A - Doors in the Center
Sign in and construct the questions directly.
Click to view the code
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;
signed main() {
int n;
cin >> n;
if (n % 2 == 1) {
for (int i = 1; i <= n / 2; i++) {
cout << '-';
}
cout << "=";
for (int i = 1; i <= n / 2; i++) {
cout << '-';
}
} else {
for (int i = 1; i <= n / 2 - 1; i++) {
cout << '-';
}
cout << "==";
for (int i = 1; i <= n / 2 - 1; i++) {
cout << '-';
}
}
}
/*
*/
B - Full House 3
If there are more than two or more identical ones, or more than three identical ones, two or more identical ones, the conditions for the gourd can be met.
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#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;
signed main() {
int n = 7;
map<int, int> mp;
set<int> se;
for (int i = 1; i <= n; i++) {
int x;
cin >> x;
mp[x]++;
(x);
}
int f = 0, ff = 0;
for (auto t : se) {
if (mp[t] >= 3) f++;
else if (mp[t] >= 2) ff++;
}
if (f >= 2 || (f == 1 && ff >= 1)) {
cout << "Yes\n";
} else {
cout << "No\n";
}
}
/*
*/
C - Uniqueness
Violence enumeration.
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#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;
signed main() {
int n;
cin >> n;
map<int, int> mp;
vector<int> a(n + 1);
for (int i = 1; i <= n; i++) {
cin >> a[i];
mp[a[i]]++;
}
int ans = -1, id = -1;
for (int i = 1; i <= n; i++) {
if (mp[a[i]] == 1) {
if (a[i] > ans) {
ans = a[i];
id = i;
}
}
}
cout << id << '\n';
}
/*
*/
D - Bonfire
\((0,0)\)The point will refresh the new smoke every time. For each move, whether the high bridge can be covered by a high bridge, just look at which refreshed smoke in front can reach the high bridge. If there is, it is 1, and if there is no, it is 0.
Two can be used in implementation\(sum\)The array records the movement amount between east and west and north and south, and searches in binary.
Of course, there are actually easier ways to implement it.
Click to view the code
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;
signed main() {
ios::sync_with_stdio(0), (0), (0);
int n, r, c;
cin >> n >> r >> c;
string s;
cin >> s;
//n r-1
//w c-1
//s r+1
//e c+1
s = ' ' + s;
vector<int> sum1(n + 1), sum2(n + 1);
for (int i = 1; i <= n; i++) {
sum1[i] = sum1[i - 1], sum2[i] = sum2[i - 1];
if (s[i] == 'N') sum1[i] -= 1;
if (s[i] == 'S') sum1[i] += 1;
if (s[i] == 'W') sum2[i] -= 1;
if (s[i] == 'E') sum2[i] += 1;
}
// for (int i = 1; i <= n; i++) {
// cout << sum1[i] << " \n"[i == n];
// }
// for (int i = 1; i <= n; i++) {
// cout << sum2[i] << " \n"[i == n];
// }
map<pair<int, int>, vector<int>> mp;
mp[ {0, 0}].push_back(1);
for (int i = 1; i <= n; i++) {
mp[ {sum1[i], sum2[i]}].push_back(i + 1);
}
string ans;
for (int i = 1; i <= n; i++) {
int x = sum1[i], y = sum2[i];
int nex = x - r, ney = y - c;
if (mp[ {nex, ney}].size() == 0 ) {
ans += "0";
continue;
}
if (mp[ {nex, ney}].back() <= i) {
ans += "1";
continue;
}
int l = upper_bound(mp[ {nex, ney}].begin(), mp[ {nex, ney}].end(), i) - mp[ {nex, ney}].begin();
if (mp[ {nex, ney}][l] <= i) ans += "1";
else {
if (l != 0) l--;
else {
ans += "0";
continue;
}
if (mp[ {nex, ney}][l] <= i) ans += "1";
}
}
cout << ans << '\n';
}
/*
10 1 2
NEESESWEES
2 2
2 1
2 0
6 -2 1
NNEEWS
*/
E - Tree Game
Interaction questions, by looking at each point\(bfs\)All $u ->v $ can be recorded through a combination of points set as even numbers. If the number of combinations is odd, then it is\(first\), otherwise\(second\). Then output the appropriate one according to the opponent's operation\((u,v)\)。
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#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;
signed main() {
ios::sync_with_stdio(0), (0), (0);
int n;
cin >> n;
vector<vector<int>> g(n + 10);
vector<int> d(n + 10), e(n + 10);
for (int i = 1; i < n; i++) {
int u, v;
cin >> u >> v;
d[u]++;
e[v]++;
g[u].push_back(v);
g[v].push_back(u);
}
queue<pair<int, int>> qq;
map<pair<int, int>, int> mp;
for (int i = 1; i <= n; i++) {
vector<int> st(n + 1), dis(n + 1, 0);
queue<int> q;
(i);
st[i] = 1;
while (()) {
int u = ();
();
for (auto v : g[u]) {
if (st[v] == 0) {
(v);
st[v] = 1;
dis[v] = dis[u] + 1;
}
}
}
for (int j = 1; j <= n; j++) {
// cout << dis[j] << " \n"[j == n];
if (j != i) {
if ((dis[j] + 1) % 2 == 0 && dis[j] + 1 != 2) {
if (mp[ {i, j}] == 0) {
({i, j});
mp[ {i, j}] = 1;
mp[ {j, i}] = 1;
}
}
}
}
}
map<pair<int, int>, int> st;
// cout << () << '\n';
if (() % 2 == 1) {
cout << "First" << endl;
while (()) {
while (() && st[()] == 1) ();
pair<int, int> t = ();
();
cout << << ' ' << << endl;
int u, v;
cin >> u >> v;
st[ {u, v}] = 1;
st[ {v, u}] = 1;
if (u == -1 && v == -1) return 0;
}
} else {
cout << "Second" << endl;
while (()) {
int u, v;
cin >> u >> v;
st[ {u, v}] = 1;
st[ {v, u}] = 1;
if (u == -1 || v == -1) return 0;
while (() && st[()] == 1) ();
if (() == 0)return 0;
pair<int, int> t = ();
();
cout << << ' ' << << endl;
}
int u, v;
cin >> u >> v;
if (u == -1 && v == -1) return 0;
}
return 0;
}
/*
6
1 2
2 3
3 4
4 5
5 6
5
1 2
1 3
2 4
2 5
*/
F - ABCBA
This question is simple, it is easy to think of the longest\([l,n]\)Pastoral string, the hash used by the code implementation.
Click to view the code
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
const int N = 1e6 + 10, base = 13331;
ull hs[N], pa[N], fhs[N], fpa[N];
ull get(int l, int r) {
return hs[r] - hs[l - 1] * pa[r - l + 1];
}
ull getf(int l, int r) {
return fhs[r] - fhs[l - 1] * fpa[r - l + 1];
}
signed main() {
ios::sync_with_stdio(0), (0), (0);
string s;
cin >> s;
string yu = s;
int n = ();
s = ' ' + s;
pa[0] = 1, fpa[0] = 1;
for (int i = 1; i <= n; i++) {
pa[i] = pa[i - 1] * base;
hs[i] = hs[i - 1] * base + s[i];
}
reverse((), ());
string fs = yu;
reverse((), ());
fs = ' ' + fs;
for (int i = 1; i <= n; i++) {
fpa[i] = fpa[i - 1] * base;
fhs[i] = fhs[i - 1] * base + fs[i];
}
int k = n;
for (int i = 1; i <= n; i++) {
ull a = get(i, n);
ull b = getf(1, n - i + 1);
if (a == b) {
k = i - 1;
break;
}
}
string ans = yu;
for (int i = k; i >= 1; i--) {
ans += s[i];
}
cout << ans << '\n';
}
/*
*/