A - tcdr
直接跳過 aeiou 輸出就好
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| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define SZ(a) (int)(a).size()
#define ALL(v) (v).begin(), (v).end()
#define X first
#define Y second
#define dbg(x) cerr << #x << " = " << x << endl;
void solve() {
string s; cin >> s;
for (int i = 0; s[i]; i++) {
if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u') continue;
else cout << s[i];
}
cout << "\n";
}
/*
CheckList:
1. long long
2. mod
3. template
4. input/output format/content
5. test sample input/output
*/
int main() {
ios::sync_with_stdio(0);cin.tie(0);
int T = 1;
// cin >> T;
while (T--) solve();
return 0;
}
|
B - The Middle Day
算出總和之後 / 2，然後一個一個月慢慢算現在有沒有超過就好
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| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define SZ(a) (int)(a).size()
#define ALL(v) (v).begin(), (v).end()
#define X first
#define Y second
#define dbg(x) cerr << #x << " = " << x << endl;
void solve() {
int m; cin >> m;
vector<int> d(m);
int sum = 0;
for (int &x : d) {
cin >> x;
sum += x;
}
sum /= 2;
for (int i = 0; i < m; i++) {
if (sum >= d[i]) sum -= d[i];
else {
cout << i + 1 << " " << sum + 1 << "\n";
return;
}
}
}
/*
CheckList:
1. long long
2. mod
3. template
4. input/output format/content
5. test sample input/output
*/
int main() {
ios::sync_with_stdio(0);cin.tie(0);
int T = 1;
// cin >> T;
while (T--) solve();
return 0;
}
|
C - Flavors
直接分兩個 case 算，如果最大的兩個不同 flavor 的話，那答案一定是他們相加，不然就去把所有 flavor 的最大拿出來，然後拿兩個最大的相加就好。
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| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define SZ(a) (int)(a).size()
#define ALL(v) (v).begin(), (v).end()
#define X first
#define Y second
#define dbg(x) cerr << #x << " = " << x << endl;
void solve() {
int n; cin >> n;
vector<pii> v(n);
for (auto &p : v) cin >> p.Y >> p.X;
sort(ALL(v));
ll res = 0;
if (v[n - 1].Y == v[n - 2].Y) {
res = v[n - 1].X + v[n - 2].X / 2;
vector<int> mxf(n + 1, -1);
for (int i = 0; i < n; i++) {
mxf[v[i].Y] = max(mxf[v[i].Y], v[i].X);
}
ll mx = v[n - 1].X, second_mx = -1;
for (int i = 1; i <= n; i++) {
if (mxf[i] > second_mx && mxf[i] < mx) second_mx = mxf[i];
}
res = max(res, second_mx + mx);
} else {
res = v[n - 1].X + v[n - 2].X;
}
cout << res << "\n";
}
/*
CheckList:
1. long long
2. mod
3. template
4. input/output format/content
5. test sample input/output
*/
int main() {
ios::sync_with_stdio(0);cin.tie(0);
int T = 1;
// cin >> T;
while (T--) solve();
return 0;
}
|
D - Magical Cookies
死亡實作 @@，可以想像像題目那樣枚舉的話會是 $O(H^2W)$ 會 TLE，所以可以利用字元集很少這個特性來做，簡單來說就是先算好每行/列的每個字元有多少，然後每次就去看現在還沒被刪掉的有沒有字元是會整行/列都一樣的，就把他移掉，然後這個過程因為只有 $H + W$ 行+列，所以只會做到 $H + W$ 次。
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| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define SZ(a) (int)(a).size()
#define ALL(v) (v).begin(), (v).end()
#define X first
#define Y second
#define dbg(x) cerr << #x << " = " << x << endl;
void solve() {
int h, w; cin >> h >> w;
vector<string> v(h);
for (auto &s : v) cin >> s;
vector<vector<int>> cnth(26, vector<int>(h, 0)), cntw(26, vector<int>(w, 0));
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
int ind = v[i][j] - 'a';
cnth[ind][i]++;
cntw[ind][j]++;
}
}
int resh = h, resw = w;
vector<bool> skiph(h, false), skipw(w, false);
for (int _ = 0; _ < h + w; _++) {
vector<pii> ch, cw;
for (int i = 0; i < h; i++) {
if (skiph[i]) continue;
for (int j = 0; j < 26; j++) {
if (cnth[j][i] == resw && resw >= 2) {
ch.push_back({i, j});
}
}
}
for (int i = 0; i < w; i++) {
if (skipw[i]) continue;
for (int j = 0; j < 26; j++) {
if (cntw[j][i] == resh && resh >= 2) {
cw.push_back({i, j});
}
}
}
for (auto p : ch) {
skiph[p.X] = true;
for (int i = 0; i < w; i++) cntw[p.Y][i]--;
resh--;
}
for (auto p : cw) {
skipw[p.X] = true;
for (int i = 0; i < h; i++) cnth[p.Y][i]--;
resw--;
}
}
cout << resh * resw << "\n";
}
/*
CheckList:
1. long long
2. mod
3. template
4. input/output format/content
5. test sample input/output
*/
int main() {
ios::sync_with_stdio(0);cin.tie(0);
int T = 1;
// cin >> T;
while (T--) solve();
return 0;
}
|
E - Prerequisites
乍看之下是簡單的拓墣排序，但是注意到他要求最小數量，所以有些根本不會在 DAG 裡的孤點我們就不管它，這時候就可以利用題目要求的是 read book 1，用一個 priority_queue 來代替平常拓墣排序用的 queue，這樣編號比較大的孤點在處理到有關 1 之前的 DAG 之前就會被先 pop 掉。
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| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define SZ(a) (int)(a).size()
#define ALL(v) (v).begin(), (v).end()
#define X first
#define Y second
#define dbg(x) cerr << #x << " = " << x << endl;
const int N = 2e5 + 5;
vector<int> g[N];
int deg[N];
void solve() {
int n; cin >> n;
for (int i = 1; i <= n; i++) {
int c; cin >> c;
for (int j = 0; j < c; j++) {
int x; cin >> x;
g[i].push_back(x);
deg[x]++;
}
}
priority_queue<int> q;
for (int i = 1; i <= n; i++) {
if (deg[i] == 0) q.push(i);
}
vector<int> res;
while (!q.empty()) {
int u = q.top(); q.pop();
res.push_back(u);
for (int v : g[u]) {
--deg[v];
if (deg[v] == 0) q.push(v);
}
}
for (int i = SZ(res) - 1; i >= 0; i--) {
if (res[i] == 1) break;
else cout << res[i] << " ";
}
}
/*
CheckList:
1. long long
2. mod
3. template
4. input/output format/content
5. test sample input/output
*/
int main() {
ios::sync_with_stdio(0);cin.tie(0);
int T = 1;
// cin >> T;
while (T--) solve();
return 0;
}
|
F - Shortcuts
待補。