BestCoder Round #65 hdu5590(水题)

ZYB's Biology

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 713    Accepted Submission(s): 522

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

 

题意：给一段DNA和RNA问是否配对

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 110
#define mod 10007
#define dian 1.000000011
char s1[MAX],s2[MAX];
using namespace std;
int judge(int x)
{
	int flag=0;
	if(s1[x]=='A'&&s2[x]=='U')
	    flag=1;
	else if(s1[x]=='C'&&s2[x]=='G')
	    flag=1;
	else if(s1[x]=='G'&&s2[x]=='C')
	    flag=1;
	else if(s1[x]=='T'&&s2[x]=='A')
	    flag=1;
	if(flag) return 1;
	else return 0;
}
int main()
{
    int t,n,m,i,j;
    scanf("%d",&t);
    while(t--)
    {
    	scanf("%d",&n);
    	scanf("%s",s1);
    	scanf("%s",s2);
    	int flag=1;
    	for(i=0;i<n;i++)
    	{
    		if(!judge(i))
    		{
    			flag=0;
    			break;
    		}
    	}
    	if(flag) printf("YES
");
    	else printf("NO
");
    }
	return 0;
}