Educational Codeforces Round 33 (Rated for Div. 2)
https://codeforces.com/contest/893
昨日vp，今日补完F
D贪心，思路值得学习；
E组合数学推式子，式子不难，关键在于模型抽象
F主席树，调了老半天，关键在于要理解查询函数的含义，确定什么时候能返回。
A. Chess For Three
居然卡了快十分钟，关键是要把模拟的过程想清楚，变量的更新到底是怎么操作的
#include <bits/stdc++.h>
using namespace std;
int find (int a, int b) {
for (int i = 1; i <= 3; i++) {
if (i != a && i != b) return i;
}
return 0;
}
int main () {
int n;
cin >> n;
int c = 3, a = 1, b = 2;
bool suc = true;
while (n --) {
int x;
cin >> x;
if (c == x) suc = false;
if (!suc) continue;
a = x, b = c, c = find (a, b);
//cout << a << ' ' << b << ' ' << c << endl;
}
if (suc) cout << "YES";
else cout << "NO";
}
B. Beautiful Divisors
暴力模拟
#include <bits/stdc++.h>
using namespace std;
bool check (int x) {
if (x == 1) return true;
string s;
//cout << x << ' ';
while (x > 0) {
s += char(x % 2 + '0');
x /= 2;
}
//cout << s << endl;
if (s.size () % 2 == 0) return false;
int k = s.size () / 2;
for (int i = 0; i < k; i++) {
if (s[i] != '0') return false;
}
for (int i = k; i < s.size (); i++) {
if (s[i] != '1') return false;
}
return true;
}
int main () {
vector<int> v;
int n, cnt = 0;
cin >> n;
for (int i = 1; i * i <= n; i++) {
if (n % i == 0) {
v.push_back (i);
if (i != n / i) v.push_back (n / i);
}
}
sort (v.begin (), v.end (), greater<int>());
//for (auto i : v) cout << i << ' ';cout << endl;
for (auto i : v) {
if (check (i)) {
cout << i << endl;
break;
}
}
}
C. Rumor
暴力模拟，累加每个连通块内的最小点
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5 + 5, M = N * 2;
ll ans, n, m, a[N];
int h[N], e[M], ne[M], idx;
bool vis[N];
void add (int a, int b) {
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
ll bfs (int st) {
//vis[st] = true;
ll res = a[st];
queue <int> q;
q.push (st);
//cout << st << ' ';
while (!q.empty ()) {
int t = q.front ();
q.pop ();
if (vis[t]) continue;
vis[t] = true;
for (int i = h[t]; ~i; i = ne[i]) {
int j = e[i];
res = min (res, a[j]);
q.push (j);
//cout << j << ' ';
}
}
//cout << endl;
return res;
}
int main () {
memset (h, -1, sizeof h);
cin >> n >> m;
for (int i = 1; i <= n; i++) cin >> a[i];
while (m --) {
int x, y;
cin >> x >> y;
add (x, y), add (y, x);
}
for (int i = 1; i <= n; i++) {
if (vis[i]) continue;
ans += bfs (i);
}
cout << ans << endl;
}
//累加每个连通块内的最小点
D. Credit Card
学习贪心思路，用两个变量模拟
上下界贪心约束
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5 + 5;
ll n, m, x, l, r, cnt;
int main () {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
cin >> x;
if (cnt == -1) continue;
l += x, r += x;
if (l > m) {
cnt = -1;
continue;
}
r = min (r, m);
if (x == 0) {
if (l < 0) l = 0;
if (r < 0) r = m, cnt++;
}
}
cout << cnt << endl;
}
//把负的最大的变为m
//上下界
E. Counting Arrays
推导过程：https://www.luogu.com.cn/blog/rated/solution-cf893e
关键在于插板法的理解
// LUOGU_RID: 115646202
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e6 + 25, mod = 1e9 + 7; //注意要稍微开大一点,因为是y+d-1
ll fact[N], infact[N], pw[N], ans;
ll qmi(ll a, ll k, ll p){
ll res = 1;
while(k){
if(k & 1)
res = (ll)res * a % p;
a = (ll)a * a % p;
k >>= 1;
}
return res;
}
ll C (int a, int b) {
if (a < b) return 0;
return fact[a] * infact[b] % mod * infact[a - b] % mod;
}
void init () {
pw[0] = 1;
for (int i = 1; i < N; i++) pw[i] = (pw[i-1] * 2) % mod;
fact[0] = infact[0] = 1;
for(int i = 1; i < N; i++){
fact[i] = (ll)fact[i-1] * i % mod;
infact[i] = (ll)infact[i-1] * qmi(i, mod - 2,mod) % mod;
}
}
int main () {
init ();
int q, x, y;
cin >> q;
while (q--) {
cin >> x >> y;
ans = 1;
for (int i = 2; i * i <= x; i++) {
if (x % i == 0) {
int d = 0;
while (x % i == 0) x /= i, d++;
(ans *= C (d + y - 1, y - 1) % mod) %= mod;
}
}
if (x > 1) (ans *= C (y, y - 1) % mod) %= mod;
(ans *= pw[y-1]) %= mod;
cout << ans << endl;
}
//cout << log2 (1e6) << endl;
}
F. Subtree Minimum Query
主席树维护限定区间的最小值
要注意的细节都在注释里面，注意query函数的含义和返回条件
调了快一天，这就是数据结构的魅力
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 100005, M = N * 2;
int a[N], n, q, ans, k, x;
int h[N], e[M], ne[M], idx;
int dep[N], dfn[N], sz[N], idx2;
int p[N], root[N], cnt;
void add (int a, int b) {
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
void dfs (int u, int fa) {
dep[u] = dep[fa] + 1; //节点深度
sz[u] = 1, dfn[u] = ++idx2; //子树大小,dfs序
for (int i = h[u]; ~i; i = ne[i]) {
int j = e[i];
if (j == fa) continue;
dfs (j, u);
sz[u] += sz[j];
}
}
struct Node {
int l, r, cnt;
}st[N * 64]; //4N + NlogN
int insert (int u, int l, int r, int x, int v) { //在版本u节点下标范围为[l,r]的树上插入下标为x,权值为v的点
int p = ++cnt;
st[p] = st[u];
if (l == r) {
st[p].cnt = v;
return p;
}
int mid = l + r >> 1;
if (x <= mid) st[p].l = insert (st[u].l, l, mid, x, v);
else st[p].r = insert (st[u].r, mid + 1, r, x, v);
st[p].cnt = min(st[st[p].l].cnt, st[st[p].r].cnt);
return p;
}
int query (int u, int l, int r, int ql, int qr) { //目前查找的是版本u,节点下标范围为[l,r]的树,要求下标在[ql,qr]范围内
if (l >= ql && r <= qr) return st[u].cnt; //注意该行的判断条件;查找的坐标范围为[ql,qr],故满足就返回,而无需l==r递归到叶节点，因为st[u].cnt已经存储了[l,r]的区间最小值了
int mid = l + r >> 1, res = 1e9;
if (ql <= mid) res = min(res, query (st[u].l, l, mid, ql, qr)); //答案落在左区间
if (qr > mid) res = min(res, query (st[u].r, mid + 1, r, ql, qr)); //答案落在右区间
return res;
}
bool cmp (int x, int y) {
return dep[x] < dep[y];
}
signed main () {
st->cnt = 1e9;
memset (h, -1, sizeof h);
scanf ("%lld%lld", &n, &q);
for (int i = 1; i <= n; i++) scanf ("%lld", &a[i]);
for (int i = 1; i < n; i++) {
int a, b;
scanf ("%lld%lld", &a, &b);
add (a, b), add (b, a);
}
dfs (q, 0);
for (int i = 1; i <= n; i++) p[i] = i; //映射下标,因为后面要根据dfs序来插入
sort (p + 1, p + n + 1, cmp);
for (int i = 1; i <= n; i++) { //按深度建树
int id = p[i]; //当前插入的点真正的下标
root[dep[id]] = insert (root[dep[p[i-1]]], 1, n, dfn[id], a[id]); //版本更新
}
scanf ("%lld", &q);
while (q--) {
scanf ("%lld%lld", &x, &k);
x = (x + ans) % n + 1, k = (k + ans) % n;
int tt = min (dep[x] + k, dep[p[n]]); //tt为当前版本
ans = query (root[tt], 1, n, dfn[x], dfn[x] + sz[x] - 1); //dep[x]版本以前一定没有深度超过dfn[x]的点所以无需使用前缀差
printf ("%lld\n", ans);
}
}
// x k
//强制在线