, the prodigal";
string line3="and perspicacious pacific suzanen";
string spaces(" \t:,\n\v\r\f");//後面都一些轉義字元
string sentence=line1+" "+line2+" "+line3;
vectorlongest_word,smallest_word;
unsigned int count=0;
string word;
string::size_type s_size_big=0,s_size_small=0;
string::size_type start_pos=0, end_pos=0;
while((start_pos=sentence.find_first_not_of(spaces,end_pos))!=string::npos)
{++count;
string::size_type size_tmp;
end_pos=sentence.find_first_of(spaces,start_pos); //這個是**的核心部分
if(end_pos==string::npos)//如果沒有匹配的話就返回npos,這是乙個非常大的數，比任何的合法的下標都大
size_tmp=sentence.size()-start_pos;
else
size_tmp=end_pos-start_pos;
word.assign(sentence,start_pos,size_tmp);
cout<<"the size of sentence: "