Hueestory

#include #include #include using namespace std;bool cmp(pair &a, pair &b) { if (a.first == b.first) return a.second b.first;}vector solution(int N, vector stages) { vector answer(N); vector> rate; for (int i = 1; i

#include #include #include #include using namespace std;bool visit[201][201][201];int A, B, C;vector ans;void BFS();int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cin >> A >> B >> C; BFS(); sort(ans.begin(), ans.end()); for (int i : ans) cout , int>> Q; Q.push(make_pair(make_pair(0, 0), C)); while (!Q.empty()) { int now_a = Q.front().fi..

#include #include #include #include #define X 1#define Y 2using namespace std;int v, e;vector> A;vector visited;void DFS(int start, int color);bool isBipartite();int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int k; cin >> k; // k 만큼 반복 for (int i = 0; i > v >> e; A.resize(v + 1); visited.resize(v + 1, 0); // 에지의 개수 만큼 반복 fo..

#include #include #include #include using namespace std;vector> A;vector visited;vector cnt;int max_visited = 0;void BFS (int node);int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n, m; cin >> n >> m; A.resize(n + 1); visited.resize(n + 1); cnt.resize(n + 1); for (int i = 0; i > a >> b; A[a].push_back(b); } for (int i = 0; i ..

#include #include #include #include using namespace std;vector> A;vector cnt;vector visited;void BFS(int node);int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n, m, k, x; cin >> n >> m >> k >> x; A.resize(n + 1); for (int i = 0; i > a >> b; A[a].push_back(b); } visited.resize(n + 1); for (int i = 0; i Q; Q.push(node); while (!..

#include #include #include using namespace std;typedef long long ll;vector> A[10];ll gcd(ll a, ll b);void DFS(int node);ll ratio[10];bool visited[10];ll lcm;int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n; cin >> n; lcm = 1; for (int i = 0; i > a >> b >> p >> q; A[a].push_back(make_tuple(b, p, q)); A[b].push_back(make_tuple(a, q, p))..

#include #include using namespace std;typedef long long ll;ll gcd(ll a, ll b) { if (a % b == 0) return b; else return gcd(b, a % b);}int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); ll a, b; cin >> a >> b; ll c; if (a >= b) c = gcd(a, b); else if (a  1. 1로 이루어진 수라고 해도, 결국 최대공약수를 구하는 문제이다ex) a = 6, b = 3인 경우6 % 3 == 0, GCD = 3111111 % 111 == 0, G..

#include using namespace std;int gcd(int a, int b);int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int t; cin >> t; int a, b; for (int i = 0; i > a >> b; if(a >= b) cout  1. 재귀함수를 이용해 최대공약수를 구하고, a * b를 최대공약수로 나누면 최소공배수가 된다