Codeforces Round #424 B. Keyboard Layouts（字符串，匹配，map）
1 #include <stdio.h> 2 #include <string.h> 3 4 char ch[3][30]; 5 char t[1005]; 6 7 int num[35]; 8 9 int main(){ 10 scanf("%s",ch[0]); 11 scanf("%s",ch[1]); 12 scanf("%s",t); 13 14 for(int i=0;i<strlen(ch[0]);i++){ 15 num[ch[0][i]-'a']=i; 16 } 17 18 for(int i=0;i<strlen(t);i++){ 19 if(0<=t[i]-'0'&&t[i]-'0'<=9){ 20 printf("%c",t[i]); 21 }else if('a'<=t[i]&&t[i]<='z'){ 22 printf("%c",ch[1][num[t[i]-'a']]); 23 }else if('A'<=t[i]&&t[i]<='Z'){ 24 printf("%c",ch[1][num[t[i]-'A']]-32); 25 } 26 27 28 29 } 30 printf("\n"); 31 return 0; 32 }
1 #include <stdio.h> 2 #include <string.h> 3 char key[300]; 4 int main () 5 { 6 char b,f[30],l[30],sent[10000]; 7 int i,j,x; 8 scanf("%s",f); 9 scanf("%s",l); 10 scanf("%s",sent); 11 for(i=0;i<26;i++) 12 { 13 key[f[i]]=l[i]; 14 key[f[i]-32]=l[i]-32; 15 } 16 for(i=0;i<strlen(sent);i++) 17 { 18 if(key[sent[i]]) 19 printf("%c",key[sent[i]]); 20 else 21 { 22 23 printf("%c",sent[i]); 24 } 25 } 26 return 0; 27 28 }
1 #include<stdio.h> 2 3 int main() 4 { 5 char a[30],b[30],ch; 6 int i=0; 7 while((a[i]=getchar())!='\n') i++; 8 i=0; 9 while((b[i]=getchar())!='\n') i++; 10 while((ch=getchar())!='\n') 11 { 12 if(ch<65) printf("%c",ch); 13 else if(ch<97) 14 { 15 ch+=32; 16 for(i=0;i<26;i++) 17 if(a[i]==ch) 18 { 19 printf("%c",b[i]-32); 20 break; 21 } 22 } 23 else 24 for(i=0;i<26;i++) 25 if(a[i]==ch) 26 { 27 printf("%c",b[i]); 28 break; 29 } 30 } 31 return 0; 32 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int main() 5 { 6 char a[10000],c; 7 int i; 8 string s1,s2,s; 9 cin>>s1>>s2>>s; 10 for(i=0;i<26;++i) 11 a[s1[i]]=s2[i]; 12 for(i=0;i<s.length();++i) 13 { 14 if(s[i]>=65&&s[i]<=90) 15 { 16 c=s[i]+32; 17 s[i]=a[c]-32; 18 } 19 else if(s[i]>=97&&s[i]<=122) 20 { 21 s[i]=a[s[i]]; 22 } 23 } 24 cout<<s; 25 return 0; 26 }
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 using namespace std; 5 6 char s1[33],s2[33],ch[222],s[1111]; 7 int len; 8 9 int main() 10 { 11 scanf("%s%s%s",s1,s2,s); 12 for(int i=0;i<=25;i++) 13 { 14 ch[s1[i]]=s2[i]; 15 } 16 len=strlen(s); 17 for(int i=0;i<len;i++) 18 { 19 if(s[i]>='A'&&s[i]<='Z') 20 { 21 s[i]+=32; 22 s[i]=ch[s[i]]-32; 23 } 24 else if(s[i]>='a'&&s[i]<='z') 25 s[i]=ch[s[i]]; 26 else 27 s[i]=s[i]; 28 } 29 printf("%s\n",s); 30 return 0; 31 }
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
Print the text if the same keys were pressed in the second layout.
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
HelloVKCup2017
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
7uduGUDUUDUgudu7
