2019 ICPC Asia Xuzhou Regional
目录
Contest Info
| Solved | A | B | C | D | E | F | G | H | I | J | K | L | M |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 8/13 | O | - | O | - | O | O | - | O | - | O | - | O | O |
- O 在比赛中通过
- Ø 赛后通过
- ! 尝试了但是失败了
- - 没有尝试
Solutions
A. Cat
题意：
每次询问给出\(L, R, S\)，要求找一个最长的连续区间\(l, r\)，满足\(l \oplus (l + 1) \oplus, \cdots, \oplus r <= S\)。
思路：
考虑\(4k \oplus (4k + 1) \oplus (4k + 2) \oplus (4k +3) = 0\)
那么我们枚举一下头，枚举一下尾，暴力判断一下即可。
代码：
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll L, R, S;
ll gao(ll l, ll r) {
if (l > r) return INF;
ll res = 0;
if (r - l + 1 <= 10) {
for (ll i = l; i <= r; ++i) {
res ^= i;
}
} else {
ll ql = l, qr = r;
while (ql % 4 != 0) {
res ^= ql;
ql++;
}
while (qr % 4 != 3) {
res ^= qr;
qr--;
}
}
return res;
}
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%lld %lld %lld", &L, &R, &S);
ll res = -1;
for (int i = 0; i <= 4; ++i) {
for (int j = 0; j <= 4; ++j) {
ll tmp = gao(L + i, R - j);
if (tmp <= S) res = max(res, (R - j) - (L + i) + 1);
}
}
printf("%lld\n", res);
}
return 0;
}
B. Cats line up
题意：
给出\(n\)个数，问有多少个排列使得任意相邻两个数的差距小于等于\(K(1 \leq K \leq 3)\)
C. < 3 numbers
题意：
令\(x\)为区间\([L, R]\)内素数个数，每次询问给出\([L, R]\)，判断下式是否成立：
\[\begin{eqnarray*}
\frac{x}{R - L + 1} < \frac{1}{3}
\end{eqnarray*}
\]
思路：
考虑素数密度，大区间直接'Yes'
小区间暴力判断
代码：
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int L, R, pri[N], check[N];
void sieve() {
memset(check, 0, sizeof check);
for (int i = 2; i < N; ++i) {
if (!check[i]) {
pri[++*pri] = i;
}
for (int j = 1; j <= *pri; ++j) {
if (1ll * i * pri[j] >= N) break;
check[i * pri[j]] = 1;
if (i % pri[j] == 0) break;
}
}
}
bool prime(int x) {
if (x < N) return !check[x];
for (int i = 2; 1ll * i * i <= x; ++i) {
if (x % i == 0)
return false;
}
return true;
}
bool ok(int l, int r) {
int tot = r - l + 1;
int p = 0;
for (int i = l; i <= r; ++i) {
if (prime(i)) {
++p;
}
}
return p * 3 < tot;
}
int main() {
sieve();
int _T; scanf("%d", &_T);
while (_T--) {
scanf("%d%d", &L, &R);
if (R - L + 1 > 60) {
puts("Yes");
} else {
puts(ok(L, R) ? "Yes" : "No");
}
}
return 0;
}
E. Multiply
题意：
给出\(n\)个数\(a_i\)，令\(Z = a_1! \times a_2! \times \cdots \times a_n!\)
现在给出\(X, Y\)，令\(b_i = Z \times X^i\)，它想要一个最大的\(i\)，使得\(b_i \;|\; Y!\)
思路：
考虑分解\(X\)得到它的所有素因子及幂次。
然后找出其每个因子在\(\frac{Y!}{Z}\)中还剩多少个。
然后贪心拼\(X\)就可以了
代码：
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template<class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << " "; err(args...); }
const int N = 2e5 + 10, INF = 0x3f3f3f3f;
int n; ll x, y, a[N], f[N];
mt19937 rd(time(0));
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
//ll mul(ll a, ll b, ll p) {
// return (a * b - (ll)(a / (long double)p * b + 1e-3) * p + p) % p;
//}
ll mul(ll a, ll b, ll p) {
return (__int128)a * b % p;
}
ll qpow(ll base, ll n, ll p) {
ll res = 1;
base %= p;
while (n) {
if (n & 1) {
res = mul(res, base, p);
}
base = mul(base, base, p);
n >>= 1;
}
return res;
}
struct Mill {
ll n, fac[22000][2], bk[22000]; int tot;
const int C = 2307;
const int S = 8;
bool check(ll a, ll n) {
ll m = n - 1, x, y = 0;
int j = 0;
while (!(m & 1)) {
m >>= 1;
++j;
}
x = qpow(a, m, n);
for (int i = 1; i <= j; x = y, ++i) {
y = mul(x, x, n);
if (y == 1 && x != 1 && x != n - 1) {
return 1;
}
}
return y != 1;
}
bool miller_rabin(ll n) {
if (n < 2) {
return 0;
} else if (n == 2) {
return 1;
} else if (!(n & 1)) {
return 0;
}
for (int i = 0; i < S; ++i) {
if (check(rd() % (n - 1) + 1, n)) {
return 0;
}
}
return 1;
}
ll pollard_rho(ll n, int c) {
ll i = 1, k = 2, x = rd() % n, y = x, d;
while (1) {
++i; x = (mul(x, x, n) + c) % n;
d = gcd(y - x, n);
if (d > 1 && d < n) {
return d;
}
if (y == x) {
return n;
}
if (i == k) {
y = x;
k <<= 1;
}
}
}
void findfac(ll n, int c) {
if (n == 1) {
return;
}
if (miller_rabin(n)) {
bk[++*bk] = n;
return;
}
ll m = n;
while (m == n) {
m = pollard_rho(n, c--);
}
findfac(m, c);
findfac(n / m, c);
}
void gao(ll _n) {
n = _n; *bk = 0;
findfac(n, C);
sort(bk + 1, bk + 1 + *bk);
fac[1][0] = bk[1];
fac[1][1] = 1;
tot = 1;
for (int i = 2; i <= *bk; ++i) {
if (bk[i] == bk[i - 1]) {
++fac[tot][1];
} else {
++tot;
fac[tot][0] = bk[i];
fac[tot][1] = 1;
}
}
}
}mill;
int main() {
int _T; cin >> _T;
while (_T--) {
scanf("%d%lld%lld", &n, &x, &y);
for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
mill.gao(x);
int tot = mill.tot;
for (int i = 1; i <= tot; ++i) f[i] = 0;
ll res = 8e18;
for (int i = 1; i <= tot; ++i) {
// dbg(mill.fac[i][0], mill.fac[i][1]);
ll now = 1;
for (int j = 1; now <= y / mill.fac[i][0]; ++j) {
now *= mill.fac[i][0];
f[i] += (y / now);
for (int o = 1; o <= n; ++o) {
f[i] -= (a[o] / now);
}
}
res = min(1ll * res, f[i] / mill.fac[i][1]);
}
printf("%lld\n", res);
}
return 0;
}
F. The Answer to the Ultimate Question of Life, The Universe, and Everything.
题意：
每次询问给出\(x(0 \leq x \leq 200)\)，需要找出\(a, b, c(|a|, |b|, |c| \leq 5000)\)满足\(a^3 + b^3 + c^3 = x\)
思路：
\(x\)只有201个，可以暴力打表。
打表的时候可以枚举两维，第三维直接算。
打表代码：
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll ok(ll need) {
ll l = -5000, r = 5000, res = -12345678;
while (r - l >= 0) {
ll mid = (l + r) >> 1;
ll tmp = mid * mid * mid;
if (tmp == need) {
return mid;
}
if (tmp > need) {
r = mid - 1;
} else {
l = mid + 1;
}
}
return res;
}
bool gao(int x) {
int limit = 5000;
for (ll a = -limit; a <= limit; ++a) {
for (ll b = -limit; b <= limit; ++b) {
ll need = 1ll * x - a * a * a - b * b * b;
ll c = ok(need);
if (abs(c) <= 5000) {
cout << a << " " << b << " " << c << endl;
return true;
}
}
}
return false;
}
int main() {
int cnt = 0;
int Y = -12345678;
for (int i = 0; i <= 200; ++i) {
if (!gao(i)) {
cout << Y << " " << Y << " " << Y << endl;
}
}
cout << cnt << " " << cnt << " " << cnt << endl;
return 0;
}
H. Yuuki and a problem
题意：
给出一个序列\(a_i\)，支持两个操作：
- 将\(a_x\)改成\(y\)
- 询问最小的不能被\(a_l \cdots a_r\)里面的数表示出来的正整数
思路：
考虑没有修改操作怎么做：
主席树权值\(i\)表示\(i\)这个数的和.
然后考虑每次递增上去，假设已经能够表示出\([1, x]\)范围的数，那么我们可以将\([1, x + 1]\)范围内还未加入的数加进去。
这个可以在主席树上查。
并且这个加入次数跟斐波那契列类似，不会很多。
那么有修改，就敲个动态主席树
注意不要把vector当参数传下去，空间要给够
代码：
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template<class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << " "; err(args...); }
const int N = 2e5 + 10;
int n, m, q, a[N], L[510][510], R[510][510], cl, cr;
inline int lowbit(int x) { return x & -x; }
struct SEG {
struct node {
int ls, rs;
ll sum;
void init() { ls = rs = sum = 0; }
}t[N * 80];
int rt[N], tot;
ll res;
int newnode() {
++tot;
t[tot].init();
return tot;
}
void init() { memset(rt, 0, sizeof rt); tot = 0; }
void update(int &rt, int l, int r, int pos, int v) {
if (!rt) rt = newnode();
t[rt].sum += v;
if (l == r) return;
int mid = (l + r) >> 1;
if (pos <= mid) update(t[rt].ls, l, mid, pos, v);
else update(t[rt].rs, mid + 1, r, pos, v);
}
void update(int x, int pos, int v) {
for (; x <= n; x += lowbit(x)) {
update(rt[x], 1, m, pos, v);
}
}
void query(int dep, int l, int r, int ql, int qr) {
if (ql > qr) return;
if (l >= ql && r <= qr) {
for (int i = 1; i <= cl; ++i) res -= t[L[dep][i]].sum;
for (int i = 1; i <= cr; ++i) res += t[R[dep][i]].sum;
return;
}
int mid = (l + r) >> 1;
if (ql <= mid) {
for (int i = 1; i <= cl; ++i) L[dep + 1][i] = t[L[dep][i]].ls;
for (int i = 1; i <= cr; ++i) R[dep + 1][i] = t[R[dep][i]].ls;
query(dep + 1, l, mid, ql, qr);
}
if (qr > mid) {
for (int i = 1; i <= cl; ++i) L[dep + 1][i] = t[L[dep][i]].rs;
for (int i = 1; i <= cr; ++i) R[dep + 1][i] = t[R[dep][i]].rs;
query(dep + 1, mid + 1, r, ql, qr);
}
}
}seg;
int main() {
m = 2e5;
while (scanf("%d%d", &n, &q) != EOF) {
for (int i = 1; i <= n; ++i) scanf("%d", a + i);
seg.init();
for (int i = 1; i <= n; ++i) {
seg.update(i, a[i], a[i]);
}
int op, x, y;
for (int i = 1; i <= q; ++i) {
scanf("%d%d%d", &op, &x, &y);
if (op == 1) {
seg.update(x, a[x], -a[x]);
a[x] = y;
seg.update(x, a[x], a[x]);
} else {
--x;
cl = cr = 0;
for (int j = x; j; j -= lowbit(j)) {
L[0][++cl] = seg.rt[j];
}
for (int j = y; j; j -= lowbit(j)) {
R[0][++cr] = seg.rt[j];
}
ll l = -1, r = 0;
while (1) {
seg.res = 0;
seg.query(0, 1, m, min(1ll * m + 1, l + 2), min(1ll * m, r + 1));
ll tot = seg.res;
// dbg(i, tot, l + 2, r + 1);
if (tot == 0) break;
l = r;
r += tot;
}
printf("%lld\n", r + 1);
}
}
}
return 0;
}
J. Loli, Yen-Jen, and a graph problem
题意：
给出一个完全图，要将这张图分成\(n - 1\)条路径，第\(i\)条路径的长度为\(i\)，并且一条边只能存在于一条路径中。
代码：
#include <bits/stdc++.h>
using namespace std;
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template<class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << " "; err(args...); }
const int N = 1e3 + 10;
int n;
int e[N][N];
int Max[N];
int now;
vector<vector<int> > vec;
void fix(int x, int y) {
e[x][y] = e[y][x] = 1;
}
int find(int x) {
while (e[Max[x]][x]) {
++Max[x];
}
return Max[x];
}
void insert(int x) {
vec[now].push_back(x);
if ((int)vec[now].size() == now + 1) {
now--;
if (now % 2 == 0) vec[now].push_back(x);
}
}
void gao(int x, int y) {
insert(x);
int ny = y + 1;
if (ny > n) return ;
insert(ny);
gao(x + 1, ny);
}
int main() {
while (scanf("%d", &n) != EOF) {
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
e[i][j] = (i == j ? 1 : 0);
}
Max[i] = 1;
}
vec.clear();
vec.resize(n + 1);
if (n & 1) {
for (int cas = n - 1; cas >= 1; --cas) {
vec[cas].push_back(n);
int sze = 1;
while (sze <= cas) {
int nxt = find(vec[cas].end()[-1]);
fix(vec[cas].end()[-1], nxt);
vec[cas].push_back(nxt);
sze++;
}
}
for (int i = 1; i < n; ++i) {
for (int j = 0, sze = vec[i].size(); j < sze; ++j) {
printf("%d%c", vec[i][j], " \n"[j == sze - 1]);
}
}
} else {
now = n - 1;
// insert(2);
// gao(1, 2);
for (int i = 2; i <= n; i += 2) {
insert(i);
gao(1, i);
}
for (int i = 1; i < n; ++i) {
for (int j = 0, sze = vec[i].size(); j < sze; ++j) {
printf("%d%c", vec[i][j], " \n"[j == sze - 1]);
}
}
}
}
return 0;
}
K. K-rectangle
题意：
给出\(n\)个点\((x_i, y_i)(x_i < x_{i + 1}, 0 < y_i)\)
现在你要找若干个矩形覆盖这些点，矩形的底边必须在\(x\)轴上，矩形之间不能有面积交，矩形的花费是\(h(w + k)\)，\(h\)为高，\(w\)为宽，\(k\)为给定参数。
求最小花费。
L. Loli, Yen-Jen, and a cool problem
题意：
给出一个Trie，每次询问给出一个\(x_i, L_i\)，问有多少个结点为起点向上跳\(L - 1\)步连成的长度为\(L\)的字符串和以\(x_i\)为起点连成的字符串相同
这里的字符在点上，不在边上
思路：
多加一个根节点，就能将点上的字符转化成边上的字符
然后Trie上建SAM，每次查找倍增跳深度最深的合法祖先，它的cnt就是答案
代码：
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10, ALP = 26, M = 20;
int n, q, trie_pos[N]; char s[N];
struct SAM {
struct node {
int maxlen, cnt, fa, nx[ALP];
void init() { maxlen = cnt = fa = 0; memset(nx, 0, sizeof nx); }
}t[N << 1];
int tot, c[N << 1], rk[N << 1], fa[N << 1][M];
vector <vector<int>> G;
int newnode() {
++tot;
t[tot].init();
return tot;
}
void init() {
tot = 0;
newnode();
}
int extend(int id, int lst, int cnt) {
int cur = newnode(), p;
t[cur].cnt = cnt;
t[cur].maxlen = t[lst].maxlen + 1;
for (p = lst; p && !t[p].nx[id]; p = t[p].fa) t[p].nx[id] = cur;
if (!p) {
t[cur].fa = 1;
} else {
int q = t[p].nx[id];
if (t[q].maxlen == t[p].maxlen + 1) {
t[cur].fa = q;
} else {
int clone = newnode();
t[clone] = t[q];
t[clone].cnt = 0;
t[clone].maxlen = t[p].maxlen + 1;
for (; p && t[p].nx[id] == q; p = t[p].fa) t[p].nx[id] = clone;
t[cur].fa = t[q].fa = clone;
}
}
return cur;
}
void dfs(int u) {
for (int i = 1; i < M; ++i)
fa[u][i] = fa[fa[u][i - 1]][i - 1];
for (auto &v : G[u]) {
fa[v][0] = u;
dfs(v);
t[u].cnt += t[v].cnt;
}
}
void build() {
memset(c, 0, sizeof c);
for (int i = 1; i <= tot; ++i) c[t[i].maxlen]++;
for (int i = 1; i <= tot; ++i) c[i] += c[i - 1];
// for (int i = 1; i <= tot; ++i) rk[c[t[i].maxlen]--] = i;
// for (int i = tot; i; --i) t[t[rk[i]].fa].cnt += t[rk[i]].cnt;
G.clear(); G.resize(tot + 1);
for (int i = 1; i <= tot; ++i) {
if (t[i].fa) {
G[t[i].fa].push_back(i);
}
}
fa[1][0] = 0;
dfs(1);
}
int query(int x, int len) {
for (int i = M - 1; i >= 0; --i) {
if (t[fa[x][i]].maxlen >= len) {
x = fa[x][i];
}
}
return t[x].cnt;
}
}sam;
vector <vector<int>> G;
struct Trie {
struct node {
int nx[ALP], cnt, sam_pos;
void init() { memset(nx, 0, sizeof nx); cnt = 0; sam_pos = 0; }
}t[N];
int rt, tot;
int newnode() {
++tot;
t[tot].init();
return tot;
}
void init() { tot = 0; rt = newnode(); }
int add(int p, int ch) {
if (!t[p].nx[ch]) {
t[p].nx[ch] = newnode();
}
int now = t[p].nx[ch];
++t[now].cnt;
return now;
}
void dfs(int u) {
for (auto &v : G[u]) {
trie_pos[v] = add(trie_pos[u], s[v] - 'A');
dfs(v);
}
}
void bfs() {
queue <int> q;
q.push(1);
t[1].sam_pos = 1;
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < ALP; ++i) {
if (!t[u].nx[i]) continue;
int now = t[u].nx[i];
t[now].sam_pos = sam.extend(i, t[u].sam_pos, t[now].cnt);
q.push(now);
}
}
}
}trie;
int main() {
while (scanf("%d%d", &n, &q) != EOF) {
G.clear(); G.resize(n + 1);
scanf("%s", s + 1);
trie.init();
trie_pos[0] = 1;
trie_pos[1] = trie.add(trie_pos[0], s[1] - 'A');
for (int u = 2, v; u <= n; ++u) {
scanf("%d", &v);
G[v].push_back(u);
}
trie.dfs(1);
sam.init(); trie.bfs(); sam.build();
int x, len;
while (q--) {
scanf("%d%d", &x, &len);
printf("%d\n", sam.query(trie.t[trie_pos[x]].sam_pos, len));
}
}
return 0;
}
M. Kill the tree
题意：
给出一棵有根树，根节点为\(1\)，定义\(d(u, v)\)为\(u\)到\(v\)简单路径的长度，\(c(w) = \sum_{v \in T} d(w, v)\)，定义结点\(w\)为一棵树\(T\)的'critical point'，当且仅当\(c(w) \leq min_{u \in T} c(u)\)
现在要对于每个\(i \in [1, n]\)，要找出以\(i\)为根节点的子树中的'critical point'
思路：
猜想'critical point'就是重心。
问题转化成找重心。
我们考虑假设\(x\)的子树中的所有重心都找出来了：
- 那么\(x\)子树的重心不可能在它的轻儿子的子树中，因为选\(x\)本身肯定更优
- 并且\(x\)的重心不可能在它重儿子重心的子树中。
- 只可能在其重儿子的重心到\(x\)的路径上
暴力判断一下即可，每个点只会被判断一次。
但是这里要求输出所有可能的点，考虑重心最多只有两个，并且是连着的，那么我们找到深度最大的一个，另一个如果存在的话，那么就是它父亲，判一下即可。
代码：
#include <bits/stdc++.h>
using namespace std;
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template<class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << " "; err(args...); }
const int N = 2e5 + 10;
int n;
vector<vector<int> >G;
int sze[N], fa[N], son[N], res[N];
void dfs(int u, int pre = 0) {
sze[u] = 1;
fa[u] = pre;
for (auto &v : G[u]) {
if (v == pre) continue;
dfs(v, u);
sze[u] += sze[v];
if (son[u] == -1 || sze[son[u]] < sze[v]) {
son[u] = v;
}
}
if (son[u] == -1) son[u] = u;
}
void gao(int u) {
if (sze[u] == 1) {
res[u] = u;
return ;
}
for (auto &v : G[u]) {
if (v == fa[u]) continue;
gao(v);
}
int now = res[son[u]];
while (now != u) {
int tmp = max(sze[u] - sze[now], sze[son[now]]);
if (tmp <= sze[u] / 2) {
break;
}
now = fa[now];
}
// assert(max(sze[u] - sze[now], sze[son[now]]) <= sze[u] / 2);
res[u] = now;
}
int main() {
while (scanf("%d", &n) != EOF) {
memset(son, -1, sizeof son);
G.clear();
G.resize(n + 1);
for (int i = 1, u, v; i < n; ++i) {
scanf("%d %d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
dfs(1);
gao(1);
for (int i = 1; i <= n; ++i) {
vector <int> vec;
int x = res[i], y = -1;
vec.push_back(x);
if (x != i) {
y = fa[x];
if (max(sze[i] - sze[y], sze[son[y]]) <= sze[i] / 2) {
vec.push_back(y);
}
}
if (vec.size() > 1 && vec[0] > vec[1]) swap(vec[0], vec[1]);
for (int j = 0, sze = vec.size(); j < sze; ++j)
printf("%d%c", vec[j], " \n"[j == sze - 1]);
}
}
return 0;
}
