$NOIP2009$ 题解报告

目录

•$Luogu\ P1071$ 潜伏者$(\ √\ )$

•$Luogu\ P1072$ $Hankson$的趣味题$(\ \ \ )$

•$Luogu\ P1073$ 最优贸易$(\ \ \ )$

•$Luogu\ P1074$ 靶形数独$(\ \ \ )$

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$Luogu\ P1071$ 潜伏者

题目传送门

题目很水，直接模拟即可，注意一下细节

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 string x,y,z;
 4 int sum[27],tot=0;
 5 bool b1[27],b2[27];
 6 int main(){
 7     cin>>x>>y>>z;
 8     for(int i=0;i<x.size();++i)
 9     {
10         if(!b1[x[i]-'A'+1]&&!b2[y[i]-'A'+1])
11         {
12             sum[x[i]-'A'+1]=y[i];
13             b1[x[i]-'A'+1]=b2[y[i]-'A'+1]=true;
14             ++tot;
15         }    
16         else if(sum[x[i]-'A'+1]!=y[i])
17         {
18             cout<<"Failed";
19             return 0;    
20         }
21     }
22     if(tot!=26)
23     {
24         cout<<"Failed";
25             return 0;    
26     }
27     for(int i=0;i<z.size();++i)
28     {
29         printf("%c",sum[z[i]-'A'+1]);    
30     }
31 }

代码戳这里

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$Luogu\ P1072$ $Hankson$的趣味题

题目传送门

 

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$Luogu\ P1073$ 最优贸易

题目传送门

 

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$Luogu\ P1074$ 靶形数独

题目传送门