DNA Consensus String UVA - 1368 模拟找规律
2017-10-19 20:04
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DNA Consensus String
Time limit: 3.000 secondsFigure 1.DNA (Deoxyribonucleic Acid) isthe 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 longstring (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:``Thymine-Adenine-Adenine-Cytosine-Thymine-Guanine-Cytosine-Cytosine-Guanine-Adenine-Thymine"Then we can represent the above DNA strand with the string ``TAACTGCCGAT." The biologist Prof. Ahn found that a gene X commonly exists in the DNA strands of five different kinds of animals, namely dogs, cats, horses, cows, and monkeys. He also discovered thatthe DNA sequences of the gene X from each animal were very alike. See Figure 2.DNA sequence of gene X | |
4000Cat: | GCATATGGCTGTGCA |
Dog: | GCAAATGGCTGTGCA |
Horse: | GCTAATGGGTGTCCA |
Cow: | GCAAATGGCTGTGCA |
Monkey: | GCAAATCGGTGAGCA |
Input
Your program is to read from standard input. The input consists of T testcases. The number of test cases T isgiven in the first line of the input. Each test case starts with a line containing two integers m and n whichare separated by a single space. The integer m (4m50) representsthe number of DNA sequences and n (4n1000) representsthe length of the DNA sequences, respectively. In each of the next m lines,each DNA sequence is given.Output
Your 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 lineof 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.SampleInput
3 5 8 TATGATAC TAAGCTAC AAAGATCC TGAGATAC TAAGATGT 4 10 ACGTACGTAC CCGTACGTAG GCGTACGTAT TCGTACGTAA 6 10 ATGTTACCAT AAGTTACGAT AACAAAGCAA AAGTTACCTT AAGTTACCAA TACTTACCAA
SampleOutput
TAAGATAC 7 ACGTACGTAA 6 AAGTTACCAA 12
题意:输入若干长度相同的字符串,然后找出一个与它们长度相同的字符串,要求这个字符串和输入的所有的字符串,不相同的字符个数最小,然后把这个字符输出,并且把不相同的字符个数也输出;
思路:先找符合的字符串,字符串就是每一竖行出现最多的那个字符;
代码如下:
//DNA Consensus String UVA - 1368#include <stdio.h>#include <string.h>char c[55][1005];char s[1005];int a[5];int main() {int N , sum;int n , m , t , p;scanf("%d",&N);while(N--) {p = 0;sum = 0;scanf("%d %d",&n,&m);for(int i = 0; i < n; i++) {scanf("%s",c[i]);}for(int i = 0; i < m; i++) {memset(a , 0 , sizeof(a));for(int j = 0; j < n; j++) {if(c[j][i] == 'A')a[1]++;else if(c[j][i] == 'C')a[2]++;else if(c[j][i] == 'G')a[3]++;elsea[4]++;}for(int i = 1; i <= 4; i++) {if(a[0] < a[i]) {a[0] = a[i];t = i;}}if(t == 1) s[p++] = 'A';else if(t == 2) s[p++] = 'C';else if(t == 3) s[p++] = 'G';else if(t == 4) s[p++] = 'T';}for(int i = 0; i < n; i++) {for(int j = 0; j < m; j++) {if(c[i][j] != s[j]) {sum++;}}}for(int i = 0; i < m; i++)printf("%c",s[i]);printf("\n");printf("%d\n",sum);}return 0;}
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