HDOJ 5590 ZYB's Biology



ZYB's Biology

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 63    Accepted Submission(s): 52

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：问DNA序列和RNA序列是否匹配，DNA和RNA匹配当且仅当每个位置上A与U,T与A,C与G,G与C匹配。

水题，代码如下：

#include<cstdio> 
#include<cstring>
char a[110],b[110];
int main()
{
	int i,n,sign,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		scanf("%s",a);
		scanf("%s",b);
		sign=1;
		for(i=0;i<n;++i)
		{
			if(a[i]=='A'&&b[i]!='U')
			{
				sign=0;
				break;
			}
			if(a[i]=='T'&&b[i]!='A')
			{
				sign=0;
				break;
			}
			if(a[i]=='C'&&b[i]!='G')
			{
				sign=0;
				break;
			}
			if(a[i]=='G'&&b[i]!='C')
			{
				sign=0;
				break;
			}
		}
		if(sign)
			printf("YES\n");
		else 
			printf("NO\n");
	}
	return 0;
}