HDU 5590 ZYB's Biology

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.

$1\le T\le 10$,$1\le N\le 100$

Output

For each testcase,print $YES$ or $NO$,describe whether the two arrays are matched.

Sample Input

Copy

24ACGTUGCA4ACGTACGU

Sample Output

YES
NO
简单题，暴力搞定
#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int maxn = 1e5 + 5;int T, n, m;char s1[maxn], s2[maxn];int main(){    scanf("%d", &T);    while (T--)    {        scanf("%d%s%s", &n, s1, s2);        int flag = 1;        for (int i = 0; i < n; i++)        {            if (s1[i] == 'A'&&s2[i] != 'U') flag = 0;            if (s1[i] == 'T'&&s2[i] != 'A') flag = 0;            if (s1[i] == 'C'&&s2[i] != 'G') flag = 0;            if (s1[i] == 'G'&&s2[i] != 'C') flag = 0;        }        if (flag) printf("YES\n"); else printf("NO\n");    }    return 0;}