zoukankan      html  css  js  c++  java
  • Blue Jeans

     Blue Jeans
    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    Description

    The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. 

    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

    Input to this problem will begin with a line containing a single integer n indicating the number of datasets. Each dataset consists of the following components:
    • 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

    For each dataset in the input, output the longest base subsequence common to all of the given base sequences. If the longest common subsequence is less than three bases in length, display the string "no significant commonalities" instead. If multiple subsequences of the same longest length exist, output only the subsequence that comes first in alphabetical order.

    Sample Input

    3
    2
    GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    3
    GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
    GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
    GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
    3
    CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
    ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

    Sample Output

    no significant commonalities
    AGATAC
    CATCATCAT
    #include<iostream>
    #include <string>
    using namespace std;
    string a[1000];
    int main()
    {int t,i,n,j,k;
    cin>>t;
    	while(t--)
    	{cin>>n;
    		string b="";
    		for(i=0;i<n;i++)
    	cin>>a[i];
    
    	for(i=0;i<=a[0].size();i++)
    	{
    		for(j=0;j<=a[0].size();j++)
    		{bool p=0;
    			string temp = a[0].substr(j, i);
    		for(k=1;k<n;k++)
    		{if (a[k].find(temp) == string::npos)
    			{p=1;
    			break;}
    		}
    	if(p==0)
    	if(temp.length()>b.length())b=temp;
    else if(temp.length()==b.length()&&b>temp)b=temp;
    		}
    
    
    	}
    
    if(b.size()<3)
    		printf("no significant commonalities
    ");
    	else
            cout<<b<<endl;
    
    
    
    
    
    	}
    
    
    return 0;}
    


  • 相关阅读:
    [开发笔记]-使用bat命令来快速安装和卸载Service服务
    [开发笔记]-多线程异步操作如何访问HttpContext?
    [开发笔记]-Windows Service服务相关注意事项
    [开发笔记]-VS2012打开解决方案崩溃或点击项目崩溃
    Chrome 开发者工具有了设备模拟器
    Mysql查看数据库表容量大小
    golang操作mysql数据库
    golang命令和VSCode配置
    golang广度优先算法-走迷宫
    golang爬取免费代理IP
  • 原文地址:https://www.cnblogs.com/cynchanpin/p/6971459.html
Copyright © 2011-2022 走看看