Codeforces Round 811 (Div. 3)
A. Everyone Loves to Sleep
#include<bits/stdc++.h>
using namespace std;
void solve() {
int n, h, m, t;
cin >> n >> h >> m;
t = h * 60 + m;
vector<int> a;
for (int i = 1, x, y; i <= n; i++)
cin >> x >> y, a.push_back(x * 60 + y);
sort( a.begin(), a.end() ) , a.push_back( a.front() + 24 * 60 );
int p = *lower_bound( a.begin(), a.end() , t );
p -= t;
cout << p / 60 << " " << p % 60 << "\n";
return;
}
int32_t main() {
ios::sync_with_stdio(0), cin.tie(0);
int t;
for (cin >> t; t; t--)
solve();
return 0;
}
B. Remove Prefix
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int inf = 1e18;
using vi = vector<int>;
const int N = 1e3;
vi val(N + 1, inf);
void solve() {
int n;
cin >> n;
vi a(n);
for( auto & i : a ) cin >> i;
set<int> cnt;
reverse(a.begin(), a.end());
for( auto & i : a )
if( cnt.insert(i).second == false ) break;
cout << n - cnt.size() << "\n";
return;
}
int32_t main() {
int t;
cin >> t;
while (t--)
solve();
return 0;
}
C. Minimum Varied Number
贪心暴搜似乎过不去，但是可以打表
#include<bits/stdc++.h>
using namespace std;
//int res, n = 0;
//vector<int> q, vis(10);
//
//void dfs(int x) {
// int cnt = 0;
// auto p = q;
// sort(p.begin(), p.end());
// for (auto i: p) cnt = cnt * 10 + i;
// if( cnt > res ) return ;
// if (x == 0) {
// res = min(res, cnt);
// return;
// }
// for (int i = 1; i <= 9; i++) {
// if (i > x) break;
// if (vis[i]) continue;
// vis[i] = 1, q.push_back(i);
// dfs(x - i);
// vis[i] = 0, q.pop_back();
// }
//}
//void solve() {
// ++ n, res = INT_MAX;
// dfs(n);
// cout << res << "\n";
//}
vector<string> res = {"1",
"2",
"3",
"4",
"5",
"6",
"7",
"8",
"9",
"19",
"29",
"39",
"49",
"59",
"69",
"79",
"89",
"189",
"289",
"389",
"489",
"589",
"689",
"789",
"1789",
"2789",
"3789",
"4789",
"5789",
"6789",
"16789",
"26789",
"36789",
"46789",
"56789",
"156789",
"256789",
"356789",
"456789",
"1456789",
"2456789",
"3456789",
"13456789",
"23456789",
"123456789"};
void solve(){
int n;
cin >> n;
cout << res[n-1]<< "\n";
}
int32_t main() {
ios::sync_with_stdio(0), cin.tie(0);
int t;
for (cin >> t; t; t--)
solve();
return 0;
}
D. Color with Occurrences
当我们确定之前覆盖的部分后，可以直接贪心的选择可以之前覆盖区间能够相连且，覆盖距离最远的。
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int inf = 1e18;
using db = long double;
using vi = vector<int>;
using pii = pair<int, int>;
void solve() {
string t, s;
int n;
cin >> t >> n;
int m = t.size();
vector<array<int, 3>> e;
for (int i = 1; i <= n; i++) {
cin >> s;
for (int l = 0, r = s.size()-1; r < m; l++, r++) {
if (t.substr(l, s.size()) == s)
e.push_back({l + 1, r + 1, i});
}
}
vector<int> res;
for (int pos = 0, left = 1, right = 0; left <= m; pos = 0) {
for (int i = 0; i < e.size(); i++)
if (e[i][0] <= left and e[i][1] > right)
right = e[i][1], pos = i;
res.push_back(pos);
left = right + 1;
if (res.size() > e.size() + 100) {
cout << "-1\n";
return;
}
}
cout << res.size() << "\n";
for (auto i: res)
cout << e[i][2] << " " << e[i][0] << "\n";
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int t;
cin >> t;
for (; t; t--)
solve();
}
E. Add Modulo 10
可以发现除了结尾是5和 0 的情况，其他的都会进入到2 4 8 6 的循环中，所以可以先把各位全部移动到循环上，然后判断差值是否都是20 的倍数
#include<bits/stdc++.h>
using namespace std;
#define int long long
using vi = vector<int>;
void solve() {
int n, f1 = 1, f2 = 1;
cin >> n;
vi a(n);
for (auto &i: a) {
cin >> i;
if (i % 10 == 0 or i % 10 == 5) {
f1 = 0;
if (i % 10 == 5) i += 5;
} else f2 = 0;
}
if (f1 + f2 == 0) {
cout << "No\n";
return;
} else if (f2) {
for (auto i: a)
if (i != a.front()) {
cout << "No\n";
return;
}
cout << "Yes\n";
return;
}
for (auto &i: a)
while (i % 10 != 2) i += i % 10;
int top = *max_element(a.begin(), a.end());
for( auto i : a )
if( (top - i) % 20 != 0 ){
cout << "No\n";
return ;
}
cout << "Yes\n";
return ;
}
int32_t main() {
ios::sync_with_stdio(0), cin.tie(0);
int t;
for (cin >> t; t; t--)
solve();
return 0;
}
F. Build a Tree and That Is It
除去不合法的情况，答案只有两种情况，一种是三个点在一条链上，此时选择中间的点作为根就好了。还有一种情况是一个 y 形的树，计算出三个点到分叉点的距离，分叉点做根就好了
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
void solve() {
int n, d1, d2, d3;
cin >> n >> d1 >> d2 >> d3;
if (max({d1, d2, d3}) * 2 > d1 + d2 + d3) {
cout << "NO\n";
return;
}
if ((d1 + d2 + d3) & 1) {
cout << "NO\n";
return;
}
if ((d1 + d2 + d3) / 2 + 1 > n) {
cout << "NO\n";
return;
}
cout << "YES\n";
vi p(n + 1);
auto solver = [&](int root, int p1, int p2, int dep1, int dep2) {
set<int> s;
for (int i = 4; i <= n; i++) s.insert(i);
int fa, dep, t;
for (fa = root, dep = 1, t; dep < dep1; dep++) {
t = *s.begin(), s.erase(s.begin());
p[t] = fa, fa = t;
}
p[p1] = fa;
for (fa = root, dep = 1, t; dep < dep2; dep++) {
t = *s.begin(), s.erase(s.begin());
p[t] = fa, fa = t;
}
p[p2] = fa;
while (!s.empty())
p[*s.begin()] = root, s.erase(s.begin());
for (int i = 1; i <= n; i++)
if (i != root) cout << i << " " << p[i] << "\n";
};
if (d1 + d2 == d3)
solver(2, 1, 3, d1, d2);
else if (d1 + d3 == d2)
solver(1, 2, 3, d1, d3);
else if (d2 + d3 == d1)
solver(3, 1, 2, d3, d2);
else {
set<int> s;
for (int i = 5; i <= n; i++) s.insert(i);
int dep1 = (d1 + d3 - d2) / 2;
int dep2 = (d1 + d2 - d3) / 2;
int dep3 = (d2 + d3 - d1) / 2;
int root = 4, fa, dep, t;
for (fa = root, dep = 1, t; dep < dep1; dep++) {
t = *s.begin(), s.erase(s.begin());
p[t] = fa, fa = t;
}
p[1] = fa;
for (fa = root, dep = 1, t; dep < dep2; dep++) {
t = *s.begin(), s.erase(s.begin());
p[t] = fa, fa = t;
}
p[2] = fa;
for (fa = root, dep = 1, t; dep < dep3; dep++) {
t = *s.begin(), s.erase(s.begin());
p[t] = fa, fa = t;
}
p[3] = fa;
while (!s.empty())
p[*s.begin()] = root, s.erase(s.begin());
for (int i = 1; i <= n; i++)
if (i != root) cout << i << " " << p[i] << "\n";
}
return;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int TC;
for (cin >> TC; TC; TC--)
solve();
return 0;
}
G. Path Prefixes
dfs 的过程记录一下到根的路径，在路径上二分一下就好
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int inf = 1e18;
using db = long double;
using vi = vector<int>;
using pii = pair<int, int>;
void solve() {
int n;
cin >> n;
vi a(n + 1), b(n + 1), q, f(n + 1);
vector<vi> e(n + 1);
for (int i = 2, p; i <= n; i++)
cin >> p >> a[i] >> b[i], e[p].push_back(i);
auto dfs = [e, a, b, &f, &q](auto &&self, int x, int cntA, int cntB, int dep) -> void {
q.push_back(cntB);
int res = -1;
for (int l = 0, r = dep, mid; l <= r;) {
mid = (l + r) / 2;
if (q[mid] <= cntA) res = mid, l = mid + 1;
else r = mid - 1;
}
f[x] = res;
for (auto y: e[x])
self(self, y, cntA + a[y], cntB + b[y], dep + 1);
q.pop_back();
};
dfs(dfs, 1, 0, 0, 0);
for( int i = 2 ; i <= n ; i ++ )
cout << f[i] << " ";
cout << "\n";
return ;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int t;
cin >> t;
for (; t; t--)
solve();
}
