POJ训练计划3080_Blue Jeans(串处理/暴力)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11542 | Accepted: 4962 |
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
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
Source
- #include <iostream>
- #include <cstdio>
- #include <cstring>
- #include <algorithm>
- using namespace std;
- char str[12][100],ch[100],ans[100];
- int main()
- {
- int t,i,j,k,n,maxl,f;
- cin>>t;
- while(t--)
- {
- memset(str,0,sizeof(str));
- memset(ans,0,sizeof(ans));
- cin>>n;
- maxl=0,f=0;
- for(i=0; i<n; i++)
- cin>>str[i];
- for(i=0; i<58; i++)
- {
- for(j=3; j<=60-i; j++)
- {
- strncpy(ch,str[0]+i,j);
- ch[j]=0;
- f=0;
- for(k=1; k<n; k++)
- {
- if(strstr(str[k],ch)==NULL)
- {
- f=1;
- break;
- }
- }
- if(!f)
- {
- if(j>maxl)
- {
- maxl=j;
- strcpy(ans,ch);
- }
- else if(j==maxl)
- {
- if(strcmp(ch,ans)<0)
- strcpy(ans,ch);
- }
- }
- }
- }
- if(maxl)
- cout<<ans<<endl;
- else cout<<"no significant commonalities"<<endl;
- }
- return 0;
- }
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