90分求调(必关)

#include<bits/stdc++.h>
using namespace std;
char s[100];
int main()
{
    int xx = 26,pl[10000] = {};
	string a,b,c;
    cin >> a >> b >> c;
    if(a.size()!=b.size()){cout << "Failed";exit(0);}
    for(int i = 1;i<=26;i++)s[i] = '0';
    for(int i = 1;i<=a.size();i++)
    {
        if(pl[a[i-1]]==0)pl[a[i-1]]++,xx--;
    }
    if(xx>0){cout << "Failed";exit(0);}
    for(int i = 0;i<a.size();i++)
    {
        if(s[a[i]-'A'+1]!='0'&&s[a[i]-'A'+1]!=b[i])
        {
            cout << "Failed";
            exit(0);
        }
        s[a[i]-'A'+1] = b[i];
    }
    for(int i = 0;i<c.size();i++)
    {
        cout<<s[c[i]-'A'+1];
    }
	return 0;
}