华为训练题：初级——合并字符串

  1 #include<iostream>
  2 #include<string>
  3 using namespace std;
  4 
  5 char daoxu(char ch)
  6 {
  7     if((ch>=‘0‘&&ch<=‘9‘)||(ch>=‘A‘&&ch<=‘F‘)||(ch>=‘a‘&&ch<=‘f‘))
  8     {
  9         if(ch>=‘0‘&&ch<=‘9‘)
 10         {
 11             switch(ch)
 12             {
 13             case ‘0‘:
 14                 return ‘0‘;
 15             case ‘1‘:
 16                 return ‘8‘;
 17             case ‘2‘:
 18                 return ‘4‘;
 19             case ‘3‘:
 20                 return ‘C‘;
 21             case ‘4‘:
 22                 return ‘2‘;
 23             case ‘5‘:
 24                 return ‘A‘;
 25             case ‘6‘:
 26                 return ‘6‘;
 27             case ‘7‘:
 28                 return ‘E‘;
 29             case ‘8‘:
 30                 return ‘1‘;
 31             case ‘9‘:
 32                 return ‘9‘;
 33             }
 34         }
 35         if(ch>=‘A‘&&ch<=‘F‘)
 36         {
 37             switch(ch)
 38             {
 39             case ‘A‘:
 40                 return ‘5‘;
 41             case ‘B‘:
 42                 return ‘D‘;
 43             case ‘C‘:
 44                 return ‘3‘;
 45             case ‘D‘:
 46                 return ‘B‘;
 47             case ‘E‘:
 48                 return ‘7‘;
 49             case ‘F‘:
 50                 return ‘F‘;
 51             }
 52         }
 53         if(ch>=‘a‘&&ch<=‘f‘)
 54         {
 55             switch(ch)
 56             {
 57             case ‘a‘:
 58                 return ‘5‘;
 59             case ‘b‘:
 60                 return ‘D‘;
 61             case ‘c‘:
 62                 return ‘3‘;
 63             case ‘d‘:
 64                 return ‘B‘;
 65             case ‘e‘:
 66                 return ‘7‘;
 67             case ‘f‘:
 68                 return ‘F‘;
 69             }
 70         }
 71     }
 72     else
 73         return ch;
 74 }
 75 
 76 void paixu(string& str)
 77 {
 78     int len=str.size();
 79     if(len==1)
 80         return ;
 81     for(int i=0;i<len-1;i++)
 82         for(int j=i+1;j<len;j++)
 83             if(str[i]>str[j])
 84             {
 85                 char temp=str[i];
 86                 str[i]=str[j];
 87                 str[j]=temp;
 88             }
 89     return;
 90 }
 91 
 92 string ProcessString1(string& str)
 93 {
 94     int i=0;
 95     string str1=str;
 96 
 97     if(str1.empty())
 98         return str1;
 99 
100     int len=str1.size();
101 
102     if(len==1)
103     {
104         str1[0]=daoxu(str1[0]);
105         return str1;
106     }
107 
108     if(len==2)
109     {
110         str1[0]=daoxu(str1[0]);
111         str1[1]=daoxu(str1[1]);
112         return str1;
113     }
114 
115     string str2,str3;
116     for(int i=0;i<len;i+=2)
117         str2.push_back(str1[i]);
118     paixu(str2);
119     for(int i=1;i<len;i+=2)
120         str3.push_back(str1[i]);
121     paixu(str3);
122 
123     if(len%2==0)
124     {
125         str1.clear();
126         int len_str2=str2.size();
127         for(int i=0;i<len_str2;i++)
128         {
129             str1.push_back(str2[i]);
130             str1.push_back(str3[i]);
131         }
132     }
133 
134     if(len%2!=0)
135     {
136         str1.clear();
137         int len_str3=str3.size();
138         for(int i=0;i<len_str3;i++)
139         {
140             str1.push_back(str2[i]);
141             str1.push_back(str3[i]);
142         }
143         str1.push_back(str2[str2.size()-1]);
144     }
145 
146     for(int i=0;i<len;i++)
147         str1[i]=daoxu(str1[i]);
148     return str1;
149 }
150 
151 int main()
152 {
153     string s;
154     getline(cin,s);
155     int i;
156     int len_s=s.size();
157     for(i=0;i<len_s;i++)
158     {
159         if(s[i]==‘ ‘)
160             break;
161     }
162     
163     for(int j=i;j<len_s-1;j++)
164         s[j]=s[j+1];
165 
166     string str;
167     for(int i=0;i<len_s-1;i++)
168         str.push_back(s[i]);
169 
170     str=ProcessString1(str);
171     cout<<str<<endl;
172 
173     system("pause");
174 }