USACO 2017 US Open Contest, Bronze Problem 3. Modern Art
原题下载 USACO2017-OPEN-B3 答案: #include <iostream> #include <fstream> #include <cmath> using namespace std; #define MAX_N 10 int N, B[MAX_N][MAX_N]; bool color_appears(int c) { ...
USACO 2017 US Open Contest, Bronze Problem 2. Bovine Genomics
原题下载 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) {...
USACO 2017 US Open Contest, Bronze Problem 1. The Lost Cow
原题下载 USACO2017-OPEN-B1 答案: #include <iostream> #include <cstdlib> using namespace std; typedef long long ll; int main() { ll x, y; cin >> x >> y; ll ans = 0; ll by = 1...
USACO 2017 US Open Contest, Silver Problem 3. Where's Bessie?
原题下载 USACO2017-OPEN-S3 答案: #include <iostream> #include <fstream> #include <cmath> #include <vector> using namespace std; int N; string img[20]; struct PCL { int i1,j1...
USACO 2016 US Open Contest, Platinum Problem 3. Landscaping
原题下载： USACO2016-OPEN-P3 答案： (Analysis by Brian Dean) Letting K=10K=10 be the maximum amount of dirt in any cell, we can ultimately solve this problem using dynamic programming in O(NK)O(NK)...
USACO 2016 US Open Contest, Platinum Problem 2. Bull in a China Shop
原题下载： USACO2016-OPEN-P2 答案： Here's my solution to this problem annotated with what each section of code aims to accomplish. #include #include #include #include #include #include #include #i...
USACO 2017 US Open Contest, Silver Problem 2. Bovine Genomics
原题下载 USACO2017-OPEN-S2 答案: #include <iostream> #include <fstream> #include <cmath> using namespace std; int N, M; string spotty[500], plain[500]; int S[500][50], P[500][50],...
USACO 2017 US Open Contest, Silver Problem 1. Paired Up
原题下载 USACO2017-OPEN-S1 答案: #include <iostream> #include <fstream> #include <cmath> #include <algorithm> #include <vector> using namespace std; typedef pair<in...
USACO 2016 US Open Contest, Platinum Problem 1. 262144
答案： Here's my solution to this problem. #include #include #include using namespace std; #define MAXN ((1 << 18) + 10) #define MAXSZ 70 int dp[MAXSZ + 1][MAXN]; int A[MAXN]; int main() { in...
