2021-2022 ICPC Northwestern European Regional Programming Contest (NWERC 2021)
A. Access Denied
先问若干次，问出长度，然后再一位一位的问即可。
#include <bits/stdc++.h>
using namespace std;
int input() {
string s;
getline(cin, s);
if (s == "ACCESS GRANTED")
exit(0);
int t = 0;
for (auto i: s) {
if (i >= '0' and i <= '9')
t = t * 10 + i - '0';
}
return t;
}
int32_t main() {
int len = -1;
for (int i = 1, t; len == -1 and i <= 20; i++) {
cout << string(i, '0') << endl << flush;
t = input();
if (t != 5) len = i;
}
vector<char> ans(len), ch;
for (char i = '0'; i <= '9'; i++) ch.push_back(i);
for (auto i = 'a'; i <= 'z'; i++) ch.push_back(i);
for (auto i = 'A'; i <= 'Z'; i++) ch.push_back(i);
for (int i = 0, t; i < len; i++) {
for (auto c: ch) {
for (int j = 0; j < i; j++)
cout << ans[j];
for (int j = i; j < len; j++)
cout << c;
cout << endl << flush;
t = input();
if (t == 5 + (i + 1) * 9) continue;
ans[i] = c;
break;
}
}
for( auto i : ans )
cout << i;
cout << endl << flush;
return 0;
}
D. Dyson Circle
首先按照题目的要求，我们可以把戴森环的凹陷部分展开，这样的戴森环就变成了一个矩形。那么对于矩形，我们让边全部在对角线方向最优，所以就要找到主副对角线的边界。如果某对角线方向长度只有 1 的话不符合题目的要求，因为题目要求内点的连通是有边相邻。
#include<bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int n, a = INT_MIN, b = INT_MIN, c = INT_MAX, d = INT_MAX;
cin >> n;
if (n == 1) cout << "4\n", exit(0);
for (int x, y; n; n--) {
cin >> x >> y;
a = max(a, x + y);
b = max(b, x - y);
c = min(c, x + y);
d = min(d, x - y);
}
cout << 4 + a + b - c - d + (a == c) + (b == d) << "\n";
return 0;
}
G. Glossary Arrangement
#include<bits/stdc++.h>
using namespace std;
using vi = vector<int>;
int main() {
ios::sync_with_stdio(0), cin.tie(0);
int n, w;
cin >> n >> w;
vector<string> s(n);
for (auto &i: s) cin >> i;
vector<int> f(n + 1);
auto dp = [n, w, s, &f](int h) {
f.assign(n + 1, w + 1);
f[0] = -1;
for (int i = 0, mx; i < n; i++) {
mx = 0;
for (int j = i; j < n and j < i + h; j++) {
mx = max(mx, int(s[j].size()));
f[j + 1] = min(f[j + 1], f[i] + mx + 1);
}
}
return f[n];
};
int l = 1, r = n, h = -1;
for (int mid; l <= r;) {
mid = (l + r) / 2;
if (dp(mid) <= w) h = mid, r = mid - 1;
else l = mid + 1;
}
cout << h << " ";
vi v;
dp(h);
for (int i = n, j, mx; i;) {
j = i, mx = 0;
do {
j--;
mx = max(mx, int(s[j].size()));
} while (f[i] != f[j] + mx + 1);
v.push_back(mx), i = j;
}
reverse(v.begin(), v.end());
cout << v.size() << "\n";
for (auto i: v) cout << i << " ";
cout << "\n";
vector<string> res(h);
for (int i = 0, k = 0; i < v.size(); i++) {
for (int j = 0; j < h; j++) {
if (i) res[j] += " ";
string t;
if (k < n and s[k].size() <= v[i]) t = s[k++];
t.resize(v[i], ' ');
res[j] += t;
}
}
for (auto i: res)
cout << i << "\n";
return 0;
}
H. Heating Up
初始的辣度满足单调性，因此我们可以用二分答案。
有了初始辣度后，要考虑的就是如何把所有的披萨吃完，首先破环成链，然后从头开始枚举，如果可以吃就直接吃，否则就把这片披萨放进栈里，并且还原耐辣值。同时每次耐辣值上升后，都重新判断一下栈顶的披萨是否可以被吃下。
#include <bits/stdc++.h>
using namespace std;
#define int __int128
using ll = long long;
using pii = pair<int, int>;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
constexpr int inf = 1e15;
int32_t main() {
int n = read();
vector<int> a(n + n + 1);
for (int i = 1; i <= n; i++)
a[i + n] = a[i] = read();
n = n * 2;
auto check = [a, n](int x) -> bool {
int tot = x;
stack<pii> stk;
for (int i = 1; i <= n; i++) {
if (a[i] <= tot) {
tot += a[i];
while (!stk.empty() and stk.top().first <= tot) {
tot += stk.top().second;
stk.pop();
}
} else {
stk.emplace(a[i], tot - x + a[i]);
tot = x;
}
}
return stk.empty();
};
int l = 0, r = inf, res = -1;
for (int mid; l <= r;) {
mid = (l + r) / 2;
if (check(mid)) res = mid, r = mid - 1;
else l = mid + 1;
}
cout << (ll) res << "\n";
return 0;
}
J. Jet Set
维度是没有用的。
如果在单次飞行中跨过的180 度，则一定是可行的。
否则，根据题目的要求，飞行过程中经过的经度实际上是确定的。我们只要标记即可。
最后判断有无没有被标记的点的即可。
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
int n;
cin >> n;
vector<int> a(n + 2), v(720);
for (int i = 1, x; i <= n; i++) {
cin >> x >> a[i];
a[i] = (a[i] + 180) * 2;
}
a[n+1] = a[1];
for (int i = 2, x, y; i <= n+1; i++) {
x = a[i - 1], y = a[i];
if (x > y) swap(x, y);
if (y - x == 360) cout << "yes\n", exit(0);
if (y - x < 360) {
for (int j = x; j <= y; j++) v[j] = 1;
} else {
for (int j = 0; j <= x; j++) v[j] = 1;
for (int j = y; j < 720; j++) v[j] = 1;
}
}
for( int i = 0 ; i < 720 ; i ++ )
if( v[i] == 0 ) cout << "no " << fixed << setprecision(1) << (0.5*i) - 180.0 << "\n", exit(0);
cout << "yes\n";
return 0;
}
K. Knitpicking
我们先尽可能的按照无法配对的情况去选择，最后在任意选一个即可配对，除非已经把所有的袜子全部选完。
如果袜子是任意的，则只有当无左无右的情况下，才能选择一个。否则要选择左右的较大值。
#include<bits/stdc++.h>
using namespace std;
#define pii pair<int, int>
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1);
const int N = 110;
void solve() {
int n;
cin >> n;
map<string, array<int, 3> > mp;
int z = 0;
for (int i = 0; i < n; ++i) {
string a, b;
int x;
cin >> a >> b >> x;
if (b == "left") {
mp[a][0] += x;
} else if (b == "right") {
mp[a][1] += x;
} else {
mp[a][2] += x;
}
}
int res = 0, mark = 0;
for (auto X: mp) {
string a = X.first;
z += mp[a][0] + mp[a][1] + mp[a][2];
if (mp[a][0] == 0 && mp[a][1] == 0 && mp[a][2] > 0) res += 1;
else res += max(mp[a][0], mp[a][1]);
}
if (res != z) {
cout << res + 1 << '\n';
} else cout << "impossible\n";
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
while (_--) {
solve();
}
return 0;
}
