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

?

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#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; }