ABC425

C. Rotate and Sum Query

如果没有第一种查询，只要直接做前缀和就行。
即使有第一种查询，也不必真的去移动元素，只要把下标整体偏移一下，记住“原序列中每个元素现在排在第几位”，就能知道想要的区间和在原序列的哪一段！

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;int main() {int n, q;cin >> n >> q;vector<int> a(n);rep(i, n) cin >> a[i];vector<ll> s(n+1);rep(i, n) s[i+1] = s[i]+a[i];int si = 0;rep(qi, q) {int type;cin >> type;if (type == 1) {int c;cin >> c;si = (si+c)%n;}else {int l, r;cin >> l >> r;--l; --r;l = (l+si)%n;r = (r+si)%n;ll ans;if (l <= r) ans = s[r+1]-s[l];else ans = s[n] - (s[l]-s[r+1]);cout << ans << '\n';}}return 0;
}

D. Ulam-Warburton Automaton

模拟
黑格子保持黑色，只有与“上一轮操作中新变黑的格子”相邻的白色格子，才可能在这一轮变成黑色，所以每次只要检查这些格子就行！而上一轮操作中新变黑的格子总数不超过 \(HW\) 个，因此整体时间复杂度为 \(O(HW)\)！

代码实现1

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)using namespace std;
using P = pair<int, int>;const int di[] = {-1, 0, 1, 0};
const int dj[] = {0, 1, 0, -1};int main() {int h, w;cin >> h >> w;vector<string> s(h);rep(i, h) cin >> s[i];const int INF = 1001001001;vector step(h, vector<int>(w, INF));queue<P> q;rep(i, h)rep(j, w) if (s[i][j] == '#') {step[i][j] = 0;q.emplace(i, j);}int ans = 0;while (q.size()) {ans++;auto [i, j] = q.front(); q.pop();int si = step[i][j];rep(v, 4) {int ni = i+di[v], nj = j+dj[v];if (ni < 0 or ni >= h or nj < 0 or nj >= w) continue;if (step[ni][nj] != INF) continue;int cnt = 0;rep(v2, 4) {int mi = ni+di[v2], mj = nj+dj[v2];if (mi < 0 or mi >= h or mj < 0 or mj >= w) continue;if (step[mi][mj] <= si) cnt++;}if (cnt == 1) {step[ni][nj] = si+1;q.emplace(ni, nj);}}}cout << ans << '\n';return 0;
}

代码实现2

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)using namespace std;
using P = pair<int, int>;const int di[] = {-1, 0, 1, 0};
const int dj[] = {0, 1, 0, -1};int main() {int h, w;cin >> h >> w;vector<string> s(h);rep(i, h) cin >> s[i];vector<P> updatedCells;rep(i, h)rep(j, w) if (s[i][j] == '#') {updatedCells.emplace_back(i, j);}int ans = 0;while (updatedCells.size()) {ans += updatedCells.size();vector<P> nextCells;for (auto [i, j] : updatedCells) {rep(v, 4) {int ni = i+di[v], nj = j+dj[v];if (ni < 0 or ni >= h or nj < 0 or nj >= w) continue;if (s[ni][nj] == '#') continue;int cnt = 0;rep(v2, 4) {int mi = ni+di[v2], mj = nj+dj[v2];if (mi < 0 or mi >= h or mj < 0 or mj >= w) continue;if (s[mi][mj] == '#') cnt++;}if (cnt == 1) {nextCells.emplace_back(ni, nj);}}}for (auto [i, j] : nextCells) {s[i][j] = '#';}swap(updatedCells, nextCells);}cout << ans << '\n';return 0;
}

E. Count Sequences 2

答案 \(= \dbinom{C_1}{C_1} \times \dbinom{C_1+C_2}{C_2} \times \dbinom{C_1+C_2+C_3}{C_3} \times \cdots\)
用杨辉三角预处理组合数即可

代码实现

#include <bits/stdc++.h>
#include <atcoder/all>
using namespace atcoder;
#define rep(i, n) for (int i = 0; i < (n); ++i)using namespace std;
using mint = modint;const int M = 5000;
mint comb[M+1][M+1];void solve() {int n;cin >> n;int s = 0;mint ans = 1;rep(i, n) {int c;cin >> c;s += c;ans *= comb[s][c];}cout << ans.val() << '\n';
}int main() {int t, mod;cin >> t >> mod;mint::set_mod(mod);comb[0][0] = 1;rep(i, M)rep(j, i+1) {comb[i+1][j] += comb[i][j];comb[i+1][j+1] += comb[i][j];}while (t--) solve();return 0;
}

F. Inserting Process

首先，这是一个统计逐字符删除操作序列数量的问题。删除方式不同却得到相同字符串，仅在连续相同字符区间删除位置不同才会发生。因此，把“在连续相同字符区间内只能选最前面的字符”作为条件，然后以当前剩余字符作为状态做状压dp即可！

代码实现

#include <bits/stdc++.h>
#include <atcoder/all>
using namespace atcoder;
#define rep(i, n) for (int i = 0; i < (n); ++i)using namespace std;
using mint = modint998244353;int main() {int n;string t;cin >> n >> t;int n2 = 1<<n;vector<mint> dp(n2);dp[n2-1] = 1;for (int s = n2-1; s; --s) {char pre = '-';rep(i, n) if (s>>i&1) {if (pre != t[i]) dp[s^1<<i] += dp[s];pre = t[i];}}mint ans = dp[0];cout << ans.val() << '\n';return 0;
}