POJ3080: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
找出n个串中最长的公共串,并且要求字典序最大
直接枚举第一串的所有子串,然后与后面的所有串进行比较即可
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; char str[15][65];
char sub[65];
char ans[65];
int len; int main()
{
int t,i,j,k,n,maxn;
scanf("%d",&t);
while(t--)
{
maxn = 0;
scanf("%d",&n);
memset(ans,'\0',sizeof(ans));
for(i = 1; i<=n; i++)
scanf("%s",str[i]);
for(i = 1; i<=60; i++)
{
int find = 0;
for(j = 0; j<=60-i; j++)
{
len = 0;
int check = 1;
for(k = j;;k++)
{
sub[len++] = str[1][k];
if(len == i)
break;
}
sub[len] = '\0';
for(k = 2; k<=n; k++)
{
if(!strstr(str[k],sub))
{
check = 0;
break;
}
}
if(check)
{
find = 1;
if(strlen(ans)<strlen(sub))
strcpy(ans,sub);
else if(strcmp(ans,sub)>0)
strcpy(ans,sub);
}
}
if(!find)
break;
}
if(strlen(ans)<3)
printf("no significant commonalities\n");
else
printf("%s\n",ans);
} return 0;
}
POJ3080:Blue Jeans的更多相关文章
- 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 ...
- POJ3080——Blue Jeans(暴力+字符串匹配)
Blue Jeans DescriptionThe Genographic Project is a research partnership between IBM and The National ...
- poj3080 Blue Jeans【KMP】【暴力】
Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions:21746 Accepted: 9653 Descri ...
- 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 ...
- (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 ...
- 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
- call()与apply()区别
一.方法的定义 call方法: 语法:call(thisObj,Object)定义:调用一个对象的一个方法,以另一个对象替换当前对象.说明:call 方法可以用来代替另一个对象调用一个方法.call ...
- C#中创建、打开、读取、写入、保存Excel的一般性代码
---转载:http://hi.baidu.com/zhaocbo/item/e840bcf941932d15fe358228 1. Excel对象微软的Excel对象模型包括了128个不同的对象,从 ...
- Mac OS X 10.9 Mavericks 修改root密码
Mac10.9忘记密码后有两种方式可以进去: 代码如下 复制代码 1.sudo passwd 重新输入密码即可,此方法修改了root的密码 代码如下 复制代码 2.sudo bash 输入当前用户 ...
- CRT detected that the application wrote to memory after end of heap buffer.
很多人的解释都不一样, 我碰到的问题是,开辟的内存空间小于操作的内存空间.也就是说,我free的内存越界了. 这是我开辟链表结构体内存的代码: PNODE Create() { int len; / ...
- [技术翻译]Guava-libraries(一): 用户指导
用户指导 本文翻译自http://code.google.com/p/guava-libraries/wiki/GuavaExplained,由十八子将翻译,发表于博客园 http://www.cnb ...
- Wireshark抓包、过滤器
查阅于http://blog.sina.com.cn/s/blog_5d527ff00100dwph.html 1.捕捉过滤器 设置捕捉过滤器的步骤是:- 选择 capture -> optio ...
- [LeetCode OJ] Single Number之一 ——Given an array of integers, every element appears twice except for one. Find that single one.
class Solution { public: int singleNumber(int A[], int n) { int i,j; ; i<n; i++) { ; j<n; j++) ...
- centos 下搭建 php环境(2) mysql 安装
CentOS下的MySQL 5.1安装 01 1.下载源码包 wget http://mysql.llarian.net/Downloads/MySQL-5.1/mysql-5.1.63.tar. ...
- python对比两个文件问题
写一个比较两个文本文件的程序. 如果不同, 给出第一个不同处的行号和 列号. 比较的时候可以使用zip()函数 a=open('test.txt','r') b=open('test2.txt','r ...
- mybatis框架搭建学习初步
mybatis框架搭建步骤:1. 拷贝jar到lib目录下,而且添加到工程中2. 创建mybatis-config.xml文件,配置数据库连接信息 <environments default=& ...