hihoCoder #1082 : 然而沼跃鱼早就看穿了一切(字符串处理)

The Marshtomp has seen it all before.
marshTomp is beaten by fjxmlhx!
AmarshtompB

The fjxmlhx has seen it all before.
fjxmlhx is beaten by fjxmlhx!
AfjxmlhxB
题目链接：https://hihocoder.com/problemset/problem/1082
分析：自己看吧！

 1 #include<iostream>   
 2 #include <string.h>   
 3 using namespace std;  
 4   
 5 typedef long long ll;  
 6   
 7 int main(int argc , const char * argv[])  
 8 {  
 9     char s[200];  
10     while (cin.getline(s,200))  
11     {  
12         char f[] = "marshtomp";  
13         char t[] = "fjxmlhx";  
14        int  sl = strlen(s);  
15        int  fl= strlen(f);  
16        int  tl = strlen(t);  
17         for (int i =0; i<strlen(s); i++)  
18         {  
19             int j = 0;  
20             int k =i;  
21             while (s[k] == f[j] || s[k] == (f[j]-32))  
22             {  
23                 k++;  
24                 j++;  
25                 if (j == fl)  
26                 {  
27                     for (int q = i; (q-i) < tl; q++)  
28                     {  
29                         s[q] = t[q-i];  
30                     }  
31                     for (int q = i+tl; q<strlen(s); q++)  
32                     {  
33                         if ( s[q+fl-tl] != '\0')  
34                         {  
35                             s[q] = s [q+fl-tl];  
36                         }  
37                         else  
38                         {  
39                              s[q] = '\0';  
40                         }                        
41                     }  
42                 }  
43             }               
44         }    
45         cout<<s<<endl;  
46     }  
47   
48 }  

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     char s[210];
 6     while(cin.getline(s,210))//连续输入,cin.getline()的用法就是这样用,具体参照我的博客
 7     {
 8         for(int i=0;s[i]!='\0';i++)
 9         {
10             if((s[i]=='M'||s[i]=='m')&&(s[i+1]=='a'||s[i+1]=='A')&&(s[i+2]=='r'||s[i+2]=='R')&&(s[i+3]=='s'||s[i+3]=='S')&&(s[i+4]=='h'||s[i+4]=='H')&&(s[i+5]=='t'||s[i+5]=='T')&&(s[i+6]=='o'||s[i+6]=='O')&&(s[i+7]=='m'||s[i+7]=='M')&&(s[i+8]=='p'||s[i+8]=='P'))
11             {
12                 s[i]='f';
13                 s[i+1]='j';
14                 s[i+2]='x';
15                 s[i+3]='m';
16                 s[i+4]='l';
17                 s[i+5]='h';
18                 s[i+6]='x';//这个if循环就是无视大小写,直接换
19                 for(int j=i+7;s[j]!='\0';j++)
20                     s[j]=s[j+2];//换完以后再挪位置
21             }
22         }
23         cout<<s<<endl;
24     }
25     return 0;
26 }