vp 赛时 AK！
A - Plural Form
考虑直接判断字符串结尾
// BLuemoon
#include <bits/stdc++.h>using namespace std;
using LL = long long;
using DB = double;const int kMaxN = 2e5 + 5;string s;void pr(bool pr) {cout << (pr ? "Yes" : "No") << '\n';
}int main() {cin >> s;cout << s << (s[s.size() - 1] == 's' ? "es" : "s") << '\n';return 0;
}
B - Go to Jail
考虑直接暴力判断
// BLuemoon
#include <bits/stdc++.h>using namespace std;
using LL = long long;
using DB = double;const int kMaxN = 1e2 + 5;int n, a[kMaxN], b[kMaxN];
bool ans;void pr(bool pr) {cout << (pr ? "Yes" : "No") << '\n';
}int main() {cin >> n;for (int i = 1; i <= n; i++) {cin >> a[i] >> b[i];}for (int i = 2; i < n; i++) {if (a[i - 1] == b[i - 1] && a[i] == b[i] && a[i + 1] == b[i + 1]) {ans = 1;break;}}pr(ans);return 0;
}
C - A x B + C
考虑只要 \(A \times B < N\)，就可以找到唯一存在的 \(C\)。
枚举 \(A\)，找满足条件的 \(B\) 的数量。
// BLuemoon
#include <bits/stdc++.h>using namespace std;
using LL = long long;
using DB = double;const int kMaxN = 2e5 + 5;int n;
LL ans;void pr(bool pr) {cout << (pr ? "Yes" : "No") << '\n';
}int main() {cin >> n;for (int i = 1; i <= n; i++) {ans += (n - 1) / i;}cout << ans << '\n';return 0;
}
D - Leaping Tak
考虑 dp。
令 \(dp_i\) 为到 \(i\) 个点的方案数。因为 \(k \le 10\)，我们可以存下 \(dp\) 数组的前缀和，对于每个区间直接调用前缀和计算和即可
// BLuemoon
#include <bits/stdc++.h>using namespace std;
using LL = long long;
using DB = double;const int kMaxN = 2e5 + 5;
const LL kP = 998244353;int n, k, l[kMaxN], r[kMaxN];
LL s[kMaxN], dp[kMaxN];
bitset<kMaxN> v;void pr(bool pr) {cout << (pr ? "Yes" : "No") << '\n';
}int main() {cin >> n >> k;for (int i = 1; i <= k; i++) {cin >> l[i] >> r[i];}dp[1] = s[1] = 1;for (int i = 2; i <= n; i++) {for (int j = 1; j <= k; j++) {(dp[i] += s[max(0, i - l[j])] - s[max(0, i - r[j] - 1)]) %= kP;(dp[i] += kP) %= kP;}s[i] = (s[i - 1] + dp[i]) % kP;}cout << dp[n] << '\n';return 0;
}
E - Sequence Sum
This
F - Simplified Reversi
考虑人类智慧
我们存下当前行/列会改到哪个点，直接更改答案即可，因为只有需要更新时才会进行 fill，所以时间复杂度为 \(O(n+q)\)
// BLuemoon
#include <bits/stdc++.h>using namespace std;
using LL = long long;
using DB = double;const int kMaxN = 2e5 + 5;int n, q, r[kMaxN], c[kMaxN], f, L, R, op, g;
LL ans;void pr(bool pr) {cout << (pr ? "Yes" : "No") << '\n';
}int main() {for (cin >> n >> q, ans = 1ll * (n - 2) * (n - 2); q; q--) {(f == 0) && (fill(r + 1, r + n + 1, n), fill(c + 1, c + n + 1, n), f = 1, L = R = n);cin >> op >> g;if (op == 1) {(g < L) && (fill(r + g, r + L + 1, R), L = g), ans -= r[g] - 2;} else {(g < R) && (fill(c + g, c + R + 1, L), R = g), ans -= c[g] - 2;}}cout << ans << '\n';return 0;
}
