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2015 Open Bronze Problem 2 Bessie Gets Even
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== Official Problem Statement == [http://www.usaco.org/index.php?page=viewproblem2&cpid=546 Bessie Gets Even] == Problem == In the problem "Bessie Gets Even" (USACO 2015 Open Bronze, Problem 2), Bessie the cow is given a list of numbers and is trying to count the number of ways she can select a subset of those numbers such that the sum of the numbers in the subset is even. The list of numbers Bessie has been given are guaranteed to be between 1 and 10, inclusive. The list may have up to 20 numbers, and each number in the list can appear multiple times. == Solution == This problem can be solved using a technique known as dynamic programming. Firstly, for each number between 1 to 10, we count the frequency of its occurrence in the list. Let's denote this frequency as freq[i]. We can then create a 2D dynamic programming table, dp[i][j], where dp[i][j] represents the number of ways to make a sum of j using the first i numbers. We initially set dp[0][0] to 1 since there is only one way to create a sum of 0, that is by taking no numbers at all. Then, for each number i from 1 to 10, we perform the following steps: * We create a temporary copy of the DP table. * We then iterate through all possible sums j from 0 to 20. For each sum, we consider adding k copies of the current number i, where k can range from 0 to freq[i]. For each valid k, we add the number of ways to form the sum j - k * i (from the previous set of numbers) to dp[i][j]. * After we've finished processing the current number i, we copy the temporary DP table back to the original one. Finally, the answer to the problem is dp[10][0], since we want to count the number of ways to form an even sum (which is 0 modulo 2). == Code == === C++ === <pre> #include <iostream> #include <cstdio> using namespace std; int num[256][2]; bool is_even(int x) { return x % 2 == 0; } int main() { freopen("geteven.in", "r", stdin); freopen("geteven.out", "w", stdout); int N; cin >> N; for (int i = 0; i < N; i++) { char letter; int val; cin >> letter >> val; num[letter][val % 2]++; } int result = 0; for(int B = 0; B < 2; B++) for(int E = 0; E < 2; E++) for(int S = 0; S < 2; S++) for(int I = 0; I < 2; I++) for(int G = 0; G < 2; G++) for(int O = 0; O < 2; O++) for(int M = 0; M < 2; M++) { int expression = (B + E + S + S + I + E) * (G + O + E + S) * (M + O + O); if (is_even(expression)) { result += num['B'][B] * num['E'][E] * num['S'][S] * num['I'][I] * num['G'][G] * num['O'][O] * num['M'][M]; } } cout << result << endl; return 0; } </pre> === Java === <pre> import java.util.*; import java.io.*; public class BessieGetsEven { public static void main(String[] args) throws IOException { BufferedReader in = new BufferedReader(new FileReader("geteven.in")); PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("geteven.out"))); int N = Integer.parseInt(in.readLine()); int[][] num = new int[256][2]; for (int i = 0; i < N; i++) { StringTokenizer st = new StringTokenizer(in.readLine()); char letter = st.nextToken().charAt(0); int val = Integer.parseInt(st.nextToken()); num[letter][val % 2]++; } int result = 0; for(int B = 0; B < 2; B++) for(int E = 0; E < 2; E++) for(int S = 0; S < 2; S++) for(int I = 0; I < 2; I++) for(int G = 0; G < 2; G++) for(int O = 0; O < 2; O++) for(int M = 0; M < 2; M++) { int expression = (B + E + S + S + I + E) * (G + O + E + S) * (M + O + O); if (expression % 2 == 0) { result += num['B'][B] * num['E'][E] * num['S'][S] * num['I'][I] * num['G'][G] * num['O'][O] * num['M'][M]; } } out.println(result); out.close(); } } </pre> === Python === <pre> def is_even(x): return x % 2 == 0 num = {c: [0, 0] for c in 'BESIGOM'} with open('geteven.in', 'r') as fin, open('geteven.out', 'w') as fout: N = int(fin.readline().strip()) for _ in range(N): letter, val = fin.readline().strip().split() val = int(val) if is_even(val): num[letter][0] += 1 else: num[letter][1] += 1 result = 0 for B in range(2): for E in range(2): for S in range(2): for I in range(2): for G in range(2): for O in range(2): for M in range(2): if is_even((B + E + S + S + I + E) * (G + O + E + S) * (M + O + O)): result += num['B'][B] * num['E'][E] * num['S'][S] * num['I'][I] * num['G'][G] * num['O'][O] * num['M'][M] fout.write(f"{result}\n") </pre> [[Category:Yearly_2014_2015]] [[Category:Bronze]] [[Category:Arrays]] [[Category:Counting]] [[Category:Brute Force]]
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