P1017潜伏者解题报告
无秒
2020-05-04 17:16:05
这题很简单,就模拟一下,开个数组来映射一下,没啥好讲的。(水题)
```cpp
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
char s1[26],s2[26];
string x1,x2,x3;
int main()
{
memset(s1,'0',sizeof(s1));
memset(s2,'0',sizeof(s2));
cin>>x1>>x2>>x3;
if(x1.size()!=x2.size())
{
printf("Failed");
return 0;
}
for(int i=0;i<x1.size();i++)
{
if(s1[x1[i]-'A']=='0'||s1[x1[i]-'A']==x2[i])
s1[x1[i]-'A']=x2[i];
else
{
printf("Failed");
return 0;
}
if(s2[x2[i]-'A']=='0'||s2[x2[i]-'A']==x1[i])
s2[x2[i]-'A']=x1[i];
else
{
printf("Failed");
return 0;
}
}
for(int i=0;i<=25;i++)
if(s1[i]=='0')
{
printf("Failed");
return 0;
}
for(int i=0;i<x3.size();i++)
printf("%c",s1[x3[i]-'A']);
return 0;
}
```