2033A - Сакурако и Косукэ Idea: FBI

~~~~~ def solve(): n = int(input()) x = 0 c = 1 while -n <= x <= n: if c % 2 == 1: x -= 2 * c — 1 else: x += 2 * c — 1 c += 1 if c % 2 == 0: print("Sakurako") else: print("Kosuke") for tc in range(int(input())): solve() ~~~~~ 2033B - Сакурако и вода Idea: FBI ~~~~~ def solve(): n = int(input()) mn = dict() for i in range(n): a = [int(x) for x in input().split()] for j in range(n): mn[i — j] = min(a[j], mn.get(i — j, 0)) ans = 0 for value in mn.values(): ans -= value print(ans) t = int(input()) for _ in range(t): solve() ~~~~~ 2033C - Экскурсия Сакурако Idea: Vladosiya ~~~~~ #include using namespace std; int main(){ int t; cin>>t; while(t--){ int n; cin>>n; int a[n+1]; for(int i=1;i<=n;i++){ cin>>a[i]; } for(int i=n/2-1;i>=1;i--){ if(a[i]==a[i+1] || a[n-i+1]==a[n-i]){ swap(a[i],a[n-i+1]); } } int re=0; for(int i=1;i 2033D - Задание Косукэ Idea: FBI ~~~~~ #include using namespace std; int main(){ int t; cin>>t; while(t--){ int n; cin>>n; int a[n+1]; mapmp; for(int i=1;i<=n;i++){ cin>>a[i]; } int p_su[n+1]; p_su[0]=0; int lst[n+1]; mp[0]=0; for(int i=1;i<=n;i++){ p_su[i]=p_su[i-1]+a[i]; if(mp.find(p_su[i])==mp.end()){ lst[i]=-1; } else{ lst[i]=mp[p_su[i]]; } mp[p_su[i]]=i; } int dp[n+1]; memset(dp,0,sizeof dp); for(int i=1;i<=n;i++){ dp[i]=max(dp[i],dp[i-1]); if(lst[i]!=-1){ dp[i]=max(dp[i],dp[lst[i]]+1); } } cout<<*max_element(dp,dp+n+1)< 2033E - Сакурако, Косукэ и перестановка Idea: FBI ~~~~~ #include using namespace std; int main(){ int t; ios_base::sync_with_stdio(false);cout.tie(nullptr);cin.tie(nullptr); cin>>t; while(t--){ int n; cin>>n; int p[n+1]; for(int i=1;i<=n;i++){ cin>>p[i]; } bool us[n+1]; memset(us,0,sizeof us); int re=0; for(int i=1;i<=n;i++){ if(!us[i]){ int cu=i; int le=0; while(us[cu]==0){ le++; us[cu]=1; cu=p[cu]; } re+=(le-1)/2; } } cout< 2033F - Лень Косукэ Idea: FBI ~~~~~ #include using namespace std; using LL = long long; #define ssize(x) (int)(x.size()) #define ALL(x) (x).begin(), (x).end() mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); int rd(int l, int r) { return uniform_int_distribution(l, r)(rng); } const LL MOD = 1e9 + 7; int bp(int a, int n) { if (n == 0) return 1; if (n % 2 == 0) return bp(1LL * a * a % MOD, n / 2); else return 1LL * bp(a, n — 1) * a % MOD; } int inv(int a) { return bp(a, MOD — 2); } void solve() { LL n, k; cin >> n >> k; n %= MOD; if (k == 1) { cout << n << "\n"; return; } vector fib(3); fib[0] = fib[1] = 1; int cnt = 0; for (int i = 2; i <= 10 * k; i++) { fib[i % 3] = (fib[(i + 2) % 3] + fib[(i + 1) % 3]) % k; if (fib[i % 3] == 0) cnt++; if (fib[i % 3] == 1 && fib[(i + 2) % 3] == 0) { cout << 1LL * i * n % MOD * inv(cnt) % MOD << "\n"; return; } } } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); int t = 1; cin >> t; while (t--) { solve(); } } ~~~~~ 2033G - Сакурако и Чефир Idea: Vladosiya ~~~~~ #include //#define int long long #define pb emplace_back #define mp make_pair #define x first #define y second #define all(a) a.begin(), a.end() #define rall(a) a.rbegin(), a.rend() typedef long double ld; typedef long long ll; using namespace std; mt19937 rnd(time(nullptr)); const int inf = 1e9; const int M = 1e9 + 7; const ld pi = atan2(0, -1); const ld eps = 1e-6; void precalc(int v, int p, vector> &sl, vector> &maxd, vector &h){ maxd[v] = {0, 0}; if (v != p) h[v] = h[p] + 1; for(int u: sl[v]){ if (u == p) continue; precalc(u, v, sl, maxd, h); if (maxd[v].y < maxd[u].x + 1) { maxd[v].y = maxd[u].x + 1; } if (maxd[v].y > maxd[v].x) { swap(maxd[v].x, maxd[v].y); } } } void calc_binups(int v, int p, vector> &sl, vector>> &binup, vector> &maxd, vector &h){ binup[v][0] = {maxd[p].x, p}; if (maxd[p].x == maxd[v].x + 1) { binup[v][0].x = maxd[p].y; } binup[v][0].x -= h[p]; for(int i = 1; i < 20; ++i){ binup[v][i].y = binup[binup[v][i — 1].y][i — 1].y; binup[v][i].x = max(binup[v][i — 1].x, binup[binup[v][i — 1].y][i — 1].x); } for(int u: sl[v]){ if (u == p) continue; calc_binups(u, v, sl, binup, maxd, h); } } int get_ans(int v, int k, vector>> &binup, vector> &maxd, vector &h){ k = min(k, h[v]); int res = maxd[v].x — h[v]; int ini = h[v]; for(int i = 19; i >= 0; --i){ if ((1 << i) <= k) { res = max(res, binup[v][i].x); v = binup[v][i].y; k -= (1 << i); } } return res + ini; } void solve(int tc){ int n; cin >> n; vector> sl(n); for(int i = 1; i < n; ++i){ int u, v; cin >> u >> v; sl[--u].emplace_back(--v); sl[v].emplace_back(u); } vector> maxd(n); vector h(n); precalc(0, 0, sl, maxd, h); vector>> binup(n, vector>(20)); calc_binups(0, 0, sl, binup, maxd, h); int q; cin >> q; for(int _ = 0; _ < q; ++_){ int v, k; cin >> v >> k; cout << get_ans(v — 1, k, binup, maxd, h) << " "; } } bool multi = true; signed main() { int t = 1; if (multi)cin >> t; for (int i = 1; i <= t; ++i) { solve(i); cout << "\n"; } return 0; } ~~~~~