【水】HDOJ ZYB's Biology 5590

ZYB's Biology

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 323    Accepted Submission(s): 269

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

24ACGTUGCA4ACGTACGU

 

Sample Output

YESNO

 

Source

BestCoder Round #65

 

AC代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int main(){    int t;    scanf("%d",&t);    while(t--){        char D[110],R[110];        int N;        scanf("%d",&N);        scanf("%s%s",D,R);        int LD=strlen(D);        int LR=strlen(R);        if(LD!=LR){            printf("NO\n");            continue;        }        int flag=1;        for(int i=0;i<N;i++){            if(D[i]=='A'&&R[i]=='U')continue;            else if(D[i]=='T'&&R[i]=='A')continue;            else if(D[i]=='C'&&R[i]=='G')continue;            else if(D[i]=='G'&&R[i]=='C')continue;            flag=0;        }        if(flag)printf("YES\n");        else printf("NO\n");    }    return 0;}