Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
cout << (i + 1) % n + 1 << " ";
}
cout << endl;
}
return 0;
}
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> cnt(n + 1), idx(n + 1);
for (int i = 0; i < n; ++i) {
int x;
cin >> x;
++cnt[x];
idx[x] = i + 1;
}
int ans = -1;
for (int i = 0; i <= n; ++i) {
if (cnt[i] == 1) {
ans = idx[i];
break;
}
}
cout << ans << endl;
}
return 0;
}
1454C - Изменение последовательности
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> a(n);
for (auto &it : a) cin >> it;
vector<int> res(n + 1, 1);
n = unique(a.begin(), a.end()) - a.begin();
a.resize(n);
for (int i = 0; i < n; ++i) {
res[a[i]] += 1;
}
res[a[0]] -= 1;
res[a[n - 1]] -= 1;
int ans = 1e9;
for (int i = 0; i < n; ++i) {
ans = min(ans, res[a[i]]);
}
cout << ans << endl;
}
return 0;
}
1454D - Число в последовательность
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
long long n;
cin >> n;
vector<pair<int, long long>> val;
for (long long i = 2; i * i <= n; ++i) {
int cnt = 0;
while (n % i == 0) {
++cnt;
n /= i;
}
if (cnt > 0) {
val.push_back({cnt, i});
}
}
if (n > 1) {
val.push_back({1, n});
}
sort(val.rbegin(), val.rend());
vector<long long> ans(val[0].first, val[0].second);
for (int i = 1; i < int(val.size()); ++i) {
for (int j = 0; j < val[i].first; ++j) {
ans.back() *= val[i].second;
}
}
cout << ans.size() << endl;
for (auto it : ans) cout << it << " ";
cout << endl;
}
return 0;
}
1454E - Количество простых путей
Идея: vovuh
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<set<int>> g(n);
for (int i = 0; i < n; ++i) {
int x, y;
cin >> x >> y;
--x, --y;
g[x].insert(y);
g[y].insert(x);
}
vector<int> val(n, 1);
queue<int> leafs;
for (int i = 0; i < n; ++i) {
if (g[i].size() == 1) {
leafs.push(i);
}
}
while (!leafs.empty()) {
int v = leafs.front();
leafs.pop();
int to = *g[v].begin();
val[to] += val[v];
val[v] = 0;
g[v].clear();
g[to].erase(v);
if (g[to].size() == 1) {
leafs.push(to);
}
}
long long ans = 0;
for (int i = 0; i < n; ++i) {
ans += val[i] * 1ll * (val[i] - 1) / 2;
ans += val[i] * 1ll * (n - val[i]);
}
cout << ans << endl;
}
return 0;
}
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
bool found;
void shift(multiset<int> &l, multiset<int> &r, int val) {
l.erase(l.find(val));
r.insert(val);
}
void check(const multiset<int> &lf, const multiset<int> &mid, const multiset<int> &rg) {
if (found || lf.empty() || mid.empty() || rg.empty()) {
return;
}
if (*lf.rbegin() == *mid.begin() && *mid.begin() == *rg.rbegin()) {
found = true;
cout << "YES" << endl;
cout << lf.size() << " " << mid.size() << " " << rg.size() << endl;
}
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> a(n);
for (auto &it : a) cin >> it;
found = false;
multiset<int> lf, mid(a.begin(), a.end()), rg;
int r = n - 1;
for (int l = 0; l < n - 2; ++l) {
shift(mid, lf, a[l]);
while (r - 1 >= l && a[r] <= *lf.rbegin()) {
shift(mid, rg, a[r]);
--r;
}
while (r - 1 < l) {
shift(rg, mid, a[r + 1]);
++r;
}
check(lf, mid, rg);
shift(lf, mid, a[l]);
check(lf, mid, rg);
shift(mid, lf, a[l]);
if (rg.empty()) continue;
shift(rg, mid, a[r + 1]);
check(lf, mid, rg);
shift(mid, rg, a[r + 1]);
}
if (!found) {
cout << "NO" << endl;
}
}
return 0;
}
Решение (Gassa)
// Author: Ivan Kazmenko (gassa@mail.ru)
module solution;
import std.algorithm;
import std.conv;
import std.range;
import std.stdio;
import std.string;
void main ()
{
auto tests = readln.strip.to !(int);
foreach (test; 0..tests)
{
auto n = readln.strip.to !(int);
auto a = readln.splitter.map !(to !(int)).array;
auto x = a.dup;
foreach (i; 1..n)
{
x[i] = max (x[i], x[i - 1]);
}
auto y = a.dup;
foreach_reverse (i; 0..n - 1)
{
y[i] = max (y[i], y[i + 1]);
}
auto v = a.maxElement;
auto maxPlaces = n.iota.filter !(i => a[i] == v).array;
int lo = maxPlaces[$ / 2];
int hi = lo + 1;
while (true)
{
if (lo == 0 || hi == n)
{
writeln ("NO");
break;
}
if (x[lo - 1] == v && y[hi] == v)
{
writeln ("YES");
writeln (lo, " ", hi - lo, " ", n - hi);
break;
}
int u = (lo - 1 == 0) ? int.min :
min (x[lo - 2], a[lo - 1]);
int w = (hi + 1 >= n) ? int.min :
min (y[hi + 1], a[hi]);
if (u > w)
{
v = min (v, a[lo - 1]);
lo -= 1;
}
else
{
v = min (v, a[hi]);
hi += 1;
}
}
}
}