Codeforces Round #421 (Div. 2), problem: (E) Mister B and Beacons on Field Solution In C/C++

Affiliate Disclosure: This post contains Amazon affiliate links. If you purchase through these links, eBlogarithm may earn a commission at no extra cost to you. Prices and availability are subject to change.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.util.TreeMap;

/**
* @author Don Li
*/
public class MisterBBeaconsField {

int N = (int) 2e6 + 10;
int SQRT = (int) (Math.sqrt(N) + 0.5);

int[] f = new int[N];
long nn, mm, ss;

long ans;
int[][] fs;
List<Long> bad = new ArrayList<>();

void solve() {
for (int i = 2; i < N; i++) {
if (f[i] == 0) {
f[i] = i;
for (int j = i * i; j < N && i <= SQRT; j += i) f[j] = i;
}
}

int T = in.nextInt();
while (T– > 0) {
int[] n = new int[3], m = new int[3], s = new int[3];
for (int i = 0; i < 3; i++) n[i] = in.nextInt();
for (int i = 0; i < 3; i++) m[i] = in.nextInt();
for (int i = 0; i < 3; i++) s[i] = in.nextInt();

s[0] *= 2;
nn = (long) n[0] * n[1] * n[2];
mm = (long) m[0] * m[1] * m[2];
ss = (long) s[0] * s[1] * s[2];

ans = 0;

fs = primeFactors(s);
dfs(0, 1);

bad.clear();
int[][] fn = primeFactors(n);
for (int i = 0; i < fn.length; i++) {
long v = fn[i][0];
for (int j = 0; j < fn[i][1]; j++) {
if (ss % v != 0) {
bad.add(v);
break;
}
v *= fn[i][0];
}
}
dfs2(0, 1, 0);

out.println(ans);
}
}

// iterate all the divisors of 2*s and count the number of divisors ≤ n
void dfs(int i, long v) {
if (i == fs.length) {
if (v <= nn) ans++;
return;
}
long mul = 1;
for (int j = 0; j <= fs[i][1]; j++) {
dfs(i + 1, v * mul);
mul *= fs[i][0];
}
}

// calc the number of k (such that gcd(n,k) | 2*s and 1≤k≤m) with inclusion-exclusion principle
void dfs2(int i, long v, int cnt) {
if (i == bad.size()) {
if (cnt == 0) ans += mm / v;
else ans -= mm / v;
return;
}
dfs2(i + 1, v, cnt);
dfs2(i + 1, v * bad.get(i), cnt ^ 1);
}

int[][] primeFactors(int[] a) {
TreeMap<Integer, Integer> map = new TreeMap<>();
for (int i = 0; i < 3; i++) {
int x = a[i];
while (x > 1 && x != f[x]) {
map.put(f[x], map.getOrDefault(f[x], 0) + 1);
x /= f[x];
}
if (x > 1) map.put(x, map.getOrDefault(x, 0) + 1);
}

int k = map.keySet().size();
int[][] cnt = new int[k][2];
int p = 0;
for (int x : map.keySet()) {
cnt[p][0] = x;
cnt[p][1] = map.get(x);
p++;
}
return cnt;
}

public static void main(String[] args) {
in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
out = new PrintWriter(System.out);
new MisterBBeaconsField().solve();
out.close();
}

static FastScanner in;
static PrintWriter out;

static class FastScanner {
BufferedReader in;
StringTokenizer st;

public FastScanner(BufferedReader in) {
this.in = in;
}

public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(in.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}

public int nextInt() {
return Integer.parseInt(nextToken());
}

public long nextLong() {
return Long.parseLong(nextToken());
}

public double nextDouble() {
return Double.parseDouble(nextToken());
}
}
}

📦 Looking for Codeforces Round 421 Div Problem Mister? Check the best deals on Amazon.

🛒 Shop Codeforces Round 421 Div Problem Mister on Amazon

As an Amazon Associate, eBlogarithm earns from qualifying purchases. Prices and availability are subject to change.

Affiliate Disclosure: This post contains Amazon affiliate links. If you purchase through these links, eBlogarithm may earn a commission at no extra cost to you. Prices and availability are subject to change.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.util.TreeMap;

/**
* @author Don Li
*/
public class MisterBBeaconsField {

int N = (int) 2e6 + 10;
int SQRT = (int) (Math.sqrt(N) + 0.5);

int[] f = new int[N];
long nn, mm, ss;

long ans;
int[][] fs;
List<Long> bad = new ArrayList<>();

void solve() {
for (int i = 2; i < N; i++) {
if (f[i] == 0) {
f[i] = i;
for (int j = i * i; j < N && i <= SQRT; j += i) f[j] = i;
}
}

int T = in.nextInt();
while (T– > 0) {
int[] n = new int[3], m = new int[3], s = new int[3];
for (int i = 0; i < 3; i++) n[i] = in.nextInt();
for (int i = 0; i < 3; i++) m[i] = in.nextInt();
for (int i = 0; i < 3; i++) s[i] = in.nextInt();

s[0] *= 2;
nn = (long) n[0] * n[1] * n[2];
mm = (long) m[0] * m[1] * m[2];
ss = (long) s[0] * s[1] * s[2];

ans = 0;

fs = primeFactors(s);
dfs(0, 1);

bad.clear();
int[][] fn = primeFactors(n);
for (int i = 0; i < fn.length; i++) {
long v = fn[i][0];
for (int j = 0; j < fn[i][1]; j++) {
if (ss % v != 0) {
bad.add(v);
break;
}
v *= fn[i][0];
}
}
dfs2(0, 1, 0);

out.println(ans);
}
}

// iterate all the divisors of 2*s and count the number of divisors ≤ n
void dfs(int i, long v) {
if (i == fs.length) {
if (v <= nn) ans++;
return;
}
long mul = 1;
for (int j = 0; j <= fs[i][1]; j++) {
dfs(i + 1, v * mul);
mul *= fs[i][0];
}
}

// calc the number of k (such that gcd(n,k) | 2*s and 1≤k≤m) with inclusion-exclusion principle
void dfs2(int i, long v, int cnt) {
if (i == bad.size()) {
if (cnt == 0) ans += mm / v;
else ans -= mm / v;
return;
}
dfs2(i + 1, v, cnt);
dfs2(i + 1, v * bad.get(i), cnt ^ 1);
}

int[][] primeFactors(int[] a) {
TreeMap<Integer, Integer> map = new TreeMap<>();
for (int i = 0; i < 3; i++) {
int x = a[i];
while (x > 1 && x != f[x]) {
map.put(f[x], map.getOrDefault(f[x], 0) + 1);
x /= f[x];
}
if (x > 1) map.put(x, map.getOrDefault(x, 0) + 1);
}

int k = map.keySet().size();
int[][] cnt = new int[k][2];
int p = 0;
for (int x : map.keySet()) {
cnt[p][0] = x;
cnt[p][1] = map.get(x);
p++;
}
return cnt;
}

public static void main(String[] args) {
in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
out = new PrintWriter(System.out);
new MisterBBeaconsField().solve();
out.close();
}

static FastScanner in;
static PrintWriter out;

static class FastScanner {
BufferedReader in;
StringTokenizer st;

public FastScanner(BufferedReader in) {
this.in = in;
}

public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(in.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}

public int nextInt() {
return Integer.parseInt(nextToken());
}

public long nextLong() {
return Long.parseLong(nextToken());
}

public double nextDouble() {
return Double.parseDouble(nextToken());
}
}
}

📦 Looking for Codeforces Round 421 Div Problem Mister? Check the best deals on Amazon.

🛒 Shop Codeforces Round 421 Div Problem Mister on Amazon

As an Amazon Associate, eBlogarithm earns from qualifying purchases. Prices and availability are subject to change.

More from author

LEAVE A REPLY

Please enter your comment!
Please enter your name here

Related posts

Advertismentspot_img

Latest posts

Paul Walker’s brother dishes on late actor’s future involvement in ‘Fast & Furious’ franchise — potentially using AI

The enduring legacy of Paul Walker continues to resonate deeply with fans across the globe, especially as the high-octane Fast & Furious saga speeds...

The Real Truth About Viral Kitchen Gadgets Before You Buy

📢 As an Amazon Associate, I earn from qualifying purchases.If you had told me six months ago that a single purchase would change how...

US suspends green-card filings by Cognizant amid visa-fraud probe: ‘Threats to American workers…’

In a major regulatory development impacting the global tech sector, the United States Department of Labor has halted green-card processing applications for IT services...

Want to stay up to date with the latest news?

We would love to hear from you! Please fill in your details and we will stay in touch. It's that simple!