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

ACGT
UGCA

ACGT
ACGU

Sample Output

YES

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\n");

    	else printf("NO\n");

    }

	return 0;

}

巴特西

BestCoder Round #65 hdu5590(水题)

ZYB's Biology

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