hdu 5590 ZYB's Biology
After getting scores in NOIP ZYB(ZJ−) 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.
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. ≤T≤,≤N≤
For each testcase,print YES or NO,describe whether the two arrays are matched.
ACGT
UGCA ACGT
ACGU
YES
NO
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 106
#define inf 1e12
int n;
char s1[N];
char s2[N];
int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
scanf("%s",s1);
scanf("%s",s2);
int flag=;
for(int i=;i<n;i++){
if(s1[i]=='A'){
if(s2[i]!='U'){
flag=;
break;
}
}
else if(s1[i]=='T'){
if(s2[i]!='A'){
flag=;
break;
}
}
else if(s1[i]=='C'){
if(s2[i]!='G'){
flag=;
break;
}
}
else if(s1[i]=='G'){
if(s2[i]!='C'){
flag=;
break;
}
}
}
if(flag){
printf("YES\n");
}else{
printf("NO\n");
} }
return ;
}
hdu 5590 ZYB's Biology的更多相关文章
- HDU 5590 ZYB's Biology 水题
ZYB's Biology Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid ...
- BestCoder Round #65 (ZYB's Biology)
ZYB's Biology Accepts: 848 Submissions: 1199 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 13 ...
- hdu 5594 ZYB's Prime 最大流
ZYB's Prime Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5 ...
- hdu 5592 ZYB's Game 树状数组
ZYB's Game Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=55 ...
- hdu 5591 ZYB's Game 博弈论
ZYB's Game Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=55 ...
- hdu 5268 ZYB loves Score 水题
ZYB loves Score Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?p ...
- HDU 6468 zyb的面试
http://acm.hdu.edu.cn/showproblem.php?pid=6468 题目 今天zyb参加一场面试,面试官听说zyb是ACMer之后立马抛出了一道算法题给zyb:有一个序列,是 ...
- Bestcoder round #65 && hdu 5593 ZYB's Tree 树形dp
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submissio ...
- Bestcoder round #65 && hdu 5592 ZYB's Premutation 线段树
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submissio ...
随机推荐
- Sql Server专题二:数据库主要对象
存储过程 (1)减少网络通信量.调用一个行数不多的存储过程与直接调用sql语句的网络通信量可能不会有很大的差别,可是如果存储过程包含上百行sql语句,那么其性能绝对比一条一条的调用sql语句要高得多. ...
- Oracle EBS-SQL (PO-14):检查报价单与成本对比.sql
select distinct msi.segment1 项目 ,msi.de ...
- windows下开发PHP扩展(无需Cygwin)
第一步:准备 1.php源码包和windows下的二进制包,以及安装Visual C++,并把Microsoft Visual Studio/Common/MSDev98/Bin的绝对路径添加到win ...
- winform,wpf全屏 还显示任务栏的解决方法
原文:winform,wpf全屏 还显示任务栏的解决方法 以wpf为例: 全屏代码: this.Topmost = true; this.WindowStyle = System.Windows.Wi ...
- linux守护进程解读
Linux系统守护进程详解 不要关闭下面这几个服务: acpid, haldaemon, messagebus, klogd, network, syslogd 1. NetworkManag ...
- 定义file input
<div class="inputFileWrapper"> <label for="inputFile"> <input typ ...
- eclipse ctrl链接设置
选择[Window]菜单 Preferences ——>General——>Editors——>Text Editors——>Hyperlinking
- 具体解释HTML中的window对象和document对象
Window -- 代表浏览器中一个打开的窗体: 对象属性 window //窗体自身 window.self //引用本窗户window=window.self window.name //为窗体命 ...
- HTML5音乐可视化
环境搭建 1,安装nodejs和Git,配置环境变量2,安装express,npm install -g express-generator3,创建项目,express -e music(项目名称)4 ...
- uva 10366 Faucet Flow
题意: 给出l和r,然后从l坐标到r坐标每隔两个位置有一个档板,给出挡板的高度,然后想(-1, 1)中间加水,问什么时候会溢出. 分析: 两边先找到距离(-1,1)最近的最大值L和R.接着比较两个L和 ...