ZYB‘s Biology

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 560    Accepted Submission(s): 413

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

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int N,length,a[100],b[100],i,j,k;
    char temp;
    cin>>N;
    while(N--)
    {
        cin>>length;
        for(i=0;i<length;i++)
        {

            cin>>temp;
            if(temp==‘A‘) a[i]=1;
            else if(temp==‘C‘)a[i]=2;
            else if(temp==‘G‘)a[i]=3;
            else if(temp==‘T‘)a[i]=4;
        }
        for(j=0;j<length;j++)
        {

            cin>>temp;
            if(temp==‘U‘) b[j]=1;
            else if(temp==‘G‘)b[j]=2;
            else if(temp==‘C‘)b[j]=3;
            else if(temp==‘A‘)b[j]=4;
        }
        for(k=0;k<length;k++)
        {
            if(a[k]!=b[k]) break;
        }
        if(k<length) cout<<"NO"<<endl;
        else if(k==length) cout<<"YES"<<endl;
    }
    return 0;
}