USACO 2017 US Open Contest, Bronze Problem 2. Bovine Genomics

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USACO2017-OPEN-B2

答案:

#include <iostream>
#include <fstream>
#include <cmath>
using namespace std;

int N, M;
string spotty[100], plain[100];

bool test_location(int j)
{
bool found_cow[2][4] = {0};
// found_cow[0] refers to spotty cows, and found_cow[1]
// refers to non-spotty cows.
for (int i=0; i<N; i++) {
if (spotty[i][j] == 'A') found_cow[0][0] = true;
if (spotty[i][j] == 'C') found_cow[0][1] = true;
if (spotty[i][j] == 'G') found_cow[0][2] = true;
if (spotty[i][j] == 'T') found_cow[0][3] = true;
}
for (int i=0; i<N; i++) {
if (plain[i][j] == 'A') found_cow[1][0] = true;
if (plain[i][j] == 'C') found_cow[1][1] = true;
if (plain[i][j] == 'G') found_cow[1][2] = true;
if (plain[i][j] == 'T') found_cow[1][3] = true;
}
for (int i = 0; i < 4; ++i) {
if (found_cow[0][i] && found_cow[1][i]) return false;
}
return true;
}

int main(void)
{
ifstream fin ("cownomics.in");
ofstream fout ("cownomics.out");
fin >> N >> M;
for (int i=0; i<N; i++) fin >> spotty[i];
for (int i=0; i<N; i++) fin >> plain[i];
int answer = 0;
for (int j=0; j<M; j++)
if (test_location(j)) answer++;
fout << answer << "\n";
return 0;
}

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