Codeforces Round #424 B

Keyboard Layouts

题意：有2个映射关系，分别是将abcd...z映射到s1和s2，现在给你经过s1映射的字符串s，求经原字符串过s2映射后的字符串

思路：映射拿map xjb搞就是了

AC代码:

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;

char s1[30],s2[30],ss[1005];
map<char,char> M1,M2;
int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>s1>>s2>>ss;
    M1.clear(),M2.clear();
    for(int i=0; i<26; ++i){
        M1[s1[i]]='a'+i;
        M1[s1[i]-32]='A'+i;
        M2['a'+i]=s2[i];
        M2['A'+i]=s2[i]-32;
    }
    for(int i=0; i<strlen(ss); ++i){
        if((ss[i]>='a' && ss[i]<='z') || (ss[i]>='A' && ss[i]<='Z')){
            cout<<M2[M1[ss[i]]];
        }
        else cout<<ss[i];
    }
    return 0;
}