Blue Jeans
Description
As an IBM researcher, you have been tasked with writing a program that will find commonalities amongst given snippets of DNA that can be correlated with individual survey information to identify new genetic markers.
A DNA base sequence is noted by listing the nitrogen bases in the order in which they are found in the molecule. There are four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). A 6-base DNA sequence could be represented as TAGACC.
Given a set of DNA base sequences, determine the longest series of bases that occurs in all of the sequences.
Input
- A single positive integer m (2 <= m <= 10) indicating the number of base sequences in this dataset.
- m lines each containing a single base sequence consisting of 60 bases.
Output
Sample Input
3
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
Sample Output
no significant commonalities
AGATAC
CATCATCAT
Hint
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
using namespace std;
int main()
{
char s[][],p[],q[];
int m,n,i,j,k,len;
scanf("%d",&m);
while(m--)
{
q[]='\0';
scanf("%d",&n);
getchar();
for(i=; i<n; i++)
scanf("%s",s[i]);//输入不解释
len=strlen(s[]);
for(i=; i<len; i++ )
{
for(j=len-i; j>; j--)
{
strncpy(p,s[]+i,j);
p[j]='\0';
for(k=; k<n; k++)
if(strstr(s[k],p)==NULL)
break;
if(k==n)
{
if(strlen(p)==strlen(q)&&strcmp(p,q)<)//字符串分段复制函数
strcpy(q,p);
if(strlen(p)>strlen(q))
strcpy(q,p);
}
}
}
if(strlen(q)>)
puts(q);
else
puts("no significant commonalities");
}
return ;
}
Blue Jeans的更多相关文章
- poj 3080 Blue Jeans
点击打开链接 Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10243 Accepted: 434 ...
- POJ3080——Blue Jeans(暴力+字符串匹配)
Blue Jeans DescriptionThe Genographic Project is a research partnership between IBM and The National ...
- POJ 3080 Blue Jeans (字符串处理暴力枚举)
Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 21078 Accepted: ...
- POJ 3080 Blue Jeans(Java暴力)
Blue Jeans [题目链接]Blue Jeans [题目类型]Java暴力 &题意: 就是求k个长度为60的字符串的最长连续公共子串,2<=k<=10 规定: 1. 最长公共 ...
- poj3080 Blue Jeans【KMP】【暴力】
Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions:21746 Accepted: 9653 Descri ...
- (字符串 KMP)Blue Jeans -- POJ -- 3080:
链接: http://poj.org/problem?id=3080 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88230#probl ...
- POJ 3080 Blue Jeans 找最长公共子串(暴力模拟+KMP匹配)
Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20966 Accepted: 9279 Descr ...
- POJ Blue Jeans [枚举+KMP]
传送门 F - Blue Jeans Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u ...
- POJ3080 Blue Jeans —— 暴力枚举 + KMP / strstr()
题目链接:https://vjudge.net/problem/POJ-3080 Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total ...
- POJ 3080 Blue Jeans (求最长公共字符串)
POJ 3080 Blue Jeans (求最长公共字符串) Description The Genographic Project is a research partnership between ...
随机推荐
- JAVA学习第四十七课 — IO流(一):文件的读写
输入流和输出流相对于内存 将外部设备的数据读取到内存中:输入 将内存中的数据写入外部设备中:输出 IO流经常使用基类 字节流的抽象基类:InputStream,OutputStream 字符的抽象基类 ...
- Map的迭代操作
Map的迭代操作 public static void main(String[] args) { Map<String, List<Integer>> map = new H ...
- android开发之service详解
service作为android的四大组件之一,其重要性可想而知,在开发中,我们经常把一些不需要与用户进行交互的工作放在service中来完成,service运行在后台,这样有些人可能会产生错觉,以为 ...
- 第六篇:python高级之网络编程
python高级之网络编程 python高级之网络编程 本节内容 网络通信概念 socket编程 socket模块一些方法 聊天socket实现 远程执行命令及上传文件 socketserver及 ...
- rabbitmq 消息持久化之receive and send
二: 任务分发 &消息持久化 启用多个接收端的时候如果某一个receive 关闭要保证消息有反馈是否收到 send端 #-*- coding: UTF-8 -*-import pika ...
- java 字符串转int
//字符转整形 String aa = "23"; //int bb = Integer.parseInt(aa);//两种方式都是可以的 int bb = new Integer ...
- LTP 分词算法实践
参考链接: https://github.com/HIT-SCIR/ltp/blob/master/doc/install.rst http://www.xfyun.cn/index.php/serv ...
- Android四大组件之一:ContentProvider(内容提供者)
Android中还提供了名为ContentProvider(内容提供者),可以向其他应用提供数据,但不常用,除非是同一公司开发的App,可以向不同应用提供数据.虽然为Android的四大组件之一,但用 ...
- Hessian(C#)介绍及使用说明
什么是Hessian? Hessian是Caucho开发的一种二进制Web Service协议.支持目前所有流行的开发平台. Hessia能干什么? hessian用来实现web服务. Hessia有 ...
- 关于ios8斯坦福公开课第二课
在这个课程中,我们遇到了这样的代码 @IBAction func oprate(sender: UIButton) { let opration = sender.currentTitle! if u ...