In this HackerRank Coprime Paths problem solution You are given an undirected, connected graph, G, with n nodes and m edges where m = n-1. Each node i is initially assigned a value, node, that has at most 3 prime divisors.
You must answer q queries in the form u v. For each query, find and print the number of (x,y) pairs of nodes on the path between u and v such that gcd(node x, node y) = 1 and the length of the path between u and v is minimal among all paths from u to v.
Problem solution in Java.
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
public class F {
InputStream is;
PrintWriter out;
String INPUT = "";
long ret;
int[] freq;
int[] pfreq;
EulerTour et;
int[] lpf = enumLowestPrimeFactors(10000005);
int[] mob = enumMobiusByLPF(10000005, lpf);
int[] a;
void solve()
{
int n = ni(), Q = ni();
a = na(n);
for(int i = 0;i < n;i++){
int pre = -1;
int mul = 1;
for(int j = a[i];j > 1;j /= lpf[j]){
if(pre != lpf[j]){
mul *= lpf[j];
pre = lpf[j];
}
}
a[i] = mul;
}
int[] from = new int[n - 1];
int[] to = new int[n - 1];
for (int i = 0; i < n - 1; i++) {
from[i] = ni() - 1;
to[i] = ni() - 1;
}
int[][] g = packU(n, from, to);
int[][] pars = parents3(g, 0);
int[] par = pars[0], ord = pars[1], dep = pars[2];
et = nodalEulerTour(g, 0);
int[][] spar = logstepParents(par);
int[][] qs = new int[Q][];
int[] special = new int[Q];
Arrays.fill(special, -1);
for(int i = 0;i < Q;i++){
int x = ni()-1, y = ni()-1;
int lca = lca2(x, y, spar, dep);
if(lca == x){
qs[i] = new int[]{et.first[x], et.first[y]};
}else if(lca == y){
qs[i] = new int[]{et.first[y], et.first[x]};
}else if(et.first[x] < et.first[y]){
qs[i] = new int[]{et.last[x], et.first[y]};
special[i] = lca;
}else{
qs[i] = new int[]{et.last[y], et.first[x]};
special[i] = lca;
}
}
long[] pqs = sqrtSort(qs, 2*n-1);
int L = 0, R = -1;
freq = new int[n];
long[] ans = new long[Q];
pfreq = new int[10000005];
for(long pa : pqs){
int ind = (int)(pa&(1<<25)-1);
int ql = qs[ind][0], qr = qs[ind][1];
while(R < qr)change(++R, 1);
while(L > ql)change(--L, 1);
while(R > qr)change(R--, -1);
while(L < ql)change(L++, -1);
if(special[ind] != -1)change(et.first[special[ind]], 1);
ans[ind] = ret;
if(special[ind] != -1)change(et.first[special[ind]], -1);
}
for(long v : ans){
out.println(v);
}
}
public static void trnz(int... o)
{
for(int i = 0;i < o.length;i++)if(o[i] != 0)System.out.print(i+":"+o[i]+" ");
System.out.println();
}
public static int[] enumMobiusByLPF(int n, int[] lpf)
{
int[] mob = new int[n+1];
mob[1] = 1;
for(int i = 2;i <= n;i++){
int j = i/lpf[i];
if(lpf[j] == lpf[i]){
// mob[i] = 0;
}else{
mob[i] = -mob[j];
}
}
return mob;
}
void dfs(int cur, int n, int d)
{
if(n == 1){
if(d > 0)ret += mob[cur] * pfreq[cur];
pfreq[cur] += d;
if(d < 0)ret -= mob[cur] * pfreq[cur];
return;
}
dfs(cur, n/lpf[n], d);
dfs(cur/lpf[n], n/lpf[n], d);
}
void change(int x, int d)
{
int ind = et.vs[x];
if(freq[ind] == 1){
dfs(a[ind], a[ind], -1);
}
freq[ind] += d;
if(freq[ind] == 1){
dfs(a[ind], a[ind], 1);
}
}
public static long[] sqrtSort(int[][] qs, int n)
{
int m = qs.length;
long[] pack = new long[m];
int S = (int)Math.sqrt(n);
for(int i = 0;i < m;i++){
pack[i] = (long)qs[i][0]/S<<50|(long)((qs[i][0]/S&1)==0?qs[i][1]:(1<<25)-1-qs[i][1])<<25|i;
}
Arrays.sort(pack);
return pack;
}
public static int lca2(int a, int b, int[][] spar, int[] depth) {
if (depth[a] < depth[b]) {
b = ancestor(b, depth[b] - depth[a], spar);
} else if (depth[a] > depth[b]) {
a = ancestor(a, depth[a] - depth[b], spar);
}
if (a == b)
return a;
int sa = a, sb = b;
for (int low = 0, high = depth[a], t = Integer.highestOneBit(high), k = Integer
.numberOfTrailingZeros(t); t > 0; t >>>= 1, k--) {
if ((low ^ high) >= t) {
if (spar[k][sa] != spar[k][sb]) {
low |= t;
sa = spar[k][sa];
sb = spar[k][sb];
} else {
high = low | t - 1;
}
}
}
return spar[0][sa];
}
protected static int ancestor(int a, int m, int[][] spar) {
for (int i = 0; m > 0 && a != -1; m >>>= 1, i++) {
if ((m & 1) == 1)
a = spar[i][a];
}
return a;
}
public static int[][] logstepParents(int[] par) {
int n = par.length;
int m = Integer.numberOfTrailingZeros(Integer.highestOneBit(n - 1)) + 1;
int[][] pars = new int[m][n];
pars[0] = par;
for (int j = 1; j < m; j++) {
for (int i = 0; i < n; i++) {
pars[j][i] = pars[j - 1][i] == -1 ? -1 : pars[j - 1][pars[j - 1][i]];
}
}
return pars;
}
public static class EulerTour
{
public int[] vs;
public int[] first;
public int[] last;
public EulerTour(int[] vs, int[] f, int[] l) {
this.vs = vs;
this.first = f;
this.last = l;
}
}
public static EulerTour nodalEulerTour(int[][] g, int root)
{
int n = g.length;
int[] vs = new int[2*n];
int[] f = new int[n];
int[] l = new int[n];
int p = 0;
Arrays.fill(f, -1);
int[] stack = new int[n];
int[] inds = new int[n];
int sp = 0;
stack[sp++] = root;
outer:
while(sp > 0){
int cur = stack[sp-1], ind = inds[sp-1];
if(ind == 0){
vs[p] = cur;
f[cur] = p;
p++;
}
while(ind < g[cur].length){
int nex = g[cur][ind++];
if(f[nex] == -1){
inds[sp-1] = ind;
stack[sp] = nex;
inds[sp] = 0;
sp++;
continue outer;
}
}
inds[sp-1] = ind;
if(ind == g[cur].length){
vs[p] = cur;
l[cur] = p;
p++;
sp--;
}
}
return new EulerTour(vs, f, l);
}
public static int[][] parents3(int[][] g, int root) {
int n = g.length;
int[] par = new int[n];
Arrays.fill(par, -1);
int[] depth = new int[n];
depth[0] = 0;
int[] q = new int[n];
q[0] = root;
for (int p = 0, r = 1; p < r; p++) {
int cur = q[p];
for (int nex : g[cur]) {
if (par[cur] != nex) {
q[r++] = nex;
par[nex] = cur;
depth[nex] = depth[cur] + 1;
}
}
}
return new int[][] { par, q, depth };
}
static int[][] packU(int n, int[] from, int[] to) {
int[][] g = new int[n][];
int[] p = new int[n];
for (int f : from)
p[f]++;
for (int t : to)
p[t]++;
for (int i = 0; i < n; i++)
g[i] = new int[p[i]];
for (int i = 0; i < from.length; i++) {
g[from[i]][--p[from[i]]] = to[i];
g[to[i]][--p[to[i]]] = from[i];
}
return g;
}
public static int[] enumLowestPrimeFactors(int n) {
int tot = 0;
int[] lpf = new int[n + 1];
int u = n + 32;
double lu = Math.log(u);
int[] primes = new int[(int) (u / lu + u / lu / lu * 1.5)];
for (int i = 2; i <= n; i++)
lpf[i] = i;
for (int p = 2; p <= n; p++) {
if (lpf[p] == p)
primes[tot++] = p;
int tmp;
for (int i = 0; i < tot && primes[i] <= lpf[p] && (tmp = primes[i] * p) <= n; i++) {
lpf[tmp] = primes[i];
}
}
return lpf;
}
void run() throws Exception
{
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
public static void main(String[] args) throws Exception { new F().run(); }
private byte[] inbuf = new byte[1024];
public int lenbuf = 0, ptrbuf = 0;
private int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private double nd() { return Double.parseDouble(ns()); }
private char nc() { return (char)skip(); }
private String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}
private int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}
private int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}
Problem solution in C++.
#include <bits/stdc++.h>
using namespace std;
const int N = 25123, LN = 15;
const int MaxVal = 10000000;
using ll = long long;
vector<int> adj[N];
int depth[N], parent[LN][N];
int ST[N], EN[N], cur_time = 0;
int vec[2 * N];
void dfs(int u = 0, int d = 0, int prev = -1) {
depth[u] = d;
parent[0][u] = prev;
ST[u] = cur_time++;
vec[ST[u]] = u;
for (int v : adj[u]) {
if (v == prev) continue;
dfs(v, d + 1, u);
}
EN[u] = cur_time++;
vec[EN[u]] = u;
}
int lca(int u, int v) {
if (depth[u] < depth[v]) swap(u, v);
int diff = depth[u] - depth[v];
for (int i = 0; i < LN; i++) {
if ((diff >> i) & 1) {
u = parent[i][u];
}
}
if (u == v) return u;
for (int i = LN - 1; i >= 0; i--) {
if (parent[i][u] != parent[i][v]) {
u = parent[i][u];
v = parent[i][v];
}
}
return parent[0][u];
}
vector<int> primes[N];
vector<pair<int, int>> upd[N];
int pr[MaxVal + 1], S;
ll ans[N];
bool used[N];
int vp[3][4 * N];
int main() {
memset(parent, -1, sizeof parent);
for (int i = 2; i <= MaxVal; i++) {
if (!pr[i]) {
for (int j = i + i; j <= MaxVal; j += i) {
if (!pr[j]) pr[j] = i;
}
}
}
int n, q;
scanf("%d %d", &n, &q);
S = sqrt(2 * n);
// printf("S = %d\n", S);
map<int, int> p1;
map<tuple<int, int>, int> p2;
map<tuple<int, int, int>, int> p3;
for (int i = 0; i < n; i++) {
int id, x;
scanf("%d", &x);
auto& v = primes[i];
while (pr[x]) {
v.push_back(pr[x]);
x /= pr[x];
}
if (x > 1) {
v.push_back(x);
}
assert(is_sorted(begin(v), end(v)));
v.resize(unique(begin(v), end(v)) - begin(v));
for (int k = 0; k < v.size(); k++) {
id = p1.size();
if (p1.count(v[k])) id = p1[v[k]];
else p1[v[k]] = id;
upd[i].emplace_back(0, id);
for (int j = k + 1; j < v.size(); j++) {
auto tmp = make_tuple(v[k], v[j]);
id = p2.size();
if (p2.count(tmp)) id = p2[tmp];
else p2[tmp] = id;
upd[i].emplace_back(1, id);
}
}
if (v.size() == 3) {
auto tmp = make_tuple(v[0], v[1], v[2]);
id = p3.size();
if (p3.count(tmp)) id = p3[tmp];
else p3[tmp] = id;
upd[i].emplace_back(2, id);
}
}
for (int i = 1; i < n; i++) {
int u, v;
scanf("%d %d", &u, &v);
u--, v--;
adj[u].push_back(v);
adj[v].push_back(u);
}
dfs();
for (int i = 1; i < LN; i++) {
for (int j = 0; j < n; j++) {
if (parent[i - 1][j] != -1) {
parent[i][j] = parent[i - 1][parent[i - 1][j]];
}
}
}
using qt = tuple<int, int, int, int>;
vector<qt> queries;
for (int i = 0; i < q; i++) {
int u, v;
scanf("%d %d", &u, &v);
u--, v--;
if (ST[u] > ST[v]) swap(u, v);
int p = lca(u, v);
if (p == u) queries.emplace_back(ST[u], ST[v], i, -1);
else {
queries.emplace_back(EN[u], ST[v], i, p);
}
}
sort(begin(queries), end(queries), [](const qt& a, const qt& b) -> bool {
int l1, r1, i1, p1;
int l2, r2, i2, p2;
tie(l1, r1, i1, p1) = a;
tie(l2, r2, i2, p2) = b;
if (l1 / S != l2 / S) return l1 / S < l2 / S;
return r1 > r2;
});
int active = 0;
ll cur = 0;
auto insert = [&](int u) {
int tmp = active;
for (auto& p : upd[u]) {
if (p.first & 1) tmp += vp[p.first][p.second];
else tmp -= vp[p.first][p.second];
}
cur += tmp;
used[u] = true;
active++;
for (auto& p : upd[u]) {
vp[p.first][p.second]++;
}
};
auto remove = [&](int u) {
used[u] = false;
active--;
for (auto& p : upd[u]) {
vp[p.first][p.second]--;
}
int tmp = active;
for (auto& p : upd[u]) {
if (p.first & 1) tmp += vp[p.first][p.second];
else tmp -= vp[p.first][p.second];
}
cur -= tmp;
};
int L = 0, R = -1;
for (auto& t : queries) {
int l, r, i, p;
tie(l, r, i, p) = t;
// printf("%d %d %d %d\n", l, r, i, p);
while (R < r) {
R++;
if (used[vec[R]]) remove(vec[R]);
else insert(vec[R]);
}
while (R > r) {
if (used[vec[R]]) remove(vec[R]);
else insert(vec[R]);
R--;
}
while (L < l) {
if (used[vec[L]]) remove(vec[L]);
else insert(vec[L]);
L++;
}
while (L > l) {
L--;
if (used[vec[L]]) remove(vec[L]);
else insert(vec[L]);
}
ans[i] = cur;
// printf("cur: %d\n", cur);
if (p != -1) {
int tmp = active;
for (auto& k : upd[p]) {
if (k.first & 1) tmp += vp[k.first][k.second];
else tmp -= vp[k.first][k.second];
}
ans[i] += tmp;
}
}
for (int i = 0; i < q; i++) {
printf("%lld\n", ans[i]);
}
return 0;
}
0 Comments