标签:turn 换行符 自动 lin div bnu span char log
练习5.9
1 int main() 2 { 3 vector<string> str; 4 string str1; 5 int cnt = 0; 6 while (cin >> str1) 7 str.push_back(str1); 8 for (auto &c : str) 9 for(auto d : c) 10 { 11 if (d == ‘a‘ || d == ‘e‘ || d == ‘i‘ || d == ‘o‘ || d == ‘u‘) 12 cnt++; 13 } 14 cout << cnt << endl; 15 system("pause"); 16 return 0; 17 }
练习5.10
1 int main() 2 { 3 unsigned aCnt = 0, eCnt = 0, iCnt = 0, oCnt = 0, uCnt = 0; 4 char ch; 5 while (cin >> ch) 6 { 7 switch (ch) 8 { 9 case ‘a‘: case ‘A‘: 10 ++aCnt; 11 break; 12 case ‘e‘: case ‘E‘: 13 ++eCnt; 14 break; 15 case ‘i‘: case ‘I‘: 16 ++iCnt; 17 break; 18 case ‘o‘: case ‘O‘: 19 ++oCnt; 20 break; 21 case ‘u‘: case ‘U‘: 22 ++uCnt; 23 break; 24 } 25 } 26 cout << aCnt << " " << eCnt << " " << iCnt << " " << oCnt << " " << uCnt << endl; 27 system("pause"); 28 return 0; 29 }
练习5.11
1 int main() 2 { 3 unsigned aCnt = 0, eCnt = 0, iCnt = 0, oCnt = 0, uCnt = 0, spaceNum = 0, tabNum = 0, n_num = 0; 4 vector<string> str1; 5 string str; 6 while ((str = cin.get()) != "#") //如果用getline函数会自动判定换行符为字符串结束条件,无法计算换行符的数量,而用cin则空格都无法计算 7 str1.push_back(str); 8 for (auto &c : str1) 9 for (auto d : c) 10 { 11 12 switch (d) 13 { 14 case ‘a‘: case ‘A‘: 15 ++aCnt; 16 break; 17 case ‘e‘: case ‘E‘: 18 ++eCnt; 19 break; 20 case ‘i‘: case ‘I‘: 21 ++iCnt; 22 break; 23 case ‘o‘: case ‘O‘: 24 ++oCnt; 25 break; 26 case ‘u‘: case ‘U‘: 27 ++uCnt; 28 break; 29 case ‘ ‘: 30 ++spaceNum; 31 break; 32 case ‘\t‘: 33 ++tabNum; 34 break; 35 case ‘\n‘: 36 ++n_num; 37 break; 38 } 39 } 40 cout << aCnt << " " << eCnt << " " << iCnt << " " << oCnt << " " << uCnt << " " << spaceNum << " " << tabNum << " " << n_num << endl; 41 system("pause"); 42 return 0; 43 }
练习5.12
1 int main() 2 { 3 unsigned aCnt = 0, eCnt = 0, iCnt = 0, oCnt = 0, uCnt = 0,ff = 0, fl = 0, fi = 0; 4 vector<char> str1; 5 char str; 6 while (cin >> str) 7 str1.push_back(str); 8 for (auto i = 0; i != str1.size(); ++i) 9 { 10 switch (str1[i]) 11 { 12 case ‘a‘: case ‘A‘: 13 ++aCnt; 14 break; 15 case ‘e‘: case ‘E‘: 16 ++eCnt; 17 break; 18 case ‘i‘: case ‘I‘: 19 ++iCnt; 20 break; 21 case ‘o‘: case ‘O‘: 22 ++oCnt; 23 break; 24 case ‘u‘: case ‘U‘: 25 ++uCnt; 26 break; 27 case ‘f‘: 28 if (str1[i + 1] == ‘f‘) 29 ++ff; 30 break; 31 } 32 } 33 cout << aCnt << " " << eCnt << " " << iCnt << " " << oCnt << " " << uCnt << " " << ff << endl; 34 system("pause"); 35 return 0; 36 }
最开始使用string类型的vector容器来储存字符串,利用双重for循环来取到每一个数的时候,再与f后面一个字符比较,这样做编译成功,但是遇到了执行上的错误,先贴上代码,等过段时间解决
1 int main() 2 { 3 unsigned aCnt = 0, eCnt = 0, iCnt = 0, oCnt = 0, uCnt = 0,ff = 0, fl = 0, fi = 0; 4 vector<string> str1; 5 string str; 6 while (cin >> str) 7 str1.push_back(str); 8 for (auto it = str1.begin();it !=str1.end(); it++) 9 for (auto i = begin(*it); i != end(*it); i++) 10 { 11 12 switch ((*i)) 13 { 14 case ‘a‘: case ‘A‘: 15 ++aCnt; 16 break; 17 case ‘e‘: case ‘E‘: 18 ++eCnt; 19 break; 20 case ‘i‘: case ‘I‘: 21 ++iCnt; 22 break; 23 case ‘o‘: case ‘O‘: 24 ++oCnt; 25 break; 26 case ‘u‘: case ‘U‘: 27 ++uCnt; 28 break; 29 case ‘f‘: 30 if (*(i + 1) == ‘f‘) 31 ++ff; 32 break; 33 } 34 } 35 cout << aCnt << " " << eCnt << " " << iCnt << " " << oCnt << " " << uCnt << " " << ff << endl; 36 system("pause"); 37 return 0; 38 }
练习5.13
a)每个case语句后少了break;
b)ix只是在1分支中定义初始化,在下面的分支中并没有定义;
c)应该写成
1 case 1 : case 3 : case 5 : case 7 : case 9 :
d)case标签必须是整形常量表达式
标签:turn 换行符 自动 lin div bnu span char log
原文地址:http://www.cnblogs.com/wuyinfenghappy/p/7270210.html