T1：First Player
模拟
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
int main() {
int n;
cin >> n;
vector<string> s(n);
vector<int> a(n);
rep(i, n) cin >> s[i] >> a[i];
int si = 0;
rep(i, n) if (a[i] < a[si]) si = i;
rep(i, n) {
int ni = (si+i)%n;
cout << s[ni] << '\n';
}
return 0;
}
T2：Subscribers
模拟
代码实现
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
string s = to_string(n);
for (int i = 3; i < s.size(); ++i) {
s[i] = '0';
}
cout << s << '\n';
return 0;
}
T3：Virus
可以将满足欧几里得距离不超过 \(d\) 的两点进行连边，在点 \(1\) 所在的连通分量内的点都会被感染
可以用 dfs/bfs/并查集 实现
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
int main() {
int n, d;
cin >> n >> d;
vector<int> x(n), y(n);
rep(i, n) cin >> x[i] >> y[i];
auto near = [&](int a, int b) {
int dx = x[a]-x[b];
int dy = y[a]-y[b];
return dx*dx+dy*dy <= d*d;
};
queue<int> q;
vector<bool> ans(n);
ans[0] = true; q.push(0);
while (q.size()) {
int v = q.front(); q.pop();
rep(u, n) {
if (near(v, u)) {
if (ans[u]) continue;
ans[u] = true;
q.push(u);
}
}
}
rep(i, n) {
puts(ans[i] ? "Yes" : "No");
}
return 0;
}
T4：A Piece of Cake
先用两次 std::lower_bound() 来确定每个草莓位于哪块区域
再用 std::map<pair<int, int>, int> 来统计每块区域中草莓的个数
然后遍历一遍 map，找到 \(m\) 和 \(M\)
注意：当 map 的大小 \(< (A+1)(B+1)\) 时，\(m = 0\)
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;
int main() {
int w, h, n;
cin >> w >> h >> n;
vector<int> p(n), q(n);
rep(i, n) cin >> p[i] >> q[i];
int A;
cin >> A;
vector<int> a(A);
rep(i, A) cin >> a[i];
int B;
cin >> B;
vector<int> b(B);
rep(i, B) cin >> b[i];
map<P, int> mp;
rep(i, n) {
int x = lower_bound(a.begin(), a.end(), p[i])-a.begin();
int y = lower_bound(b.begin(), b.end(), q[i])-b.begin();
mp[P(x, y)]++;
}
int m = n, M = 0;
for (auto e : mp) {
m = min(m, e.second);
M = max(M, e.second);
}
if (mp.size() < (ll)(A+1)*(B+1)) m = 0;
cout << m << ' ' << M << '\n';
return 0;
}