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    # Code Ton round2 題解 先聲明這篇題解不是我寫的,我只負責把他從簡體中文翻譯成繁體中文XD (~~基本上就是Google翻譯完檢查一下而已~~) (UPD: 我錯了,latex不能複製貼上,要重打好累QQ) ## A Two 0-1 sequences 第一個字串後$m-1$位必須和第二個字串一樣,前面的位只要有一個第二個字串的第一個字符就行。 後來加的題,沒代碼 ## B Luke is a foodie 直接順著掃到第一個不存在合法的 $v$ 的區間給答案 +1,然後從這裡再開始做就行。 :::spoiler Code ```cpp= #include <cstdio> #include <iostream> using namespace std; int n, d; void solve() { cin >> n >> d; int ans = 0, l = 0, r = 2e9; while (n--) { int x; cin >> x; int a = x - d, b = x + d; if (b < l || a > r) { ans++; l = a, r = b; } else l = max(l, a), r = min(r, b); } cout << ans << '\n'; } int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); int t; cin >> t; while (t--) solve(); } ``` ::: ## C Virus 把所有未被感染的連續段拉出來,一定是從長的開始,堵上兩邊,模擬就行。 :::spoiler Code ```cpp= #include <algorithm> #include <cstdio> #include <iostream> #include <vector> using namespace std; const int N = 100005; int n, m, a[N], b[N]; void solve() { cin >> n >> m; for (int i = 1; i <= m; i++) cin >> a[i]; sort(a + 1, a + m + 1); b[1] = a[1] - 1 + n - a[m]; for (int i = 2; i <= m; i++) b[i] = a[i] - a[i - 1] - 1; sort(b + 1, b + m + 1, greater<int>()); int sum = 0; for (int i = 1; i <= m; i++) { b[i] -= 4 * (i - 1); if (b[i] == 1) sum += 1; if (b[i] > 1) sum += b[i] - 1; } cout << n - sum << '\n'; } int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); int t; cin >> t; while (t--) solve(); } ``` ::: ## D Magical Array 感覺挺妙的,Anton 4min 就秒了,外國老哥這方面都挺強。我想了很長時間(至少20min),智商不太夠。 發現對於$s = \sum_{i=1}^{k} ia_i$,操作一不會改變,而操作二會使$s$加上$1$,把每個數組的$s$算一下就行。 :::spoiler Code ```cpp= #include <cstdio> #include <iostream> #include <vector> using namespace std; int n, m; void solve() { cin >> n >> m; long long mn = 1ll << 60, mx = 0; int p = 0; for (int i = 1; i <= n; i++) { long long s = 0; for (int j = 1; j <= m; j++) { long long x; cin >> x; s += x * j; } if (s < mn) mn = s; if (s > mx) mx = s, p = i; } cout << p << ' ' << mx - mn << '\n'; } int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); int t; cin >> t; while (t--) solve(); } ``` ::: ## E Count Seconds 先跑遍拓撲,然後發現$n$輪之後如果一個點有,那麼它能到達的所有點一定有。所以暴力模擬前$n$輪,然後算出最後到達匯點的值就行。 :::spoiler Code ```cpp= #include <cstdio> #include <iostream> #include <vector> using namespace std; const int N = 1005, mod = 998244353; int n, m, a[N], in[N]; bool u[N]; vector<int> to[N]; int add(int x, int y) { return x + y < mod ? x + y : x + y - mod; } bool check() { for (int i = 1; i <= n; i++) if (a[i] > 0) return 1; return 0; } void solve() { static int q[N]; cin >> n >> m; for (int i = 1; i <= n; i++) cin >> a[i]; while (m--) { int x, y; cin >> x >> y; to[x].push_back(y); in[y]++; } int t = 0; while ((++t) <= n) { for (int i = 1; i <= n; i++) if (in[i] == 0 && u[i] == 0) { u[i] = 1, q[t] = i; for (int j : to[i]) in[j]--; break; } } for (int i = 1; i <= n; i++) u[i] = 0; int ans = 0; while (ans <= n && check()) { ans++; for (int i = n; i; i--) if (a[q[i]]) { a[q[i]]--; for (int j : to[q[i]]) a[j]++; } } if (!check()) { cout << ans << '\n'; return; } for (int i = 1; i <= n; i++) { a[q[i]] %= mod; for (int j : to[q[i]]) a[j] = add(a[j], a[q[i]]); } cout << add(ans, a[q[n]]) << '\n'; } void clear() { for (int i = 1; i <= n; i++) to[i].clear(); } int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); int t; cin >> t; while (t--) { solve(); clear(); } } ``` ::: ## F Colouring Game 也挺妙的,完全不會了。 可以發現如果顏色個數不相等,那麼誰的多誰就贏,因為剛開始一定是雙方將一個RB 或BR 刪掉,然後沒了之後再每次刪掉自己的一個顏色,那一定是個數少的人先輸。 不然兩個人個數相等,就是比每次刪一個RB 或BR,看誰先不能操作,就只能刪自己的一個,然後輸了。 把每個RB 相間的段提出來,因為刪的是兩個不同顏色,所以雙方操作集合相同,是個公平博弈。直接對所有長度算出所有sg 就行。 暴力sg 是$O(n^2)$的,但是打表之後能發現一個有個在三十左右的$len$,sg 在$3 \times len$及以後有長為$len$的循環節,所以打前面一些就行。 ~~這份代碼是錯的,因為沒有對$3 \times len$取min,但之前數據太水了過了,懶得改了。~~ :::spoiler Code ```cpp= #include <cstdio> #include <iostream> using namespace std; int n, f[103]; char a[500005]; void prep() { for (int i = 2; i <= 102; i++) { bool u[10] = {0}; for (int j = 1; j != i; j++) u[f[j - 1] ^ f[i - j - 1]] = 1; for (int x = 0;; x++) if (!u[x]) { f[i] = x; break; } } } void solve() { cin >> n >> (a + 1); int s = 0, ans = 0; for (int i = 1, j; i <= n; i = j + 1) { for (j = i; j < n && a[j] != a[j + 1]; j++) ; ans ^= f[j - i + 1]; } for (int i = 1; i <= n; i++) if (a[i] == 'R') s++; else s--; if (s > 0 || (s == 0 && ans)) cout << "Alice\n"; else cout << "Bob\n"; } int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); prep(); int t; cin >> t; while (t--) solve(); } ``` ::: ## G Mio and Lucky Array 忽略一些細節,只講大致思路。 將$a$和$b$的偶數位取反,於是操作變成根據操作開始位置的奇偶性給後面加或減$1,2,...$。然後做二階差分,枚舉一個開始位置,就要求從這里之後$a$和$b$對位相差不超過一,且根據奇偶性是$a_i \leq b_j \leq a_i+1$或$b_j \leq a_i \leq b_j+1$。這個是類似經典的fft 做字符串匹配,可能要設計一個權值,不然平方的話值大於$998244353$可能在模意義下被判為合法。 還有差分之前的兩位是要特殊判的,大概就是開始位置前面操作的帶符號和。$i+1$位和$i$的差告訴了奇數位操作次數- 偶數位操作次數,而告訴了奇數位操作位置和- 偶數位操作位置和。由於操作是連續的,也就是說是$1,2,3,4,...i$都可以操作,所以最後確定了操作次數的差,就可以貪心算出在這個差下$a_i$可能的最小值和最大值,而在這之間奇偶性相同應該都可以取到,判一下$a_i$是否在這個區間裡即可。 ~~這題是後來加的,所以口胡,沒代碼。~~ ## H Game of AI 建立一張有向圖,$i$指向$a_i$,那麼就得到了一個基環樹森林。考慮計算這個基環樹森林的答案,那麼對於環之外的每個點,如果它沒有被孩子佔領,那麼它的孩子一定會它孩子的孩子佔領。於是可以$F_0(x), F_1(x)$設分別表示自己被孩子佔領/不被孩子佔領的樹的生成函數(EGF),那麼則有: $F_0(x) = x(F_0(x) + F_1(x))exp(F_0(x) + F_1(x))$ ($1$) $F_1(x) = xexpF_0(x)$ ($2$) $F_0(x)$遞推即為確定被哪個孩子佔領,$F_1(x)$就是孩子全部被孩子的孩子佔領。 將($2$)帶入($1$)即可得到$F_0(x)$的方程,牛頓迭代即可得到$F_0(x)$,進而得到$F_1(x)$。 現在考慮把樹拼起來變成一棵基環樹。我們設在環上的一個點和它的子樹,它被佔領/不被佔領的生成函數為$G_0(x), G_1(x)$,那麼考慮一個點是被自己的孩子佔領還是被環上的上一個點佔領,生成函數就是$G_0(x)=F_0(x)+F_1(x)$。 考慮環上每一個點,如果某個點沒有被別人佔領那麼它在環上的後一個點一定會被佔領。那麼如果一個點它沒有被別的點佔領,我們就在後面強制拼上一個被佔領的點,即$G_1(x)=F_1(x)+G_0(x)$。 對於環還有一個限制,就是不能環上的每一個點都被它在環上的前一個點佔領。可以證明一個環合法當且僅當這兩個條件成立。 設$G_2(x)=G_0(x)+G_1(x)$。先不考慮第二個限制,那麼一個環的生成函數就是: $\sum_{i=1}^{+\infty} \frac{G_2(x)^i}{i} - G_0(x)$ = $-ln(1-G_2(x)) - G_0(x)$ 減掉$G_0(x)$是因為環上不能只有一個點。 考慮第二個限制,類似上面可以得到違反第二條限制的生成函數是$-ln(1-F_1(x))-F_1(x)$。 兩部分相減得到基環樹的生成函數之後再exp一下就能得到原問題(基環樹森林)的答案了。 時間複雜度$O(nlogn)$。 :::spoiler Code ```cpp= #include <algorithm> #include <cstdio> #include <cstring> #include <iostream> using namespace std; typedef unsigned long long u64; typedef __uint128_t u128; const int o = 18, len = 1 << o, B = 16; int n, fac[len], ifac[len], iv[len], mod; int f[len], ef[len], xef[len], a[len], b[len], h[len]; struct fastmod { u64 b; int m; fastmod(int mod) : b(((u128)1 << 64) / mod), m(mod) {} int reduce(u64 a) { u64 q = ((u128)a * b) >> 64; int r = a - q * m; return r < m ? r : r - m; } } z(998244353); int add(int x, int y) { return x + y < mod ? x + y : x + y - mod; } int sub(int x, int y) { return x < y ? x + mod - y : x - y; } int power(int a, int n) { int tp = 1; while (n) { if (n & 1) tp = z.reduce(1ll * tp * a); a = z.reduce(1ll * a * a), n >>= 1; } return tp; } void prep(int n) { fac[0] = 1; for (int i = 1; i <= n; i++) fac[i] = z.reduce(1ll * fac[i - 1] * i); ifac[n] = power(fac[n], mod - 2); for (int i = n - 1; i != -1; i--) ifac[i] = z.reduce(1ll * ifac[i + 1] * (i + 1)); for (int i = 1; i <= n; i++) iv[i] = z.reduce(1ll * ifac[i] * fac[i - 1]); iv[0] = 1; } namespace poly { int I[len], w[len], r[len], up, l; int findg(int n) { static int a[101]; int cnt = 0, x = n - 1; for (int i = 2; i * i <= x; i++) if (x % i == 0) { a[++cnt] = i; while (x % i == 0) x /= i; } if (x > 1) a[++cnt] = x; for (int g = 2;; g++) { bool ok = 1; for (int i = 1; i <= cnt; i++) if (power(g, (n - 1) / a[i]) == 1) { ok = 0; break; } if (ok) return g; } } void init() { I[0] = 1; const int w0 = power(findg(mod), (mod - 1) >> o); w[len >> 1] = 1; for (int i = (len >> 1) + 1; i != len; i++) w[i] = z.reduce(1ll * w[i - 1] * w0); for (int i = (len >> 1) - 1; i; i--) w[i] = w[i << 1]; for (int i = 0; i != len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (o - 1)); } void ntt(int *a, int n, bool op) { static u64 t[len], x, y; for (int i = 0; i != n; i += 2) { x = a[r[i] >> (o - l)], y = a[r[i + 1] >> (o - l)]; t[i] = x + y, t[i + 1] = x + mod - y; } for (int l = 2; l != n; l <<= 1) { int *k = w + l; for (u64 *f = t; f != t + n; f += l) for (int *j = k; j != k + l; j++, f++) { u64 x = *f, y = z.reduce(f[l] * *j); f[l] = x + mod - y, *f += y; } } if (op) { for (int i = 0, x = mod - (mod >> l); i != n; i++) a[i] = z.reduce(t[i] * x); reverse(a + 1, a + n); } else for (int i = 0; i != n; i++) a[i] = z.reduce(t[i]); } void pre(int n) { l = 32 - __builtin_clz(n), up = 1 << l; } void mul(int *f, int n, int *g, int m, int *h, int q) { static int x[len], y[len]; memcpy(x, f, (n + 1) << 2), memcpy(y, g, (m + 1) << 2); pre(n + m), ntt(x, up, 0), ntt(y, up, 0); for (int i = 0; i < up; i++) h[i] = z.reduce(1ll * x[i] * y[i]); ntt(h, up, 1); memset(x, 0, up << 2), memset(y, 0, up << 2), fill(h + q + 1, h + up, 0); } void div(int *a, int *b, int n, int *f) { static int iv[len], x[len], tmp[len]; static int nb[len << 2], nf[len << 2]; static u64 s0[len], s1[len]; if (n <= 16) { int x = power(b[0], mod - 2); for (int i = 0; i <= n; i++) { u64 s = 0; for (int j = 0; j != i; j++) s += 1ll * f[j] * b[i - j]; f[i] = z.reduce(1ll * x * (a[i] + mod - z.reduce(s))); } return; } int m = 1 << (32 - __builtin_clz(n)); int k = m >> 4, z = k << 1; div(I, b, k - 1, tmp), memcpy(x, a, k << 2); memcpy(iv, tmp, k << 2), memset(tmp, 0, k << 2); pre(z - 1); ntt(iv, up, 0), ntt(x, up, 0); for (int i = 0; i != up; i++) x[i] = ::z.reduce(1ll * x[i] * iv[i]); ntt(x, up, 1); memcpy(f, x, k << 2); memset(x, 0, up << 2); memcpy(nb, b, k << 2); ntt(nb, up, 0); for (int i = 1;; i++) { if (i * k > n) { memset(iv, 0, up << 2); memset(nb, 0, i * z * 4); memset(nf, 0, (i - 1) * z * 4); fill(f + n + 1, f + i * k, 0); break; } memcpy(nb + i * z, b + i * k, k << 2); memcpy(nf + (i - 1) * z, f + (i - 1) * k, k << 2); ntt(nb + i * z, up, 0), ntt(nf + (i - 1) * z, up, 0); for (int l1 = 0; l1 != i; l1++) for (int j = 0; j != up; j++) s0[j] += 1ll * nf[l1 * z + j] * nb[(i - l1) * z + j]; for (int l1 = 0; l1 != i; l1++) for (int j = 0; j != up; j++) s1[j] += 1ll * nf[l1 * z + j] * nb[(i - l1 - 1) * z + j]; for (int j = 0; j != up; j += 2) { x[j] = ::z.reduce(s0[j] + ::z.reduce(s1[j])); x[j + 1] = ::z.reduce(s0[j + 1] + mod - ::z.reduce(s1[j + 1])); s0[j] = s1[j] = s0[j + 1] = s1[j + 1] = 0; } ntt(x, up, 1); memset(x + k, 0, k << 2); for (int j = 0; j != k; j++) x[j] = sub(a[i * k + j], x[j]); ntt(x, up, 0); for (int j = 0; j != up; j++) x[j] = ::z.reduce(1ll * x[j] * iv[j]); ntt(x, up, 1); memcpy(f + i * k, x, k << 2); memset(x, 0, up << 2); } } void dcexp(int *a, int l, int r, int n, int *f, int *g, int *h) { static u64 s[len]; static int tp[len]; if (r - l + 1 <= 32) { for (int i = l; i <= r && i <= n; i++) { u64 s = 0; for (int j = l; j < i; j++) { s += 1ll * f[j] * a[i - j]; if (!(j & 15)) s = z.reduce(s); } f[i] = z.reduce(((u64)f[i] + z.reduce(s)) * iv[i]); } return; } int *tg[B], *th[B]; int len = (r - l + 1) / B, k = 2 * len; for (int i = 0; i < B - 1; i++) tg[i] = g + i * k, th[i] = h + i * k; if (!l) { pre(k - 1); for (int i = 0; i < B - 1; i++) { if ((i + 1) * len > n) break; memcpy(th[i], a + i * len, k << 2); ntt(th[i], k, 0); } } for (int i = 0; i < B; i++) { if (l + i * len > n) break; memset(s, 0, k << 3); for (int j = 0; j != i; j++) for (int t = 0; t != k; t++) s[t] += 1ll * tg[j][t] * th[i - j - 1][t]; for (int t = 0; t != k; t++) tp[t] = z.reduce(s[t]); pre(k - 1), ntt(tp, k, 1); for (int t = 0; t < len; t++) f[l + i * len + t] = add(f[l + i * len + t], tp[t + len]); dcexp(a, l + i * len, l + (i + 1) * len - 1, n, f, g + k * B, h + k * B); if (i != B - 1) { memcpy(tg[i], f + l + i * len, len << 2); pre(k - 1), ntt(tg[i], k, 0); } } memset(tg[0], 0, (k * B) << 2); } void exp(int *a, int n, int *f) { static int x[len << 1], v1[len << 2], v2[len << 2]; for (int i = 1; i <= n; i++) x[i] = z.reduce(1ll * a[i] * i); f[0] = 1, fill(f + 1, f + n + 1, 0); int m = 1 << (32 - __builtin_clz(n)); dcexp(x, 0, m - 1, n, f, v1, v2); memset(x, 0, (n + 1) << 2), fill(f + n + 1, f + m, 0); } } // namespace poly void calc(int n) { static int xef[len], eef[len], a[len], b[len], c[len]; if (n == 2) { f[2] = 1; return; } calc((n + 1) / 2); poly::exp(f, n - 1, xef); for (int i = n; i; i--) xef[i] = xef[i - 1]; xef[0] = 0; for (int i = 1; i <= n; i++) a[i] = add(xef[i], f[i]); poly::exp(a, n - 1, eef); for (int i = n; i; i--) eef[i] = eef[i - 1]; eef[0] = 0; poly::pre(2 * n); int up = poly::up; poly::ntt(a, up, 0), poly::ntt(eef, up, 0); for (int i = 0; i != up; i++) c[i] = z.reduce(1ll * a[i] * eef[i]); poly::ntt(c, up, 1), fill(c + n + 1, c + up, 0); for (int i = 1; i <= n; i++) c[i] = sub(f[i], c[i]); xef[0] = 1; poly::ntt(xef, up, 0); for (int i = 0; i != up; i++) b[i] = z.reduce(1ll * xef[i] * (a[i] + 1)); poly::ntt(b, up, 1); fill(b + n + 1, b + up, 0); poly::ntt(b, up, 0); for (int i = 0; i != up; i++) b[i] = z.reduce(1ll * b[i] * eef[i]); poly::ntt(b, up, 1); fill(b + n + 1, b + up, 0); b[0] = 1; for (int i = 1; i <= n; i++) b[i] = sub(0, b[i]); memset(a, 0, up << 2); poly::div(c, b, n, a); for (int i = 2; i <= n; i++) f[i] = sub(f[i], a[i]); memset(xef, 0, up << 2), memset(eef, 0, up << 2); memset(a, 0, (n + 1) << 2), memset(b, 0, (n + 1) << 2), memset(c, 0, (n + 1) << 2); } int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); cin >> n >> mod, z = fastmod(mod); if (n == 1) { cout << 0; return 0; } poly::init(), prep(n); calc(n); poly::exp(f, n, ef); for (int i = n; i; i--) a[i] = add(f[i], ef[i - 1]); poly::exp(a, n, h); for (int i = n; i; i--) h[i] = h[i - 1]; h[0] = 0; for (int i = 1; i <= n; i++) a[i] = add(f[i], h[i]); for (int i = n; i; i--) xef[i] = ef[i - 1]; xef[0] = 1; poly::mul(a, n, xef, n, a, n); a[0] = 1; for (int i = 1; i <= n; i++) a[i] = sub(0, a[i]); poly::mul(a, n, ef, n, a, n); h[0] = 1; for (int i = 1; i <= n; i++) h[i] = sub(0, h[i]); poly::div(h, a, n, b); for (int i = 1; i <= n; i++) cout << z.reduce(1ll * b[i] * fac[i]) << '\n'; } ``` :::

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