POJ - 3080 Blue Jeans
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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
- 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
用头文件《cstring》里面的一些库函数即可。
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 6 using namespace std; 7 8 char s[15][65]; 9 char goal[65],now[65]; 10 11 int main() 12 { 13 int T,n; 14 scanf("%d",&T); 15 while(T--) 16 { 17 //int flag=0; 18 memset(goal,‘\0‘,sizeof(goal)); 19 memset(now,‘\0‘,sizeof(now)); 20 memset(s,‘\0‘,sizeof(s)); 21 scanf("%d",&n); 22 for(int i=0;i<n;i++) 23 scanf("%s",s[i]); 24 for(int i=1;i<=60;i++) 25 { 26 for(int j=0;j<=60-i;j++) 27 { 28 int len =0; 29 int check=1; 30 for(int k=j;;k++) 31 { 32 now[len++]=s[0][k]; 33 if(len==i) 34 break; 35 } 36 now[len]=‘\0‘; 37 for(int k=1;k<n;k++) 38 if(!strstr(s[k],now)) 39 { 40 check=0; 41 break; 42 } 43 if(check==1) 44 { 45 if(strlen(goal)<strlen(now)) 46 strcpy(goal,now); 47 else if(strcmp(goal,now)>0) 48 strcpy(goal,now); 49 } 50 } 51 } 52 if(strlen(goal)<3) 53 printf("no significant commonalities\n"); 54 else 55 printf("%s\n",goal); 56 } 57 58 return 0; 59 }
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