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UVA - 1368 DNA Consensus String
DNA Consensus String
Description Figure 1.
DNA (Deoxyribonucleic Acid) is the molecule which contains the genetic instructions. It consists of four different nucleotides, namely Adenine, Thymine, Guanine, and Cytosine as shown in Figure 1. If we represent a nucleotide by its initial character, a DNA
strand can be regarded as a long string (sequence of characters) consisting of the four characters A, T, G, and C. For example, assume we are given some part of a DNA strand which is composed of the following sequence of nucleotides:
Figure 2. DNA sequences of gene X in five animals.
InputYour program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing two integers m and n which are separated by a single space. The integer m(4m50) represents the number of DNA sequences and n(4n1000) represents the length of the DNA sequences, respectively. In each of the next m lines, each DNA sequence is given.OutputYour program is to write to standard output. Print the consensus string in the first line of each case and the consensus error in the second line of each case. If there exists more than one consensus string, print the lexicographically smallest consensus string. The following shows sample input and output for three test cases.Sample Input3 5 8 TATGATAC TAAGCTAC AAAGATCC TGAGATAC TAAGATGT 4 10 ACGTACGTAC CCGTACGTAG GCGTACGTAT TCGTACGTAA 6 10 ATGTTACCAT AAGTTACGAT AACAAAGCAA AAGTTACCTT AAGTTACCAA TACTTACCAA Sample OutputTAAGATAC 7 ACGTACGTAA 6 AAGTTACCAA 12 #include<iostream> #include<cstring> #include<string> #include<cstdio> using namespace std; char s[100][1005]; int main() { int casen; cin >> casen; while (casen--) { int m, n; cin >> m >> n; for (int i = 0; i < m; i++) cin >> s[i]; int A, C, G, T; char ans[1005]; for (int i = 0; i < n; i++) { A = C = G = T = 0; for (int j = 0; j < m; j++) { switch (s[j][i]) { case ‘A‘: A++; break; case ‘C‘: C++; break; case ‘G‘: G++; break; case ‘T‘: T++; break; } } if (A >= C && A >= G && A >= T) ans[i] = ‘A‘; else if (C > A && C >= G && C >= T) ans[i] = ‘C‘; else if (G > C && G > A && G >= T) ans[i] = ‘G‘; else if (T > G && T > A && T > C) ans[i] = ‘T‘; } ans[n] = ‘\0‘; int sum = 0; for (int i = 0; i < m;i++) for (int j = 0; j < n;j++) if (s[i][j] != ans[j]) sum++; cout << ans << endl; cout << sum << endl; } } |
UVA - 1368 DNA Consensus String