Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 10083   Accepted: 4262

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 题意:寻找m个序列中共有的最长的子串,长度小于3的输出no significant commonalities,长度相等时输出字典序小的;
解题思路:枚举第一个序列的所有子串,每枚举出一个检查该子串是否是m个序列共有的,若是,比较该串与ans[]的长度取较长的,若长度相等取字典序小的;当枚举完所有子串后,ans[]保存的就是m个序列共有的最长的串。
 #include<stdio.h>
#include<string.h> int main()
{
int i,j,k,n,t;
char DNA[][],tmp[],ans[];
scanf("%d",&t);
while(t--)
{
ans[] = '\0';
scanf("%d",&n);
for(i = ; i < n; i++)
scanf("%s",DNA[i]);
for(i = ; i < ; i++)//枚举子串的起点
{
for(j = i+; j < ; j++)//枚举子串的终点
{
int cnt = ;
for(k = i; k <= j; k++)
{
tmp[cnt++] = DNA[][k];
}
tmp[cnt] = '\0';//得到一个子串
for(k = ; k < n; k++)
{
if(strstr(DNA[k],tmp) == NULL)
break;
}
if(k < n) continue;
if(strlen(ans) == strlen(tmp))
{
if(strcmp(ans,tmp) > )
strcpy(ans,tmp);
}
else
{
if(strlen(tmp) > strlen(ans))
strcpy(ans,tmp);
}
}
}
if(strlen(ans) < ) printf("no significant commonalities\n");
else printf("%s\n",ans);
}
return ;
}

 

Blue Jeans(串)的更多相关文章

  1. POJ 3080 Blue Jeans(串)

    题目网址:http://poj.org/problem?id=3080 思路: 以第一个DNA序列s为参考序列,开始做以下的操作. 1.将一个字母s[i]作为匹配串.(i为当前遍历到的下标) 2.遍历 ...

  2. 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 ...

  3. poj3080 Blue Jeans【KMP】【暴力】

    Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:21746   Accepted: 9653 Descri ...

  4. (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 ...

  5. 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 ...

  6. POJ Blue Jeans [枚举+KMP]

    传送门 F - Blue Jeans Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u ...

  7. 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 ...

  8. poj 3080 Blue Jeans

    点击打开链接 Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10243   Accepted: 434 ...

  9. POJ3080——Blue Jeans(暴力+字符串匹配)

    Blue Jeans DescriptionThe Genographic Project is a research partnership between IBM and The National ...

随机推荐

  1. iOS中__block 关键字的底层实现原理

    在 <iOS面试题集锦(附答案)> 中有这样一道题目: 在block内如何修改block外部变量?(38题)答案如下: 默认情况下,在block中访问的外部变量是复制过去的,即:写操作不对 ...

  2. java文件处理工具类

    import java.io.BufferedInputStream; import java.io.BufferedOutputStream; import java.io.BufferedRead ...

  3. unlocker208安装之后看不到Apple macos选项,解决办法.

    前段时间升级了win10,最新的unlocker208安装以后看不到mac os的选项,为什么呢?我们在管理员允许win-install.cmd的过程中,会在cmd中看到这么一句话:LookupErr ...

  4. Weex 初始

    1.一旦数据和模板绑定,数据的变化会立即体现在前台的变化 <template> <container> <text style="font-size: {{si ...

  5. IK分词算法设计总结

    IK分词算法设计思考 加载词典 IK分词算法初始化时加载了“敏感词”.“主词典”.“停词”.“量词”,如果这些词语的数量很多,怎么保证加载的时候内存不溢出 分词缓冲区 在分词缓冲区中进行分词操作,怎么 ...

  6. (转)Smarty Foreach 使用说明

    foreach 是除 section 之外处理循环的另一种方案(根据不同需要选择不同的方案). foreach 用于处理简单数组(数组中的元素的类型一致),它的格式比 section 简单许多,缺点是 ...

  7. Activity和Fragment生命周期变化

    情形一:启动应用加载Activity和Fragment Activity::onCreate Fragment::onAttach Fragment::onCreate Fragment::onCre ...

  8. Windows7添加SSD硬盘后重启卡住正在启动

    楼主办公电脑,原来只配置了一块机械硬盘,用着总很不顺心,于是说服领导给加了块SSD固态硬盘. 操作如下: 1.在PE下分区格式化新固态硬盘,将原来机械硬盘的C盘GHOST备份后还原到新固态硬盘: 2. ...

  9. 分享最近写的 两条sql语句

    1. 搭建基本环境 插入测试数据 insert into jgdm (jgdm,jgmc)  values('12300000000','河南省');insert into jgdm (jgdm,jg ...

  10. ios8及以前的特性

    目前最新系统为ios8.以下为历代系统的回顾: iOS 1 关键词:iPhone的诞生 也许放在现在来看,当时的情景很难想象.当第一代iPhone正式发布时,在某些功能和方面其实是要远远落后于当时的竞 ...