We were waiting several weeks to setting this contest and hope the problem was good enough.↵
#### Events↵
↵
There was a difficulty with [problem:805E]/[problem:804C]. A little bug in the checker fortunately yields accepting an incorrect solution of only one person during the contest. I should apologize all of you because of this.↵
↵
For this sentence, `Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph.` You can find the meaning of "connected subgraph" with [connected](http://mathworld.wolfram.com/ConnectedGraph.html) and [subgraph](http://www.edmath.org/MATtours/discrete/concepts/csubgr.html), thus it can be empty as it is more logical. How ever, I should apologize all of the participants because of weak sample tests in the statement.↵
↵
I will write the full editorial in the few next days, now some hints and short solutions exist here.↵
Hints↵
------------------↵
[problem:805A]↵
↵
<spoiler summary="hint1">↵
Almost half of numbers are divisible by two.↵
</spoiler>↵
↵
[problem:805B]↵
↵
<spoiler summary="hint1">↵
For the string shouldn't be any $i \leq n-2$, $s_i = s_{i+2}$. Make some examples of little sizes.↵
</spoiler>↵
↵
[problem:805C] / [problem:804A]↵
↵
<spoiler summary="hint1">↵
Find minimum weighted edges.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
There is some edges with weight 0, try to connect them in a carefully way.↵
</spoiler>↵
↵
[problem:805D] / [problem:804B]↵
↵
<spoiler summary="hint1">↵
The last situation is some $'a'$ characters after some $'b'$ ones.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
The last situation is unique.↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
The number of steps is also unique.↵
</spoiler>↵
↵
↵
<spoiler summary="hint4">↵
Each $'b'$ character makes a number of $'b'$ characters in the last situation according to the number of $'a'$ characters before it. ↵
</spoiler>↵
↵
[problem:805E] / [problem:804C]↵
↵
<spoiler summary="hint1">↵
We want to color the ice creams in a way that all of ice creams in a vertex would be different with this condition that if two vertices $u$ and $v$ have ice cream $i$, then all of vertices in the unique path of $u$ and $v$ would have it.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
Let's have a trivial lower-bound for the answer.↵
↵
<spoiler summary="answer">↵
The trivial lower-bound should be maximum size of vertices' sets.↵
</spoiler>↵
↵
</spoiler>↵
↵
<spoiler summary="hint3">↵
You can prove that by induction on leaves of the tree. And code it how you proved it. ↵
</spoiler>↵
↵
[problem:805F] / [problem:804D]↵
↵
<spoiler summary="hint1">↵
Let's solve the problem for just two tree. ↵
↵
<spoiler summary="Trivail solution">↵
$d_t$ is diameter of $t$-th tree and $f_i$ is the maximum path starting from vertex $i$, for all valid $i$ and $j$, assume that the edge is between them and find the diameter with $\max(f_i + f_j + 1, \max(d_1, d_2))$.↵
</spoiler>↵
↵
↵
consider this trivial $\mathcal{O}{(n^2)}$ solution and try to improve it to $\mathcal{O}{(n\cdot \log{(n)})}$.↵
</spoiler>↵
↵
<spoiler summary="hint2">↵
$$\sum_{i=1}^{n} sz_i = n$$↵
↵
<spoiler summary="Actually">↵
seems SQRT. ↵
↵
<spoiler summary="How?">↵
For every new query, try to do it with the complexity of $\mathcal{O}{( \min(sz_u, sz_v) \cdot log{(\max{( sz_{u} , sz_{v} )}}) )}$ , which $sz_i$ means size of the components of $i$-th vertex. Prove its complexity is $\mathcal{O}{(q \cdot \sqrt{(n)} \cdot \log{(n)})}$.↵
</spoiler>↵
↵
</spoiler>↵
↵
</spoiler>↵
↵
[problem:804E]↵
↵
<spoiler summary="hint1">↵
You can prove, we need even number of swaps to reach the same permutation. So ${n \choose 2} \mod 2 = 0$ and the answer for $4 \cdot k + 3$ and $4 \cdot k+4$ is -1.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
The solution almost is constructive.↵
↵
<spoiler summary="Try to solve it for n = 4">↵
And then solve $n=8$ by $n=4$.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
Let's partition $n=4\cdot k$ numbers to classes, each class contains a $4$ consecutive numbers. And when $n=4\cdot k+1$ there is a class with $5$ consecutive numbers.↵
</spoiler>↵
↵
[problem:804F]↵
↵
<spoiler summary="hint1">↵
The problem consists two parts, the first part is to find $mn_i$ and $mx_i$, the minimum and the maximum score $i$-th vertex may get after the score distribution. The second part is to finding the answer and all your need to solve this part is those two arrays. It is trivial $mn_i$ is number of real bullion of golds in $i$-th vertex.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
Let's find the scores for two vertices $u$ and $v$, $u$ have a directed edge to $v$.↵
↵
<spoiler summary="u -> v">↵
$$g = \gcd(s_u, s_v)$$↵
If $i$-th element from $u$-th vertex was painted and $j$-th element from $v$-th vertex was unpainted such that $i \mod g \equiv j \mod g$ then $j$-th element from $v$-th vertex will be painted quickly. For two vertex, we can do it in complexity $\mathcal{O}{(s_u+s_v)}$.↵
</spoiler>↵
↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
If we have this directed walk ${w_1, w_2, \... , w_k}$, you should affect $v_1$ on $v_k$ how I said in the previous hint with $g = \gcd( s_{w_1}, s_{w_2}, ... , s_{w_k} )$.↵
↵
<spoiler summary="What about strogly connected components.">↵
$$g = \gcd( s_{1} , s_{2} , ... , s_{n} )$$↵
In this components if $i$-th element from $u$-th vertex was painted then for all of $j$ such that $i \mod g \equiv j \mod g$, $j$-th element from all vertices will be painted quickly. For this components, we can do it with complexity $\mathcal{O}{({\sum_{i=1}^{n} s_i})}$. In the main problem, we can suppose all of a strongly conected component a vertex with size $g$ in this way and we can calculate $mx_i$ with answer of its component's vertex multiplied by $\frac{s_i}{g}$.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint4">↵
SCC of the tournament yields a Hamiltonian path of strongly connected components( if we consider each component a vertex ). The problem decreased to solve this path. we can solve the path with the complexity of sum of vertices' sizes.↵
</spoiler>↵
↵
↵
<spoiler summary="hint5">↵
Now, for every vertex we have minimum and maximum score it may get. For B known vertices, to check if they can be the wanted set, we can assume their score be maximum possible and the other scores be minimum possible. If they were top, then they can be a sample of the wanted set.↵
</spoiler>↵
↵
↵
<spoiler summary="hint6">↵
Consider a sorted sequence of combination of two array $mn$ and $mx$. ↵
↵
For every vertex, consider the number of ways it(with its maximum score) can be the minimum score of the set of B vertices.↵
</spoiler>↵
↵
↵
Solutions↵
==================↵
[tutorial:805A]↵
Time Complexity: $\mathcal{O}{(1)}$↵
↵
Memory complexity: $\mathcal{O}{(1)}$↵
↵
<spoiler summary="python3">↵
~~~~~↵
l,r=list(map(int,input().split()))↵
print(2 if l<r else l)↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
int main(){↵
int l, r;↵
scanf("%d%d", &l, &r);↵
printf("%d", l == r ? l : 2);↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805B]↵
Time Complexity: $\mathcal{O}{(n)}$↵
↵
Memory complexity: $\mathcal{O}{(n)}$↵
↵
<spoiler summary="python3">↵
~~~~~↵
print(("aabb" * 50000)[:int(input())])↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
int N;↵
int main()↵
{↵
scanf("%d", &N);↵
for (int i = 0; i < N; i++)↵
putchar(i & 2 ? 'b' : 'a');↵
puts("");↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805C]↵
Time Complexity: $\mathcal{O}{(1)}$↵
↵
Memory complexity: $\mathcal{O}{(1)}$↵
↵
<spoiler summary="python3">↵
~~~~~↵
print((int(input())-1)//2)↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
int main(){↵
int n;↵
scanf("%d", &n);↵
printf("%d\n", (n-1)/2);↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805D]↵
↵
Time Complexity: $\mathcal{O}{(n)}$↵
↵
Memory complexity: $\mathcal{O}{(n)}$↵
↵
<spoiler summary="python3">↵
~~~~~↵
s = input()↵
cnt = 0↵
m=10**9 + 7↵
t = 0↵
↵
for i in range(len(s)):↵
if s[~i] == 'a':↵
cnt = (cnt+t)%m↵
t = (t*2)%m↵
else:↵
t += 1↵
print(cnt)↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
int c,d,i,n,m,k,x,j=1000000007;↵
string s;↵
main(){↵
cin>>s;↵
for(i=s.size()-1;i>=0;i--){↵
if(s[i]=='b')c++;else{↵
k+=c;c*=2;k%=j;c%=j;↵
}↵
}↵
cout<<k;↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805E]↵
Time Complexity: $\mathcal{O}{(n + \sum_i s_i \cdot \log(\sum_i s_i))}$↵
↵
Memory complexity: $\mathcal{O}{(n + \sum_i si)}$↵
↵
<spoiler summary="C++">↵
~~~~~↵
// +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+ \\↵
↵
int const N = 3e5 + 10 ;↵
int n , m , ans = 1 , res[N] ;↵
vector <int> g[N] , vec[N] ;↵
↵
inline bool cmp (int a , int b) {↵
return res[a] < res[b] ;↵
}↵
↵
void dfs (int v , int par = -1) {↵
int col = ans , p = (int)vec[v].size() - 1 ;↵
↵
sort(vec[v].begin() , vec[v].end() , cmp) ;↵
↵
for (int x : vec[v]) {↵
if (res[x]) {↵
continue ;↵
}↵
↵
while (p >= 0 && col == res[vec[v][p]]) { ↵
col -- ;↵
p -- ;↵
}↵
↵
res[x] = col ;↵
col -- ;↵
↵
}↵
↵
for (int u : g[v]) {↵
if (u != par) {↵
dfs(u , v) ;↵
}↵
}↵
}↵
↵
int main(){↵
ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;↵
↵
cin >> n >> m ;↵
↵
for (int i = 0 ; i < n ; i ++) {↵
int sz ;↵
cin >> sz ;↵
↵
ans = max(ans , sz) ;↵
↵
while (sz --) {↵
int x ;↵
cin >> x ;↵
x -- ;↵
vec[i].push_back(x) ;↵
}↵
}↵
↵
for (int i = 0 ; i < n - 1 ; i ++) {↵
int u , v ;↵
cin >> u >> v ;↵
u -- , v -- ;↵
↵
g[u].push_back(v) ;↵
g[v].push_back(u) ;↵
}↵
↵
dfs(0) ;↵
↵
cout << ans << '\n' ;↵
↵
for (int i = 0 ; i < m ; i ++) {↵
cout << max(res[i] , 1) << ' ' ;↵
}↵
cout << '\n' ;↵
}↵
~~~~~↵
</spoiler>↵
↵
Time Complexity: $\mathcal{O}{(n + \sum_i s_i)}$↵
↵
Memory complexity: $\mathcal{O}{(n + \sum_i si)}$↵
↵
<spoiler summary="C++">↵
~~~~~↵
vector <int> Vs[300050];↵
vector <int> conn[300050];↵
↵
int ans[300050];↵
bool chk[500050];↵
bool dchk[300050];↵
vector <int> Vu;↵
void DFS(int n) {↵
dchk[n] = true;↵
↵
for (auto it : Vs[n]) {↵
if (ans[it]) chk[ans[it]] = true;↵
else Vu.push_back(it);↵
}↵
↵
int a = 1;↵
for (auto it : Vu) {↵
while (chk[a]) a++;↵
ans[it] = a++;↵
}↵
for (auto it : Vs[n]) chk[ans[it]] = false;↵
Vu.clear();↵
↵
for (auto it : conn[n]) {↵
if (dchk[it]) continue;↵
DFS(it);↵
}↵
}↵
int main() {↵
int N, M, i;↵
scanf("%d %d", &N, &M);↵
for (i = 1; i <= N; i++) {↵
int t1, t2;↵
scanf("%d", &t1);↵
while (t1--) {↵
scanf("%d", &t2);↵
Vs[i].push_back(t2);↵
}↵
}↵
for (i = 1; i < N; i++) {↵
int t1, t2;↵
scanf("%d %d", &t1, &t2);↵
conn[t1].push_back(t2);↵
conn[t2].push_back(t1);↵
}↵
DFS(1);↵
↵
int mx = 0;↵
for (i = 1; i <= M; i++) {↵
mx = max(mx, ans[i]);↵
if (ans[i] == 0) ans[i] = 1;↵
}↵
mx = max(mx, 1);↵
printf("%d\n", mx);↵
for (i = 1; i <= M; i++) printf("%d ", ans[i]);↵
↵
return !printf("\n");↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805F]↵
↵
Time Complexity: $\mathcal{O}{(q \cdot \sqrt(n) \cdot \log(n) + n \cdot \sqrt(n) \cdot \log(n) )}$↵
↵
Memory complexity: $\mathcal{O}{(n)}$↵
↵
<spoiler summary="C++">↵
~~~~~↵
using namespace std ;↵
↵
#define int long long↵
↵
const int MAXN = 1e5 + 100 ;↵
↵
vector<int>ver[MAXN] , com[MAXN] , ps[MAXN] ; ↵
int vis[MAXN] , dis[MAXN] ; ↵
int mxr , mxid ; ↵
void dfs(int v , int col = -1 , int h = 0 , int par = -1){↵
vis[v] = max(vis[v] , col) ; ↵
if(mxr < h)mxid = v , mxr = h ; ↵
dis[v] = max(h , dis[v]) ; ↵
if(col > 0)com[col] . push_back(dis[v]) ; ↵
for(auto u : ver[v]){↵
if(u == par)continue ; ↵
dfs(u , col , h + 1 , v) ; ↵
}↵
}↵
map<pair<int , int> , int> check ;↵
int32_t main(){↵
ios_base::sync_with_stdio(0) ;↵
cin . tie(0) ; cout . tie(0) ;↵
int n , m , q ; cin >> n >> m >> q ; ↵
for(int i = 0 ; i < m ; i ++){↵
int x , y ; cin >> x >> y ; ↵
x -- , y -- ; ↵
ver[x] . push_back(y) ; ↵
ver[y] . push_back(x) ;↵
}↵
int cnt = 0 ; ↵
for(int i = 0 ; i < n ; i ++){↵
if(vis[i])continue ; ↵
mxr = 0 , mxid = i ; ↵
dfs(i) ; ↵
mxr = 0 ; ↵
dfs(mxid) ;↵
dfs(mxid , ++ cnt) ; sort(com[cnt] . begin() , com[cnt] . end()) ; ↵
ps[cnt] . push_back(0) ; ↵
for(auto v : com[cnt])ps[cnt] . push_back(ps[cnt] . back() + v) ; ↵
}↵
cout << fixed << setprecision(10) ;↵
while(q --){↵
int x , y ; cin >> x >> y ; ↵
x -- , y -- ; x = vis[x] ; y = vis[y] ; ↵
if(com[x] . size() > com[y] . size())swap(x , y) ; ↵
if(x == y){cout << -1 << '\n' ; continue ; }↵
if(check[{x , y}] > 0){cout << double(check[{x , y}]) / (1ll * com[x] . size() * com[y] . size()) << '\n' ; continue ; }↵
int ans = 0 , mxd = max(com[x] . back() , com[y] . back()) ; ↵
for(auto v : com[x]){↵
int num = lower_bound(com[y] . begin() , com[y] . end() , mxd - v - 1) - com[y] . begin() ; ↵
ans += num * mxd + (com[y] . size() - num) * (v + 1) + ps[y] . back() - ps[y][num] ; ↵
}↵
check[{x , y}] = ans ; ↵
cout << double(check[{x , y}]) / (1ll * com[x] . size() * com[y] . size()) << '\n' ; ↵
}↵
}↵
↵
~~~~~↵
</spoiler>↵
↵
[tutorial:804E]↵
↵
Time Complexity: $\mathcal{O}{(n^2)}$↵
↵
Memory complexity: $\mathcal{O}{(1)}$↵
↵
<spoiler summary="C++">↵
~~~~~↵
int n;↵
↵
int main()↵
{↵
scanf("%d",&n);↵
if(n%4>1)return printf("NO\n"),0;↵
printf("YES\n");↵
for(int i=1;i<n;i+=4)↵
{↵
if(n%4)printf("%d %d\n%d %d\n%d %d\n",i+2,n,i+2,i+3,i+3,n);↵
else printf("%d %d\n",i+2,i+3);↵
printf("%d %d\n",i,i+2);↵
printf("%d %d\n",i+1,i+3);↵
printf("%d %d\n",i+1,i+2);↵
printf("%d %d\n",i,i+3);↵
if(n%4)printf("%d %d\n%d %d\n%d %d\n",i,n,i,i+1,i+1,n);↵
else printf("%d %d\n",i,i+1);↵
}↵
for(int i=1;i<n;i+=4)↵
for(int j=i+4;j<n;j+=4)↵
{↵
printf("%d %d\n",i+3,j+2);↵
printf("%d %d\n",i+2,j+2);↵
printf("%d %d\n",i+3,j+1);↵
printf("%d %d\n",i,j+1);↵
printf("%d %d\n",i+1,j+3);↵
printf("%d %d\n",i+2,j+3);↵
printf("%d %d\n",i+1,j+2);↵
printf("%d %d\n",i+1,j+1);↵
printf("%d %d\n",i+3,j);↵
printf("%d %d\n",i+3,j+3);↵
printf("%d %d\n",i,j+2);↵
printf("%d %d\n",i,j+3);↵
printf("%d %d\n",i+2,j);↵
printf("%d %d\n",i+2,j+1);↵
printf("%d %d\n",i+1,j);↵
printf("%d %d\n",i,j);↵
}↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:804F]↵
Time Complexity: $\mathcal{O}{(\sum_i s_i + n \cdot b)}$↵
↵
Memory complexity: $\mathcal{O}{(n^2 + \sum_i s_i)}$↵
↵
<spoiler summary="Java8">↵
~~~~~↵
import java.io.OutputStream;↵
import java.io.IOException;↵
import java.io.InputStream;↵
import java.io.OutputStream;↵
import java.io.PrintWriter;↵
import java.io.BufferedWriter;↵
import java.util.InputMismatchException;↵
import java.io.IOException;↵
import java.util.Stack;↵
import java.util.ArrayList;↵
import java.util.List;↵
import java.util.stream.Stream;↵
import java.util.Vector;↵
import java.io.Writer;↵
import java.io.OutputStreamWriter;↵
import java.util.BitSet;↵
import java.io.InputStream;↵
↵
/**↵
* Built using CHelper plug-in↵
* Actual solution is at the top↵
*/↵
public class Main {↵
public static void main(String[] args) {↵
InputStream inputStream = System.in;↵
OutputStream outputStream = System.out;↵
InputReader in = new InputReader(inputStream);↵
OutputWriter out = new OutputWriter(outputStream);↵
TaskF solver = new TaskF();↵
solver.solve(1, in, out);↵
out.close();↵
}↵
↵
static class TaskF {↵
public int mod = 1000000007;↵
public int n;↵
public int a;↵
public int b;↵
public char[][] adj;↵
public List<Integer>[] graph;↵
↵
public void solve(int testNumber, InputReader in, OutputWriter out) {↵
n = in.nextInt();↵
a = in.nextInt();↵
b = in.nextInt();↵
adj = new char[n][n];↵
graph = Stream.generate(ArrayList::new).limit(n).toArray(List[]::new);↵
for (int i = 0; i < n; i++) adj[i] = in.next().toCharArray();↵
for (int i = 0; i < n; i++) {↵
for (int j = 0; j < n; j++) {↵
if (adj[i][j] == '1') {↵
graph[i].add(j);↵
}↵
}↵
}↵
List<List<Integer>> comp = new SCCTarjan().scc(graph);↵
boolean[][] can = new boolean[n][];↵
int[] low = new int[n];↵
for (int i = 0; i < n; i++) {↵
int q = in.nextInt();↵
char[] w = in.next().toCharArray();↵
can[i] = new boolean[q];↵
for (int j = 0; j < q; j++) {↵
can[i][j] = w[j] == '1';↵
if (can[i][j]) low[i]++;↵
}↵
}↵
int[] wcomp = new int[n];↵
int idx = 0;↵
List<BitSet> bb = new ArrayList<>();↵
List<Integer> bs = new ArrayList<>();↵
for (List<Integer> gg : comp) {↵
int gcd = 0;↵
for (int x : gg) {↵
gcd = Utils.gcd(gcd, can[x].length);↵
wcomp[x] = idx;↵
}↵
idx++;↵
BitSet d = new BitSet(gcd);↵
bs.add(gcd);↵
bb.add(d);↵
for (int x : gg) {↵
for (int i = 0; i < can[x].length; i++) {↵
if (can[x][i]) {↵
d.set(i % gcd, true);↵
}↵
}↵
}↵
}↵
↵
for (int kk = comp.size() - 1; kk > 0; kk--) {↵
int g = Utils.gcd(bs.get(kk), bs.get(kk - 1));↵
BitSet ww = new BitSet(g);↵
for (int i = 0; i < bs.get(kk); i += g) {↵
ww.or(bb.get(kk).get(i, i + g));↵
}↵
for (int i = 0; i < bs.get(kk - 1); i++) {↵
if (ww.get(i % g)) {↵
bb.get(kk - 1).set(i, true);↵
}↵
}↵
}↵
int[] high = new int[n];↵
for (int i = 0; i < n; i++) {↵
for (int j = 0; j < can[i].length; j++) {↵
if (bb.get(wcomp[i]).get(j % bs.get(wcomp[i])))↵
high[i]++;↵
}↵
}↵
long ret = 0;↵
int[][] comb = Utils.getComb(n + 1, mod);↵
for (int lowest = 0; lowest < n; lowest++) {↵
int countabove = 0;↵
for (int other = 0; other < n; other++) {↵
if (other == lowest) continue;↵
if (low[other] > high[lowest]) {↵
countabove++;↵
}↵
}↵
if (countabove >= a) {↵
continue;↵
}↵
int canbig = 0;↵
for (int other = 0; other < n; other++) {↵
if (other == lowest) continue;↵
if (low[other] <= high[lowest] && (high[other] > high[lowest] || (high[other] == high[lowest] && other > lowest))) {↵
canbig++;↵
}↵
}↵
↵
for (int take = 0; take < b && take <= canbig && take + countabove < a; take++) {↵
ret = (ret + 1L * comb[canbig][take] * comb[countabove][b - take - 1]) % mod;↵
}↵
}↵
out.println(ret);↵
}↵
↵
}↵
↵
static class OutputWriter {↵
private final PrintWriter writer;↵
↵
public OutputWriter(OutputStream outputStream) {↵
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));↵
}↵
↵
public OutputWriter(Writer writer) {↵
this.writer = new PrintWriter(writer);↵
}↵
↵
public void close() {↵
writer.close();↵
}↵
↵
public void println(long i) {↵
writer.println(i);↵
}↵
↵
}↵
↵
static class Utils {↵
public static int gcd(int a, int b) {↵
return b == 0 ? a : gcd(b, a % b);↵
}↵
↵
public static int[][] getComb(int sz, int mod) {↵
int[][] comb = new int[sz][sz];↵
for (int i = 0; i < sz; i++) {↵
comb[i][0] = 1;↵
for (int j = 1; j <= i; j++) {↵
comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];↵
if (comb[i][j] >= mod) comb[i][j] -= mod;↵
}↵
}↵
return comb;↵
}↵
↵
}↵
↵
static class InputReader {↵
private InputStream stream;↵
private byte[] buf = new byte[1024];↵
private int curChar;↵
private int numChars;↵
↵
public InputReader(InputStream stream) {↵
this.stream = stream;↵
}↵
↵
public int read() {↵
if (this.numChars == -1) {↵
throw new InputMismatchException();↵
} else {↵
if (this.curChar >= this.numChars) {↵
this.curChar = 0;↵
↵
try {↵
this.numChars = this.stream.read(this.buf);↵
} catch (IOException var2) {↵
throw new InputMismatchException();↵
}↵
↵
if (this.numChars <= 0) {↵
return -1;↵
}↵
}↵
↵
return this.buf[this.curChar++];↵
}↵
}↵
↵
public int nextInt() {↵
int c;↵
for (c = this.read(); isSpaceChar(c); c = this.read()) {↵
;↵
}↵
↵
byte sgn = 1;↵
if (c == 45) {↵
sgn = -1;↵
c = this.read();↵
}↵
↵
int res = 0;↵
↵
while (c >= 48 && c <= 57) {↵
res *= 10;↵
res += c - 48;↵
c = this.read();↵
if (isSpaceChar(c)) {↵
return res * sgn;↵
}↵
}↵
↵
throw new InputMismatchException();↵
}↵
↵
public String next() {↵
int c;↵
while (isSpaceChar(c = this.read())) {↵
;↵
}↵
↵
StringBuilder result = new StringBuilder();↵
result.appendCodePoint(c);↵
↵
while (!isSpaceChar(c = this.read())) {↵
result.appendCodePoint(c);↵
}↵
↵
return result.toString();↵
}↵
↵
public static boolean isSpaceChar(int c) {↵
return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;↵
}↵
↵
}↵
↵
static class SCCTarjan {↵
List<Integer>[] graph;↵
boolean[] visited;↵
Stack<Integer> stack;↵
int time;↵
int[] lowlink;↵
List<List<Integer>> components;↵
↵
public List<List<Integer>> scc(List<Integer>[] graph) {↵
int n = graph.length;↵
this.graph = graph;↵
visited = new boolean[n];↵
stack = new Stack<>();↵
time = 0;↵
lowlink = new int[n];↵
components = new ArrayList<>();↵
↵
for (int u = 0; u < n; u++)↵
if (!visited[u])↵
dfs(u);↵
↵
return components;↵
}↵
↵
void dfs(int u) {↵
lowlink[u] = time++;↵
visited[u] = true;↵
stack.add(u);↵
boolean isComponentRoot = true;↵
↵
for (int v : graph[u]) {↵
if (!visited[v])↵
dfs(v);↵
if (lowlink[u] > lowlink[v]) {↵
lowlink[u] = lowlink[v];↵
isComponentRoot = false;↵
}↵
}↵
↵
if (isComponentRoot) {↵
List<Integer> component = new ArrayList<>();↵
while (true) {↵
int x = stack.pop();↵
component.add(x);↵
lowlink[x] = Integer.MAX_VALUE;↵
if (x == u)↵
break;↵
}↵
components.add(component);↵
}↵
}↵
↵
}↵
}↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
↵
#define pb push_back↵
#define mp make_pair↵
#define all(x) (x).begin(),(x).end()↵
#define fi first↵
#define se second↵
↵
const int maxn = 5005;↵
const int maxelem = 2000006;↵
const int MOD = 1000000007;↵
↵
char gr[maxn][maxn];↵
string init[maxn];↵
int mincnt[maxn], maxcnt[maxn];↵
int c[maxn][maxn];↵
int n, m, A, B;↵
bool was[maxn];↵
vector<int> order;↵
vector<bool> can[maxn];↵
bool nowcan[maxelem];↵
char s[maxelem];↵
int gcd[maxn];↵
vector<int> comp[maxn];↵
↵
void preorder(int cur)↵
{↵
if (was[cur]) return;↵
was[cur] = true;↵
for (int i = 0; i < n; i++) if (gr[cur][i] == '1') preorder(i);↵
order.pb(cur);↵
}↵
↵
void color(int cur, int cc)↵
{↵
if (was[cur]) return;↵
was[cur] = true;↵
gcd[cc] = __gcd(gcd[cc], (int)init[cur].size());↵
comp[cc].pb(cur);↵
for (int i = 0; i < n; i++) if (gr[i][cur] == '1') color(i, cc);↵
}↵
↵
int getc(int n, int k)↵
{↵
if (k < 0 || k > n) return 0;↵
return c[n][k];↵
}↵
↵
int main()↵
{↵
scanf("%d%d%d", &n, &A, &B);↵
for (int i = 0; i < n; i++)↵
{↵
scanf("%s", gr[i]);↵
}↵
for (int i = 0; i < n; i++)↵
{↵
scanf("%*d");↵
scanf("%s", s);↵
init[i] = string(s);↵
for (int j = 0; j < (int)init[i].size(); j++) mincnt[i] += init[i][j] == '1';↵
maxcnt[i] = mincnt[i];↵
}↵
for (int i = 0; i < n; i++) if (!was[i]) preorder(i);↵
memset(was, 0, sizeof(was));↵
int cc = 0;↵
reverse(all(order));↵
for (auto t : order) if (!was[t])↵
{↵
color(t, cc);↵
cc++;↵
}↵
for (int i = 0; i < cc; i++)↵
{↵
can[i].resize(gcd[i]);↵
for (int j = 0; j < gcd[i]; j++) can[i][j] = false;↵
for (auto t : comp[i])↵
{↵
for (int j = 0; j < (int)init[t].size(); j++) if (init[t][j] == '1') can[i][j % gcd[i]] = true;↵
}↵
}↵
for (int i = 0; i < cc - 1; i++)↵
{↵
int g = __gcd(gcd[i], gcd[i + 1]);↵
for (int j = 0; j < g; j++) nowcan[j] = false;↵
for (int j = 0; j < gcd[i]; j++) nowcan[j % g] |= can[i][j];↵
for (int j = 0; j < gcd[i + 1]; j++) can[i + 1][j] = (can[i + 1][j] | nowcan[j % g]);↵
}↵
for (int i = 0; i < cc; i++)↵
{↵
for (auto t : comp[i])↵
{↵
for (int j = 0; j < (int)init[t].size(); j++) if (init[t][j] == '0' && can[i][j % gcd[i]]) maxcnt[t]++;↵
}↵
}↵
↵
c[0][0] = 1;↵
for (int i = 1; i <= n; i++)↵
{↵
c[i][0] = 1;↵
for (int j = 1; j <= i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;↵
}↵
int answer = 0;↵
for (int i = 0; i < n; i++)↵
{↵
int cntbigger = 0;↵
int cntupper = 0;↵
for (int j = 0; j < n; j++) if (i != j)↵
{↵
if (mincnt[j] > maxcnt[i]) cntbigger++;↵
else if (maxcnt[j] > maxcnt[i] || (maxcnt[j] == maxcnt[i] && j > i)) cntupper++;↵
}↵
if (cntbigger >= A) continue;↵
for (int j = max(0, B - 1 - cntbigger); j <= cntupper && j + cntbigger + 1 <= A && j + 1 <= B; j++)↵
{↵
answer = (answer + (ll)getc(cntupper, j) * getc(cntbigger, B - 1 - j)) % MOD;↵
}↵
}↵
cout << answer << endl;↵
return 0;↵
} ↵
↵
~~~~~↵
</spoiler>↵
↵
<spoiler summary="C++">↵
↵
~~~~~↵
#include <bits/stdc++.h>↵
#define xx first↵
#define yy second↵
#define mp make_pair↵
#define pb push_back↵
#define fill( x, y ) memset( x, y, sizeof x )↵
#define copy( x, y ) memcpy( x, y, sizeof x )↵
using namespace std;↵
↵
typedef long long LL;↵
typedef pair < int, int > pa;↵
↵
inline int read()↵
{↵
int sc = 0, f = 1; char ch = getchar();↵
while( ch < '0' || ch > '9' ) { if( ch == '-' ) f = -1; ch = getchar(); }↵
while( ch >= '0' && ch <= '9' ) sc = sc * 10 + ch - '0', ch = getchar();↵
return sc * f;↵
}↵
↵
const int MAXN = 5005;↵
const int MAXM = 2000005;↵
const int mod = 1e9 + 7;↵
↵
int n, a, b, len[MAXN], can[MAXM], id_cnt, L[MAXN], R[MAXN], C[MAXN][MAXN];↵
int dfn[MAXN], scc[MAXN], low[MAXN], tim, num, st[MAXN], top, w[MAXN], ans;↵
vector < int > G[MAXN], id[MAXN], scc_bit[MAXN], v[MAXN];↵
char ch[MAXM];↵
↵
inline void inc(int &x, int y) { x += y; if( x >= mod ) x -= mod; }↵
↵
inline void dfs(int x)↵
{↵
dfn[ x ] = low[ x ] = ++tim; st[ ++top ] = x;↵
for( auto y : G[ x ] )↵
if( !dfn[ y ] ) dfs( y ), low[ x ] = min( low[ x ], low[ y ] );↵
else if( !scc[ y ] ) low[ x ] = min( low[ x ], dfn[ y ] );↵
if( dfn[ x ] == low[ x ] )↵
{↵
num++; int t = 0;↵
while( t ^ x )↵
{↵
scc[ t = st[ top-- ] ] = num;↵
w[ num ] = __gcd( w[ num ], len[ t ] );↵
v[ num ].pb( t );↵
}↵
scc_bit[ num ].resize( w[ num ] );↵
for( auto y : v[ num ] )↵
for( int d = 0 ; d < len[ y ] ; d++ ) if( can[ id[ y ][ d ] ] ) scc_bit[ num ][ d % w[ num ] ] = 1;↵
}↵
}↵
↵
int main()↵
{↵
#ifdef wxh010910↵
freopen( "data.in", "r", stdin );↵
#endif↵
n = read(), a = read(), b = read();↵
for( int i = 0 ; i <= n ; i++ )↵
{↵
C[ i ][ 0 ] = 1;↵
for( int j = 1 ; j <= i ; j++ ) inc( C[ i ][ j ] = C[ i - 1 ][ j ], C[ i - 1 ][ j - 1 ] );↵
}↵
for( int i = 1 ; i <= n ; i++ )↵
{↵
scanf( "%s", ch + 1 );↵
for( int j = 1 ; j <= n ; j++ ) if( ch[ j ] == '1' ) G[ i ].pb( j );↵
}↵
for( int i = 1 ; i <= n ; i++ )↵
{↵
len[ i ] = read(); scanf( "%s", ch );↵
for( int j = 0 ; j < len[ i ] ; j++ ) id[ i ].pb( ++id_cnt ), can[ id_cnt ] = ch[ j ] == '1', L[ i ] += can[ id_cnt ];↵
}↵
for( int i = 1 ; i <= n ; i++ ) if( !dfn[ i ] ) dfs( i );↵
for( int i = num ; i > 1 ; i-- )↵
{↵
int g = __gcd( w[ i ], w[ i - 1 ] );↵
w[ i - 1 ] = g;↵
for( int j = 0 ; j < w[ i ] ; j++ ) scc_bit[ i - 1 ][ j % g ] |= scc_bit[ i ][ j ];↵
}↵
for( int i = 1 ; i <= n ; i++ )↵
{↵
int x = scc[ i ];↵
for( int j = 0 ; j < len[ i ] ; j++ )↵
R[ i ] += scc_bit[ x ][ j % w[ x ] ];↵
}↵
for( int i = 1 ; i <= n ; i++ )↵
{↵
int t1 = 0, t2 = 0;↵
for( int j = 1 ; j <= n ; j++ ) if( L[ j ] > R[ i ] ) t1++;↵
if( t1 >= a ) continue;↵
for( int j = 1 ; j <= n ; j++ ) if( L[ j ] <= R[ i ] && mp( R[ j ], j ) > mp( R[ i ], i ) ) t2++;↵
for( int j = 0 ; j < b && j <= t2 && j + t1 < a ; j++ )↵
inc( ans, 1LL * C[ t2 ][ j ] * C[ t1 ][ b - j - 1 ] % mod ); ↵
}↵
return printf( "%d\n", ans ), 0;↵
}↵
~~~~~↵
↵
↵
</spoiler>↵
↵
#### Events↵
↵
There was a difficulty with [problem:805E]/[problem:804C]. A little bug in the checker fortunately yields accepting an incorrect solution of only one person during the contest. I should apologize all of you because of this.↵
↵
For this sentence, `Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph.` You can find the meaning of "connected subgraph" with [connected](http://mathworld.wolfram.com/ConnectedGraph.html) and [subgraph](http://www.edmath.org/MATtours/discrete/concepts/csubgr.html), thus it can be empty as it is more logical. How ever, I should apologize all of the participants because of weak sample tests in the statement.↵
↵
I will write the full editorial in the few next days, now some hints and short solutions exist here.↵
Hints↵
------------------↵
[problem:805A]↵
↵
<spoiler summary="hint1">↵
Almost half of numbers are divisible by two.↵
</spoiler>↵
↵
[problem:805B]↵
↵
<spoiler summary="hint1">↵
For the string shouldn't be any $i \leq n-2$, $s_i = s_{i+2}$. Make some examples of little sizes.↵
</spoiler>↵
↵
[problem:805C] / [problem:804A]↵
↵
<spoiler summary="hint1">↵
Find minimum weighted edges.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
There is some edges with weight 0, try to connect them in a carefully way.↵
</spoiler>↵
↵
[problem:805D] / [problem:804B]↵
↵
<spoiler summary="hint1">↵
The last situation is some $'a'$ characters after some $'b'$ ones.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
The last situation is unique.↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
The number of steps is also unique.↵
</spoiler>↵
↵
↵
<spoiler summary="hint4">↵
Each $'b'$ character makes a number of $'b'$ characters in the last situation according to the number of $'a'$ characters before it. ↵
</spoiler>↵
↵
[problem:805E] / [problem:804C]↵
↵
<spoiler summary="hint1">↵
We want to color the ice creams in a way that all of ice creams in a vertex would be different with this condition that if two vertices $u$ and $v$ have ice cream $i$, then all of vertices in the unique path of $u$ and $v$ would have it.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
Let's have a trivial lower-bound for the answer.↵
↵
<spoiler summary="answer">↵
The trivial lower-bound should be maximum size of vertices' sets.↵
</spoiler>↵
↵
</spoiler>↵
↵
<spoiler summary="hint3">↵
You can prove that by induction on leaves of the tree. And code it how you proved it. ↵
</spoiler>↵
↵
[problem:805F] / [problem:804D]↵
↵
<spoiler summary="hint1">↵
Let's solve the problem for just two tree. ↵
↵
<spoiler summary="Trivail solution">↵
$d_t$ is diameter of $t$-th tree and $f_i$ is the maximum path starting from vertex $i$, for all valid $i$ and $j$, assume that the edge is between them and find the diameter with $\max(f_i + f_j + 1, \max(d_1, d_2))$.↵
</spoiler>↵
↵
↵
consider this trivial $\mathcal{O}{(n^2)}$ solution and try to improve it to $\mathcal{O}{(n\cdot \log{(n)})}$.↵
</spoiler>↵
↵
<spoiler summary="hint2">↵
$$\sum_{i=1}^{n} sz_i = n$$↵
↵
<spoiler summary="Actually">↵
seems SQRT. ↵
↵
<spoiler summary="How?">↵
For every new query, try to do it with the complexity of $\mathcal{O}{( \min(sz_u, sz_v) \cdot log{(\max{( sz_{u} , sz_{v} )}}) )}$ , which $sz_i$ means size of the components of $i$-th vertex. Prove its complexity is $\mathcal{O}{(q \cdot \sqrt{(n)} \cdot \log{(n)})}$.↵
</spoiler>↵
↵
</spoiler>↵
↵
</spoiler>↵
↵
[problem:804E]↵
↵
<spoiler summary="hint1">↵
You can prove, we need even number of swaps to reach the same permutation. So ${n \choose 2} \mod 2 = 0$ and the answer for $4 \cdot k + 3$ and $4 \cdot k+4$ is -1.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
The solution almost is constructive.↵
↵
<spoiler summary="Try to solve it for n = 4">↵
And then solve $n=8$ by $n=4$.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
Let's partition $n=4\cdot k$ numbers to classes, each class contains a $4$ consecutive numbers. And when $n=4\cdot k+1$ there is a class with $5$ consecutive numbers.↵
</spoiler>↵
↵
[problem:804F]↵
↵
<spoiler summary="hint1">↵
The problem consists two parts, the first part is to find $mn_i$ and $mx_i$, the minimum and the maximum score $i$-th vertex may get after the score distribution. The second part is to finding the answer and all your need to solve this part is those two arrays. It is trivial $mn_i$ is number of real bullion of golds in $i$-th vertex.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
Let's find the scores for two vertices $u$ and $v$, $u$ have a directed edge to $v$.↵
↵
<spoiler summary="u -> v">↵
$$g = \gcd(s_u, s_v)$$↵
If $i$-th element from $u$-th vertex was painted and $j$-th element from $v$-th vertex was unpainted such that $i \mod g \equiv j \mod g$ then $j$-th element from $v$-th vertex will be painted quickly. For two vertex, we can do it in complexity $\mathcal{O}{(s_u+s_v)}$.↵
</spoiler>↵
↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
If we have this directed walk ${w_1, w_2, \... , w_k}$, you should affect $v_1$ on $v_k$ how I said in the previous hint with $g = \gcd( s_{w_1}, s_{w_2}, ... , s_{w_k} )$.↵
↵
<spoiler summary="What about strogly connected components.">↵
$$g = \gcd( s_{1} , s_{2} , ... , s_{n} )$$↵
In this components if $i$-th element from $u$-th vertex was painted then for all of $j$ such that $i \mod g \equiv j \mod g$, $j$-th element from all vertices will be painted quickly. For this components, we can do it with complexity $\mathcal{O}{({\sum_{i=1}^{n} s_i})}$. In the main problem, we can suppose all of a strongly conected component a vertex with size $g$ in this way and we can calculate $mx_i$ with answer of its component's vertex multiplied by $\frac{s_i}{g}$.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint4">↵
SCC of the tournament yields a Hamiltonian path of strongly connected components( if we consider each component a vertex ). The problem decreased to solve this path. we can solve the path with the complexity of sum of vertices' sizes.↵
</spoiler>↵
↵
↵
<spoiler summary="hint5">↵
Now, for every vertex we have minimum and maximum score it may get. For B known vertices, to check if they can be the wanted set, we can assume their score be maximum possible and the other scores be minimum possible. If they were top, then they can be a sample of the wanted set.↵
</spoiler>↵
↵
↵
<spoiler summary="hint6">↵
Consider a sorted sequence of combination of two array $mn$ and $mx$. ↵
↵
For every vertex, consider the number of ways it(with its maximum score) can be the minimum score of the set of B vertices.↵
</spoiler>↵
↵
↵
Solutions↵
==================↵
[tutorial:805A]↵
Time Complexity: $\mathcal{O}{(1)}$↵
↵
Memory complexity: $\mathcal{O}{(1)}$↵
↵
<spoiler summary="python3">↵
~~~~~↵
l,r=list(map(int,input().split()))↵
print(2 if l<r else l)↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
int main(){↵
int l, r;↵
scanf("%d%d", &l, &r);↵
printf("%d", l == r ? l : 2);↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805B]↵
Time Complexity: $\mathcal{O}{(n)}$↵
↵
Memory complexity: $\mathcal{O}{(n)}$↵
↵
<spoiler summary="python3">↵
~~~~~↵
print(("aabb" * 50000)[:int(input())])↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
int N;↵
int main()↵
{↵
scanf("%d", &N);↵
for (int i = 0; i < N; i++)↵
putchar(i & 2 ? 'b' : 'a');↵
puts("");↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805C]↵
Time Complexity: $\mathcal{O}{(1)}$↵
↵
Memory complexity: $\mathcal{O}{(1)}$↵
↵
<spoiler summary="python3">↵
~~~~~↵
print((int(input())-1)//2)↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
int main(){↵
int n;↵
scanf("%d", &n);↵
printf("%d\n", (n-1)/2);↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805D]↵
↵
Time Complexity: $\mathcal{O}{(n)}$↵
↵
Memory complexity: $\mathcal{O}{(n)}$↵
↵
<spoiler summary="python3">↵
~~~~~↵
s = input()↵
cnt = 0↵
m=10**9 + 7↵
t = 0↵
↵
for i in range(len(s)):↵
if s[~i] == 'a':↵
cnt = (cnt+t)%m↵
t = (t*2)%m↵
else:↵
t += 1↵
print(cnt)↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
int c,d,i,n,m,k,x,j=1000000007;↵
string s;↵
main(){↵
cin>>s;↵
for(i=s.size()-1;i>=0;i--){↵
if(s[i]=='b')c++;else{↵
k+=c;c*=2;k%=j;c%=j;↵
}↵
}↵
cout<<k;↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805E]↵
Time Complexity: $\mathcal{O}{(n + \sum_i s_i \cdot \log(\sum_i s_i))}$↵
↵
Memory complexity: $\mathcal{O}{(n + \sum_i si)}$↵
↵
<spoiler summary="C++">↵
~~~~~↵
// +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+ \\↵
↵
int const N = 3e5 + 10 ;↵
int n , m , ans = 1 , res[N] ;↵
vector <int> g[N] , vec[N] ;↵
↵
inline bool cmp (int a , int b) {↵
return res[a] < res[b] ;↵
}↵
↵
void dfs (int v , int par = -1) {↵
int col = ans , p = (int)vec[v].size() - 1 ;↵
↵
sort(vec[v].begin() , vec[v].end() , cmp) ;↵
↵
for (int x : vec[v]) {↵
if (res[x]) {↵
continue ;↵
}↵
↵
while (p >= 0 && col == res[vec[v][p]]) { ↵
col -- ;↵
p -- ;↵
}↵
↵
res[x] = col ;↵
col -- ;↵
↵
}↵
↵
for (int u : g[v]) {↵
if (u != par) {↵
dfs(u , v) ;↵
}↵
}↵
}↵
↵
int main(){↵
ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;↵
↵
cin >> n >> m ;↵
↵
for (int i = 0 ; i < n ; i ++) {↵
int sz ;↵
cin >> sz ;↵
↵
ans = max(ans , sz) ;↵
↵
while (sz --) {↵
int x ;↵
cin >> x ;↵
x -- ;↵
vec[i].push_back(x) ;↵
}↵
}↵
↵
for (int i = 0 ; i < n - 1 ; i ++) {↵
int u , v ;↵
cin >> u >> v ;↵
u -- , v -- ;↵
↵
g[u].push_back(v) ;↵
g[v].push_back(u) ;↵
}↵
↵
dfs(0) ;↵
↵
cout << ans << '\n' ;↵
↵
for (int i = 0 ; i < m ; i ++) {↵
cout << max(res[i] , 1) << ' ' ;↵
}↵
cout << '\n' ;↵
}↵
~~~~~↵
</spoiler>↵
↵
Time Complexity: $\mathcal{O}{(n + \sum_i s_i)}$↵
↵
Memory complexity: $\mathcal{O}{(n + \sum_i si)}$↵
↵
<spoiler summary="C++">↵
~~~~~↵
vector <int> Vs[300050];↵
vector <int> conn[300050];↵
↵
int ans[300050];↵
bool chk[500050];↵
bool dchk[300050];↵
vector <int> Vu;↵
void DFS(int n) {↵
dchk[n] = true;↵
↵
for (auto it : Vs[n]) {↵
if (ans[it]) chk[ans[it]] = true;↵
else Vu.push_back(it);↵
}↵
↵
int a = 1;↵
for (auto it : Vu) {↵
while (chk[a]) a++;↵
ans[it] = a++;↵
}↵
for (auto it : Vs[n]) chk[ans[it]] = false;↵
Vu.clear();↵
↵
for (auto it : conn[n]) {↵
if (dchk[it]) continue;↵
DFS(it);↵
}↵
}↵
int main() {↵
int N, M, i;↵
scanf("%d %d", &N, &M);↵
for (i = 1; i <= N; i++) {↵
int t1, t2;↵
scanf("%d", &t1);↵
while (t1--) {↵
scanf("%d", &t2);↵
Vs[i].push_back(t2);↵
}↵
}↵
for (i = 1; i < N; i++) {↵
int t1, t2;↵
scanf("%d %d", &t1, &t2);↵
conn[t1].push_back(t2);↵
conn[t2].push_back(t1);↵
}↵
DFS(1);↵
↵
int mx = 0;↵
for (i = 1; i <= M; i++) {↵
mx = max(mx, ans[i]);↵
if (ans[i] == 0) ans[i] = 1;↵
}↵
mx = max(mx, 1);↵
printf("%d\n", mx);↵
for (i = 1; i <= M; i++) printf("%d ", ans[i]);↵
↵
return !printf("\n");↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:805F]↵
↵
Time Complexity: $\mathcal{O}{(q \cdot \sqrt(n) \cdot \log(n) + n \cdot \sqrt(n) \cdot \log(n) )}$↵
↵
Memory complexity: $\mathcal{O}{(n)}$↵
↵
<spoiler summary="C++">↵
~~~~~↵
using namespace std ;↵
↵
#define int long long↵
↵
const int MAXN = 1e5 + 100 ;↵
↵
vector<int>ver[MAXN] , com[MAXN] , ps[MAXN] ; ↵
int vis[MAXN] , dis[MAXN] ; ↵
int mxr , mxid ; ↵
void dfs(int v , int col = -1 , int h = 0 , int par = -1){↵
vis[v] = max(vis[v] , col) ; ↵
if(mxr < h)mxid = v , mxr = h ; ↵
dis[v] = max(h , dis[v]) ; ↵
if(col > 0)com[col] . push_back(dis[v]) ; ↵
for(auto u : ver[v]){↵
if(u == par)continue ; ↵
dfs(u , col , h + 1 , v) ; ↵
}↵
}↵
map<pair<int , int> , int> check ;↵
int32_t main(){↵
ios_base::sync_with_stdio(0) ;↵
cin . tie(0) ; cout . tie(0) ;↵
int n , m , q ; cin >> n >> m >> q ; ↵
for(int i = 0 ; i < m ; i ++){↵
int x , y ; cin >> x >> y ; ↵
x -- , y -- ; ↵
ver[x] . push_back(y) ; ↵
ver[y] . push_back(x) ;↵
}↵
int cnt = 0 ; ↵
for(int i = 0 ; i < n ; i ++){↵
if(vis[i])continue ; ↵
mxr = 0 , mxid = i ; ↵
dfs(i) ; ↵
mxr = 0 ; ↵
dfs(mxid) ;↵
dfs(mxid , ++ cnt) ; sort(com[cnt] . begin() , com[cnt] . end()) ; ↵
ps[cnt] . push_back(0) ; ↵
for(auto v : com[cnt])ps[cnt] . push_back(ps[cnt] . back() + v) ; ↵
}↵
cout << fixed << setprecision(10) ;↵
while(q --){↵
int x , y ; cin >> x >> y ; ↵
x -- , y -- ; x = vis[x] ; y = vis[y] ; ↵
if(com[x] . size() > com[y] . size())swap(x , y) ; ↵
if(x == y){cout << -1 << '\n' ; continue ; }↵
if(check[{x , y}] > 0){cout << double(check[{x , y}]) / (1ll * com[x] . size() * com[y] . size()) << '\n' ; continue ; }↵
int ans = 0 , mxd = max(com[x] . back() , com[y] . back()) ; ↵
for(auto v : com[x]){↵
int num = lower_bound(com[y] . begin() , com[y] . end() , mxd - v - 1) - com[y] . begin() ; ↵
ans += num * mxd + (com[y] . size() - num) * (v + 1) + ps[y] . back() - ps[y][num] ; ↵
}↵
check[{x , y}] = ans ; ↵
cout << double(check[{x , y}]) / (1ll * com[x] . size() * com[y] . size()) << '\n' ; ↵
}↵
}↵
↵
~~~~~↵
</spoiler>↵
↵
[tutorial:804E]↵
↵
Time Complexity: $\mathcal{O}{(n^2)}$↵
↵
Memory complexity: $\mathcal{O}{(1)}$↵
↵
<spoiler summary="C++">↵
~~~~~↵
int n;↵
↵
int main()↵
{↵
scanf("%d",&n);↵
if(n%4>1)return printf("NO\n"),0;↵
printf("YES\n");↵
for(int i=1;i<n;i+=4)↵
{↵
if(n%4)printf("%d %d\n%d %d\n%d %d\n",i+2,n,i+2,i+3,i+3,n);↵
else printf("%d %d\n",i+2,i+3);↵
printf("%d %d\n",i,i+2);↵
printf("%d %d\n",i+1,i+3);↵
printf("%d %d\n",i+1,i+2);↵
printf("%d %d\n",i,i+3);↵
if(n%4)printf("%d %d\n%d %d\n%d %d\n",i,n,i,i+1,i+1,n);↵
else printf("%d %d\n",i,i+1);↵
}↵
for(int i=1;i<n;i+=4)↵
for(int j=i+4;j<n;j+=4)↵
{↵
printf("%d %d\n",i+3,j+2);↵
printf("%d %d\n",i+2,j+2);↵
printf("%d %d\n",i+3,j+1);↵
printf("%d %d\n",i,j+1);↵
printf("%d %d\n",i+1,j+3);↵
printf("%d %d\n",i+2,j+3);↵
printf("%d %d\n",i+1,j+2);↵
printf("%d %d\n",i+1,j+1);↵
printf("%d %d\n",i+3,j);↵
printf("%d %d\n",i+3,j+3);↵
printf("%d %d\n",i,j+2);↵
printf("%d %d\n",i,j+3);↵
printf("%d %d\n",i+2,j);↵
printf("%d %d\n",i+2,j+1);↵
printf("%d %d\n",i+1,j);↵
printf("%d %d\n",i,j);↵
}↵
return 0;↵
}↵
~~~~~↵
</spoiler>↵
↵
[tutorial:804F]↵
Time Complexity: $\mathcal{O}{(\sum_i s_i + n \cdot b)}$↵
↵
Memory complexity: $\mathcal{O}{(n^2 + \sum_i s_i)}$↵
↵
<spoiler summary="Java8">↵
~~~~~↵
import java.io.OutputStream;↵
import java.io.IOException;↵
import java.io.InputStream;↵
import java.io.OutputStream;↵
import java.io.PrintWriter;↵
import java.io.BufferedWriter;↵
import java.util.InputMismatchException;↵
import java.io.IOException;↵
import java.util.Stack;↵
import java.util.ArrayList;↵
import java.util.List;↵
import java.util.stream.Stream;↵
import java.util.Vector;↵
import java.io.Writer;↵
import java.io.OutputStreamWriter;↵
import java.util.BitSet;↵
import java.io.InputStream;↵
↵
/**↵
* Built using CHelper plug-in↵
* Actual solution is at the top↵
*/↵
public class Main {↵
public static void main(String[] args) {↵
InputStream inputStream = System.in;↵
OutputStream outputStream = System.out;↵
InputReader in = new InputReader(inputStream);↵
OutputWriter out = new OutputWriter(outputStream);↵
TaskF solver = new TaskF();↵
solver.solve(1, in, out);↵
out.close();↵
}↵
↵
static class TaskF {↵
public int mod = 1000000007;↵
public int n;↵
public int a;↵
public int b;↵
public char[][] adj;↵
public List<Integer>[] graph;↵
↵
public void solve(int testNumber, InputReader in, OutputWriter out) {↵
n = in.nextInt();↵
a = in.nextInt();↵
b = in.nextInt();↵
adj = new char[n][n];↵
graph = Stream.generate(ArrayList::new).limit(n).toArray(List[]::new);↵
for (int i = 0; i < n; i++) adj[i] = in.next().toCharArray();↵
for (int i = 0; i < n; i++) {↵
for (int j = 0; j < n; j++) {↵
if (adj[i][j] == '1') {↵
graph[i].add(j);↵
}↵
}↵
}↵
List<List<Integer>> comp = new SCCTarjan().scc(graph);↵
boolean[][] can = new boolean[n][];↵
int[] low = new int[n];↵
for (int i = 0; i < n; i++) {↵
int q = in.nextInt();↵
char[] w = in.next().toCharArray();↵
can[i] = new boolean[q];↵
for (int j = 0; j < q; j++) {↵
can[i][j] = w[j] == '1';↵
if (can[i][j]) low[i]++;↵
}↵
}↵
int[] wcomp = new int[n];↵
int idx = 0;↵
List<BitSet> bb = new ArrayList<>();↵
List<Integer> bs = new ArrayList<>();↵
for (List<Integer> gg : comp) {↵
int gcd = 0;↵
for (int x : gg) {↵
gcd = Utils.gcd(gcd, can[x].length);↵
wcomp[x] = idx;↵
}↵
idx++;↵
BitSet d = new BitSet(gcd);↵
bs.add(gcd);↵
bb.add(d);↵
for (int x : gg) {↵
for (int i = 0; i < can[x].length; i++) {↵
if (can[x][i]) {↵
d.set(i % gcd, true);↵
}↵
}↵
}↵
}↵
↵
for (int kk = comp.size() - 1; kk > 0; kk--) {↵
int g = Utils.gcd(bs.get(kk), bs.get(kk - 1));↵
BitSet ww = new BitSet(g);↵
for (int i = 0; i < bs.get(kk); i += g) {↵
ww.or(bb.get(kk).get(i, i + g));↵
}↵
for (int i = 0; i < bs.get(kk - 1); i++) {↵
if (ww.get(i % g)) {↵
bb.get(kk - 1).set(i, true);↵
}↵
}↵
}↵
int[] high = new int[n];↵
for (int i = 0; i < n; i++) {↵
for (int j = 0; j < can[i].length; j++) {↵
if (bb.get(wcomp[i]).get(j % bs.get(wcomp[i])))↵
high[i]++;↵
}↵
}↵
long ret = 0;↵
int[][] comb = Utils.getComb(n + 1, mod);↵
for (int lowest = 0; lowest < n; lowest++) {↵
int countabove = 0;↵
for (int other = 0; other < n; other++) {↵
if (other == lowest) continue;↵
if (low[other] > high[lowest]) {↵
countabove++;↵
}↵
}↵
if (countabove >= a) {↵
continue;↵
}↵
int canbig = 0;↵
for (int other = 0; other < n; other++) {↵
if (other == lowest) continue;↵
if (low[other] <= high[lowest] && (high[other] > high[lowest] || (high[other] == high[lowest] && other > lowest))) {↵
canbig++;↵
}↵
}↵
↵
for (int take = 0; take < b && take <= canbig && take + countabove < a; take++) {↵
ret = (ret + 1L * comb[canbig][take] * comb[countabove][b - take - 1]) % mod;↵
}↵
}↵
out.println(ret);↵
}↵
↵
}↵
↵
static class OutputWriter {↵
private final PrintWriter writer;↵
↵
public OutputWriter(OutputStream outputStream) {↵
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));↵
}↵
↵
public OutputWriter(Writer writer) {↵
this.writer = new PrintWriter(writer);↵
}↵
↵
public void close() {↵
writer.close();↵
}↵
↵
public void println(long i) {↵
writer.println(i);↵
}↵
↵
}↵
↵
static class Utils {↵
public static int gcd(int a, int b) {↵
return b == 0 ? a : gcd(b, a % b);↵
}↵
↵
public static int[][] getComb(int sz, int mod) {↵
int[][] comb = new int[sz][sz];↵
for (int i = 0; i < sz; i++) {↵
comb[i][0] = 1;↵
for (int j = 1; j <= i; j++) {↵
comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];↵
if (comb[i][j] >= mod) comb[i][j] -= mod;↵
}↵
}↵
return comb;↵
}↵
↵
}↵
↵
static class InputReader {↵
private InputStream stream;↵
private byte[] buf = new byte[1024];↵
private int curChar;↵
private int numChars;↵
↵
public InputReader(InputStream stream) {↵
this.stream = stream;↵
}↵
↵
public int read() {↵
if (this.numChars == -1) {↵
throw new InputMismatchException();↵
} else {↵
if (this.curChar >= this.numChars) {↵
this.curChar = 0;↵
↵
try {↵
this.numChars = this.stream.read(this.buf);↵
} catch (IOException var2) {↵
throw new InputMismatchException();↵
}↵
↵
if (this.numChars <= 0) {↵
return -1;↵
}↵
}↵
↵
return this.buf[this.curChar++];↵
}↵
}↵
↵
public int nextInt() {↵
int c;↵
for (c = this.read(); isSpaceChar(c); c = this.read()) {↵
;↵
}↵
↵
byte sgn = 1;↵
if (c == 45) {↵
sgn = -1;↵
c = this.read();↵
}↵
↵
int res = 0;↵
↵
while (c >= 48 && c <= 57) {↵
res *= 10;↵
res += c - 48;↵
c = this.read();↵
if (isSpaceChar(c)) {↵
return res * sgn;↵
}↵
}↵
↵
throw new InputMismatchException();↵
}↵
↵
public String next() {↵
int c;↵
while (isSpaceChar(c = this.read())) {↵
;↵
}↵
↵
StringBuilder result = new StringBuilder();↵
result.appendCodePoint(c);↵
↵
while (!isSpaceChar(c = this.read())) {↵
result.appendCodePoint(c);↵
}↵
↵
return result.toString();↵
}↵
↵
public static boolean isSpaceChar(int c) {↵
return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;↵
}↵
↵
}↵
↵
static class SCCTarjan {↵
List<Integer>[] graph;↵
boolean[] visited;↵
Stack<Integer> stack;↵
int time;↵
int[] lowlink;↵
List<List<Integer>> components;↵
↵
public List<List<Integer>> scc(List<Integer>[] graph) {↵
int n = graph.length;↵
this.graph = graph;↵
visited = new boolean[n];↵
stack = new Stack<>();↵
time = 0;↵
lowlink = new int[n];↵
components = new ArrayList<>();↵
↵
for (int u = 0; u < n; u++)↵
if (!visited[u])↵
dfs(u);↵
↵
return components;↵
}↵
↵
void dfs(int u) {↵
lowlink[u] = time++;↵
visited[u] = true;↵
stack.add(u);↵
boolean isComponentRoot = true;↵
↵
for (int v : graph[u]) {↵
if (!visited[v])↵
dfs(v);↵
if (lowlink[u] > lowlink[v]) {↵
lowlink[u] = lowlink[v];↵
isComponentRoot = false;↵
}↵
}↵
↵
if (isComponentRoot) {↵
List<Integer> component = new ArrayList<>();↵
while (true) {↵
int x = stack.pop();↵
component.add(x);↵
lowlink[x] = Integer.MAX_VALUE;↵
if (x == u)↵
break;↵
}↵
components.add(component);↵
}↵
}↵
↵
}↵
}↵
~~~~~↵
</spoiler>↵
↵
↵
<spoiler summary="C++">↵
~~~~~↵
↵
#define pb push_back↵
#define mp make_pair↵
#define all(x) (x).begin(),(x).end()↵
#define fi first↵
#define se second↵
↵
const int maxn = 5005;↵
const int maxelem = 2000006;↵
const int MOD = 1000000007;↵
↵
char gr[maxn][maxn];↵
string init[maxn];↵
int mincnt[maxn], maxcnt[maxn];↵
int c[maxn][maxn];↵
int n, m, A, B;↵
bool was[maxn];↵
vector<int> order;↵
vector<bool> can[maxn];↵
bool nowcan[maxelem];↵
char s[maxelem];↵
int gcd[maxn];↵
vector<int> comp[maxn];↵
↵
void preorder(int cur)↵
{↵
if (was[cur]) return;↵
was[cur] = true;↵
for (int i = 0; i < n; i++) if (gr[cur][i] == '1') preorder(i);↵
order.pb(cur);↵
}↵
↵
void color(int cur, int cc)↵
{↵
if (was[cur]) return;↵
was[cur] = true;↵
gcd[cc] = __gcd(gcd[cc], (int)init[cur].size());↵
comp[cc].pb(cur);↵
for (int i = 0; i < n; i++) if (gr[i][cur] == '1') color(i, cc);↵
}↵
↵
int getc(int n, int k)↵
{↵
if (k < 0 || k > n) return 0;↵
return c[n][k];↵
}↵
↵
int main()↵
{↵
scanf("%d%d%d", &n, &A, &B);↵
for (int i = 0; i < n; i++)↵
{↵
scanf("%s", gr[i]);↵
}↵
for (int i = 0; i < n; i++)↵
{↵
scanf("%*d");↵
scanf("%s", s);↵
init[i] = string(s);↵
for (int j = 0; j < (int)init[i].size(); j++) mincnt[i] += init[i][j] == '1';↵
maxcnt[i] = mincnt[i];↵
}↵
for (int i = 0; i < n; i++) if (!was[i]) preorder(i);↵
memset(was, 0, sizeof(was));↵
int cc = 0;↵
reverse(all(order));↵
for (auto t : order) if (!was[t])↵
{↵
color(t, cc);↵
cc++;↵
}↵
for (int i = 0; i < cc; i++)↵
{↵
can[i].resize(gcd[i]);↵
for (int j = 0; j < gcd[i]; j++) can[i][j] = false;↵
for (auto t : comp[i])↵
{↵
for (int j = 0; j < (int)init[t].size(); j++) if (init[t][j] == '1') can[i][j % gcd[i]] = true;↵
}↵
}↵
for (int i = 0; i < cc - 1; i++)↵
{↵
int g = __gcd(gcd[i], gcd[i + 1]);↵
for (int j = 0; j < g; j++) nowcan[j] = false;↵
for (int j = 0; j < gcd[i]; j++) nowcan[j % g] |= can[i][j];↵
for (int j = 0; j < gcd[i + 1]; j++) can[i + 1][j] = (can[i + 1][j] | nowcan[j % g]);↵
}↵
for (int i = 0; i < cc; i++)↵
{↵
for (auto t : comp[i])↵
{↵
for (int j = 0; j < (int)init[t].size(); j++) if (init[t][j] == '0' && can[i][j % gcd[i]]) maxcnt[t]++;↵
}↵
}↵
↵
c[0][0] = 1;↵
for (int i = 1; i <= n; i++)↵
{↵
c[i][0] = 1;↵
for (int j = 1; j <= i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;↵
}↵
int answer = 0;↵
for (int i = 0; i < n; i++)↵
{↵
int cntbigger = 0;↵
int cntupper = 0;↵
for (int j = 0; j < n; j++) if (i != j)↵
{↵
if (mincnt[j] > maxcnt[i]) cntbigger++;↵
else if (maxcnt[j] > maxcnt[i] || (maxcnt[j] == maxcnt[i] && j > i)) cntupper++;↵
}↵
if (cntbigger >= A) continue;↵
for (int j = max(0, B - 1 - cntbigger); j <= cntupper && j + cntbigger + 1 <= A && j + 1 <= B; j++)↵
{↵
answer = (answer + (ll)getc(cntupper, j) * getc(cntbigger, B - 1 - j)) % MOD;↵
}↵
}↵
cout << answer << endl;↵
return 0;↵
} ↵
↵
~~~~~↵
</spoiler>↵
↵
<spoiler summary="C++">↵
↵
~~~~~↵
#include <bits/stdc++.h>↵
#define xx first↵
#define yy second↵
#define mp make_pair↵
#define pb push_back↵
#define fill( x, y ) memset( x, y, sizeof x )↵
#define copy( x, y ) memcpy( x, y, sizeof x )↵
using namespace std;↵
↵
typedef long long LL;↵
typedef pair < int, int > pa;↵
↵
inline int read()↵
{↵
int sc = 0, f = 1; char ch = getchar();↵
while( ch < '0' || ch > '9' ) { if( ch == '-' ) f = -1; ch = getchar(); }↵
while( ch >= '0' && ch <= '9' ) sc = sc * 10 + ch - '0', ch = getchar();↵
return sc * f;↵
}↵
↵
const int MAXN = 5005;↵
const int MAXM = 2000005;↵
const int mod = 1e9 + 7;↵
↵
int n, a, b, len[MAXN], can[MAXM], id_cnt, L[MAXN], R[MAXN], C[MAXN][MAXN];↵
int dfn[MAXN], scc[MAXN], low[MAXN], tim, num, st[MAXN], top, w[MAXN], ans;↵
vector < int > G[MAXN], id[MAXN], scc_bit[MAXN], v[MAXN];↵
char ch[MAXM];↵
↵
inline void inc(int &x, int y) { x += y; if( x >= mod ) x -= mod; }↵
↵
inline void dfs(int x)↵
{↵
dfn[ x ] = low[ x ] = ++tim; st[ ++top ] = x;↵
for( auto y : G[ x ] )↵
if( !dfn[ y ] ) dfs( y ), low[ x ] = min( low[ x ], low[ y ] );↵
else if( !scc[ y ] ) low[ x ] = min( low[ x ], dfn[ y ] );↵
if( dfn[ x ] == low[ x ] )↵
{↵
num++; int t = 0;↵
while( t ^ x )↵
{↵
scc[ t = st[ top-- ] ] = num;↵
w[ num ] = __gcd( w[ num ], len[ t ] );↵
v[ num ].pb( t );↵
}↵
scc_bit[ num ].resize( w[ num ] );↵
for( auto y : v[ num ] )↵
for( int d = 0 ; d < len[ y ] ; d++ ) if( can[ id[ y ][ d ] ] ) scc_bit[ num ][ d % w[ num ] ] = 1;↵
}↵
}↵
↵
int main()↵
{↵
#ifdef wxh010910↵
freopen( "data.in", "r", stdin );↵
#endif↵
n = read(), a = read(), b = read();↵
for( int i = 0 ; i <= n ; i++ )↵
{↵
C[ i ][ 0 ] = 1;↵
for( int j = 1 ; j <= i ; j++ ) inc( C[ i ][ j ] = C[ i - 1 ][ j ], C[ i - 1 ][ j - 1 ] );↵
}↵
for( int i = 1 ; i <= n ; i++ )↵
{↵
scanf( "%s", ch + 1 );↵
for( int j = 1 ; j <= n ; j++ ) if( ch[ j ] == '1' ) G[ i ].pb( j );↵
}↵
for( int i = 1 ; i <= n ; i++ )↵
{↵
len[ i ] = read(); scanf( "%s", ch );↵
for( int j = 0 ; j < len[ i ] ; j++ ) id[ i ].pb( ++id_cnt ), can[ id_cnt ] = ch[ j ] == '1', L[ i ] += can[ id_cnt ];↵
}↵
for( int i = 1 ; i <= n ; i++ ) if( !dfn[ i ] ) dfs( i );↵
for( int i = num ; i > 1 ; i-- )↵
{↵
int g = __gcd( w[ i ], w[ i - 1 ] );↵
w[ i - 1 ] = g;↵
for( int j = 0 ; j < w[ i ] ; j++ ) scc_bit[ i - 1 ][ j % g ] |= scc_bit[ i ][ j ];↵
}↵
for( int i = 1 ; i <= n ; i++ )↵
{↵
int x = scc[ i ];↵
for( int j = 0 ; j < len[ i ] ; j++ )↵
R[ i ] += scc_bit[ x ][ j % w[ x ] ];↵
}↵
for( int i = 1 ; i <= n ; i++ )↵
{↵
int t1 = 0, t2 = 0;↵
for( int j = 1 ; j <= n ; j++ ) if( L[ j ] > R[ i ] ) t1++;↵
if( t1 >= a ) continue;↵
for( int j = 1 ; j <= n ; j++ ) if( L[ j ] <= R[ i ] && mp( R[ j ], j ) > mp( R[ i ], i ) ) t2++;↵
for( int j = 0 ; j < b && j <= t2 && j + t1 < a ; j++ )↵
inc( ans, 1LL * C[ t2 ][ j ] * C[ t1 ][ b - j - 1 ] % mod ); ↵
}↵
return printf( "%d\n", ans ), 0;↵
}↵
~~~~~↵
↵
↵
</spoiler>↵
↵
