【字符串】HDU5590ZYB's Biology【BestCoder Round #65】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5590

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

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
using namespace std;
char a[1101],b[1100],c[110];
char change(char s)
{
char ss='0';
if(s=='A') ss='U';
else if(s=='C') ss='G';
else if(s=='G') ss='C';
else if(s=='T') ss='A';
return ss;
}
int main()
{
int t,n;
scanf("%d",&t);
//getchar();
while(t--){
getchar();
scanf("%d",&n);
getchar();
for(int i=0;i<n;i++){
scanf("%c",&a[i]);
c[i]=change(a[i]);
//cout<<c[i];
}
getchar();
//cout<<endl;
int flag=0;
for(int i=0;i<n;i++){
scanf("%c",&b[i]);
if(c[i]!=b[i]) flag=1;
}
if(flag) cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}
return 0;
}