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NatanGarciasNatanGarcias
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more training
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‎Atcoder/AtCoder Regular Contest 159/A.cpp

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#include<bits/stdc++.h>
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using namespace std;
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template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { os << "{"; for (typename vector<T>::const_iterator vi = v.begin(); vi != v.end(); ++vi) { if (vi != v.begin()) os << ", "; os << *vi; } os << "}"; return os; }
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template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { os << '(' << p.first << ", " << p.second << ')'; return os; }
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typedef long long ll;
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typedef long double ld;
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typedef pair<int,int> pii;
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#define optimize ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
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#define endl "\n"
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#define fi first
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#define se second
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#define pb push_back
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#define all(x) x.begin(),x.end()
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#define ms(x,a) memset(x,a,sizeof(x))
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#define INF 0x3f3f3f3f
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#define INFLL 0x3f3f3f3f3f3f3f3f
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#define mod 1000000007LL
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#define MAXN 110
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mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
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//#define _DEBUG
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// #ifdef _DEBUG
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// #endif
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/* -------------------------------- Solution starts below -------------------------------- */
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ll T,N,M,K;
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int A[MAXN][MAXN];
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int floyd[MAXN][MAXN];
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void solve() {
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cin >> N >> M;
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ms(floyd, 0x3f);
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for(int i = 0; i < N; i++) {
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for(int j = 0; j < N; j++) {
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cin >> A[i][j];
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if(A[i][j]) floyd[i][j] = 1;
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}
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}
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for(int k = 0; k < N; k++) {
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for(int i = 0; i < N; i++) {
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for(int j = 0; j < N; j++) {
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floyd[i][j] = min(floyd[i][j], floyd[i][k] + floyd[k][j]);
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}
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}
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}
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cin >> K;
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for(ll i = 0, u, v; i < K; i++) {
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cin >> u >> v;
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u--, v--;
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if(u == v) cout << 0 << endl;
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else {
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u = u % N;
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v = v % N;
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if(floyd[u][v] == INF) cout << -1 << endl;
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else cout << floyd[u][v] << endl;
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}
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}
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}
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int main() {
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optimize;
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T = 1;
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//cin >> T;
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while(T--) {
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solve();
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}
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return 0;
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}

‎Atcoder/AtCoder Regular Contest 159/B.cpp

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#include<bits/stdc++.h>
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using namespace std;
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template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { os << "{"; for (typename vector<T>::const_iterator vi = v.begin(); vi != v.end(); ++vi) { if (vi != v.begin()) os << ", "; os << *vi; } os << "}"; return os; }
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template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { os << '(' << p.first << ", " << p.second << ')'; return os; }
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typedef long long ll;
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typedef long double ld;
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typedef pair<int,int> pii;
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#define optimize ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
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#define endl "\n"
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#define fi first
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#define se second
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#define pb push_back
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#define all(x) x.begin(),x.end()
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#define ms(x,a) memset(x,a,sizeof(x))
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#define INF 0x3f3f3f3f
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#define INFLL 0x3f3f3f3f3f3f3f3f
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#define mod 1000000007LL
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#define MAXN 200010
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mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
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//#define _DEBUG
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// #ifdef _DEBUG
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// #endif
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/* -------------------------------- Solution starts below -------------------------------- */
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ll T,N,M,K;
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void solve() {
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cin >> N >> M;
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ll ans = 0;
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while(N && M) {
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ll g = __gcd(N, M);
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if(g > 1) {
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ans += 1;
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N = N / g - 1;
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M = M / g - 1;
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continue;
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}
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if(N < M) swap(N, M);
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ll d = N - M;
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ll b = 0;
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for(ll i = 1; i * i <= d; i++) {
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if(d % i != 0) continue;
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// para todos os divisores de i
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ll x = d / i;
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// maior valor de x < N que tenha o gcd diferente de 1 com d
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// dado que N > M e N = M + d
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// gcd(N, M) = gcd(d, M)
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if((M / i) * i < M) b = max(b, (M / i) * i);
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if((M / x) * x < M) b = max(b, (M / x) * x);
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}
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ans += M - b;
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N = b + d;
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M = b;
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}
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cout << ans << endl;
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}
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int main() {
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optimize;
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T = 1;
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while(T--) {
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solve();
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}
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return 0;
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}

‎Atcoder/AtCoder Regular Contest 160/A.cpp

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#include<bits/stdc++.h>
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using namespace std;
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template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { os << "{"; for (typename vector<T>::const_iterator vi = v.begin(); vi != v.end(); ++vi) { if (vi != v.begin()) os << ", "; os << *vi; } os << "}"; return os; }
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template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { os << '(' << p.first << ", " << p.second << ')'; return os; }
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typedef long long ll;
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typedef long double ld;
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typedef pair<int,int> pii;
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#define optimize ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
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#define endl "\n"
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#define fi first
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#define se second
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#define pb push_back
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#define all(x) x.begin(),x.end()
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#define ms(x,a) memset(x,a,sizeof(x))
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#define INF 0x3f3f3f3f
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#define INFLL 0x3f3f3f3f3f3f3f3f
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#define mod 1000000007LL
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#define MAXN 7070
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mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
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//#define _DEBUG
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// #ifdef _DEBUG
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// #endif
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/* -------------------------------- Solution starts below -------------------------------- */
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ll T,N,M,K;
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int A[MAXN];
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void f(int l, int r) {
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while(l < r) swap(A[l++], A[r--]);
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}
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void g() {
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vector<tuple<int,int,int,int>> m1, m2, m3;
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for(int i = 0 ; i < N; i++) {
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for(int j = i; j < N; j++) {
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if(A[i] == A[j]) m2.emplace_back(-i, A[j], -j, -A[i]);
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else if(A[i] < A[j]) m3.emplace_back(-i, A[j], -j, -A[i]);
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else m1.emplace_back( i, A[j], j, -A[i]);
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}
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}
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M--;
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if(M < m1.size()) {
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sort(m1.begin(), m1.end());
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int l = get<0> (m1[M]);
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int r = get<2> (m1[M]);
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f(l, r);
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}
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else if(M < m1.size() + m2.size()) {
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return;
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}
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else {
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sort(m3.begin(), m3.end());
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M -= m1.size() + m2.size();
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int l = -get<0> (m3[M]);
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int r = -get<2> (m3[M]);
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f(l, r);
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}
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}
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void solve() {
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cin >> N >> M;
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for(int i = 0; i < N; i++) cin >> A[i];
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g();
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for(int i = 0; i < N; i++) cout << A[i] << " \n"[i == N - 1];
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}
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int main() {
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optimize;
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T = 1;
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//cin >> T;
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while(T--) {
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solve();
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}
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return 0;
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}
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‎Atcoder/UNIQUE VISION Programming Contest 2023 Autumn(AtCoder Beginner Contest 323)/A.cpp

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#include<bits/stdc++.h>
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using namespace std;
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template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { os << "{"; for (typename vector<T>::const_iterator vi = v.begin(); vi != v.end(); ++vi) { if (vi != v.begin()) os << ", "; os << *vi; } os << "}"; return os; }
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template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { os << '(' << p.first << ", " << p.second << ')'; return os; }
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typedef long long ll;
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typedef long double ld;
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typedef pair<int,int> pii;
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#define optimize ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
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#define endl "\n"
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#define fi first
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#define se second
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#define pb push_back
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#define all(x) x.begin(),x.end()
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#define ms(x,a) memset(x,a,sizeof(x))
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#define INF 0x3f3f3f3f
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#define INFLL 0x3f3f3f3f3f3f3f3f
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#define mod 1000000007LL
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#define MAXN 200010
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mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
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//#define _DEBUG
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// #ifdef _DEBUG
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// #endif
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/* -------------------------------- Solution starts below -------------------------------- */
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ll T,N,M,K;
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void solve() {
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string s;
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cin >> s;
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int ans = 1;
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for(int i = 1; i <= 15; i+= 2) {
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if(s[i] != '0') {
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ans = 0;
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}
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}
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if(ans) cout << "Yes\n";
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else cout << "No\n";
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}
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int main() {
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optimize;
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T = 1;
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//cin >> T;
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while(T--) {
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solve();
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}
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return 0;
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}

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