【字符串】HDU5590ZYB's Biology【BestCoder Round #65】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5590

Problem Description

After getting  600  scores in  NOIP   ZYB(ZJ−267)  begins to work with biological questions.Now he give you a simple biological questions:
he gives you a  DNA  sequence and a  RNA  sequence,then he asks you whether the  DNA  sequence and the  RNA  sequence are 
matched.

The  DNA  sequence is a string consisted of  A,C,G,T ;The  RNA  sequence is a string consisted of  A,C,G,U .

DNA  sequence and  RNA  sequence are matched if and only if  A  matches  U , T  matches  A , C  matches  G , G  matches  C  on each position. 

 

Input

In the first line there is the testcase  T .

For each teatcase:

In the first line there is one number  N .

In the next line there is a string of length  N ,describe the  DNA  sequence.

In the third line there is a string of length  N ,describe the  RNA  sequence.

1≤T≤10 , 1≤N≤100

 

Output

For each testcase,print  YES  or  NO ,describe whether the two arrays are matched.

 

Sample Input

   
   
   
   

    
    
    
    2
4
ACGT
UGCA
4
ACGT
ACGU
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    YES
NO
   
   
   
   
   
   
   
   


   
   
   
   

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
using namespace std;
char a[1101],b[1100],c[110];
char change(char s)
{
    char ss='0';
    if(s=='A') ss='U';
    else if(s=='C') ss='G';
    else if(s=='G') ss='C';
    else if(s=='T') ss='A';
    return ss;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    //getchar();
    while(t--){
        getchar();
        scanf("%d",&n);
        getchar();
        for(int i=0;i<n;i++){
            scanf("%c",&a[i]);
            c[i]=change(a[i]);
            //cout<<c[i];
        }
        getchar();
        //cout<<endl;
        int flag=0;
        for(int i=0;i<n;i++){
            scanf("%c",&b[i]);
            if(c[i]!=b[i]) flag=1;
        }
        if(flag) cout<<"NO"<<endl;
        else cout<<"YES"<<endl;
    }
    return 0;
}