hdoj 5590 ZYB's Biology 【水题】

最新推荐文章于 2017-12-04 14:17:28 发布

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本文介绍了一道关于DNA与RNA序列匹配的问题，探讨了如何判断一个DNA序列与一个RNA序列是否匹配，并提供了一个简单的模拟算法实现。

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ZYB's Biology

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 9 Accepted Submission(s): 8

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

T matches

C matches

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

水题，直接模拟。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define INF 0x3f3f3f
#define eps 1e-8
#define MAXN (100+10)
#define MAXM (100000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
bool judge(char a, char b)
{
    if(a == 'A' && b == 'U') return true;
    if(a == 'T' && b == 'A') return true;
    if(a == 'C' && b == 'G') return true;
    if(a == 'G' && b == 'C') return true;
    return false;
}
int main()
{
    int t; Ri(t);
    W(t)
    {
        int n;
        char a[110], b[110];
        Ri(n);
        Rs(a); Rs(b);
        bool flag = true;
        for(int i = 0; i < n; i++)
        {
            if(!judge(a[i], b[i]))
            {
                flag = false;
                break;
            }
        }
        printf(flag ? "YES\n" : "NO\n");
    }
    return 0;
}