GCD Matrix
Problem Statement :
Alex has two arrays defined as A = [a0,a1,...,an-1] and B = [b0,b1,...,bm-1]. He created an n*m matrix, M, where Mij = gcd(ai,bj) for each i,j in M. Recall that gcd(a,b) is the greatest common divisor of a and b. For example, if A= [2,3] and b = [5,6], he builds M = [[1,2], [1,3]] like so: Alex's friend Kiara loves matrices, so he gives her q questions about matrix M where each question is in the form of some submatrix of M with its upper-left corner at M(r1,c1) and its bottom-right corner at M(r2,c2). For each question, find and print the number of distinct integers in the given submatrix on a new line. Input Format The first line contains three space-separated integers describing the respective values of n (the size of array A), m(the size of array B), and q (Alex's number of questions). The second line contains nspace-separated integers describing a0,a1,...,an-1. The third line contains m space-separated integers describing b0,b1,...,bm-1. Each line i of the q subsequent lines contains four space-separated integers describing the respective values of r1, c1, r2, and c2 for the ith question (i.e., defining a submatrix with upper-left corner (r1,c1) and bottom-right corner (r2,c2)). Constraints 1 <= n,m <= 10^5 1 <= ai,bi <= 10^5 1 <= q <= 10 0 <= r1,r2 < n 0 <= c1,c2 < m
Solution :
Solution in C :
In C++ :
#define _CRT_SECURE_NO_WARNINGS
#pragma comment(linker, "/stack:16777216")
#include <string>
#include <vector>
#include <map>
#include <list>
#include <iterator>
#include <set>
#include <queue>
#include <iostream>
#include <sstream>
#include <stack>
#include <deque>
#include <cmath>
#include <memory.h>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <utility>
#include <assert.h>
#include <time.h>
#include <complex.h>
#include <fstream>
#include <sys/stat.h>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
#define FOR(i, a, b) for(int i=(a);i<(b);i++)
#define RFOR(i, b, a) for(int i=(b)-1;i>=(a);--i)
#define FILL(A,value) memset(A,value,sizeof(A))
#define ALL(V) V.begin(), V.end()
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define Pi 3.14159265358979
typedef long long Int;
typedef unsigned long long UInt;
typedef vector<int> VI;
typedef pair<Int, Int> PII;
const int INF = 1000000000;
const int MAX = 100007;
const int MAXD = 20;
const int MOD = 1000000007;
int a[MAX];
int b[MAX];
int ca[MAX];
int cb[MAX];
Int A[MAX];
int main()
{
//freopen("in.txt" , "r" , stdin);
//freopen("out.txt" , "w" , stdout);
int n , m;
cin >> n >> m;
int q;
cin >> q;
FOR(i,0,n) cin >> a[i];
FOR(i,0,m) cin >> b[i];
FOR(i,0,q)
{
int r1,c1,r2,c2;
cin >> r1 >> c1 >> r2 >> c2;
FILL(ca , 0);
FILL(cb , 0);
FOR(i,r1,r2 + 1)
ca[a[i]] ++;
FOR(i,c1,c2 + 1)
cb[b[i]] ++;
FILL(A, 0);
FOR(i,1,MAX)
{
int cntA = 0;
int cntB = 0;
for(int j = i; j < MAX; j += i)
{
cntA += ca[j];
cntB += cb[j];
}
A[i] = 1LL * cntA * cntB;
}
int res = 0;
RFOR(i,MAX,1)
{
for(int j = 2 * i; j < MAX; j += i)
A[i] -= A[j];
if (A[i]) ++ res;
}
cout << res << endl;
}
return 0;
}
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 C2 {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve()
{
int n = ni(), m = ni(), Q = ni();
int[] a = na(n);
int[] b = na(m);
for(int z = 0;z < Q;z++){
int dist = 0;
int r1 = ni(), c1 = ni(), r2 = ni(), c2 = ni();
int[] fr = make(a, r1, r2);
int[] fc = make(b, c1, c2);
long[] f = new long[100005];
for(int i = 0;i < 100005;i++)f[i] = (long)fr[i] * fc[i];
for(int i = 100004;i >= 1;i--){
for(int j = 2*i;j < 100005;j+=i){
f[i] -= f[j];
}
}
for(int i = 0;i < 100005;i++){
assert f[i] >= 0;
if(f[i] > 0)dist++;
}
out.println(dist);
}
}
int[] make(int[] a, int l, int r)
{
int[] f = new int[100005];
for(int i = l;i <= r;i++){
for(int d = 1;d*d <= a[i];d++){
if(a[i] % d == 0){
f[d]++;
if(d*d < a[i])f[a[i]/d]++;
}
}
}
return f;
}
public static long gcd3(long a, long b) {
if(a == 0)return b;
if(b == 0)return a;
int ashift = Long.numberOfTrailingZeros(a);
int bshift = Long.numberOfTrailingZeros(b);
a >>>= ashift;
b >>>= bshift;
while(b != a){
if(a > b){
long t = a; a = b; b = t;
}
b -= a;
b >>>= Long.numberOfTrailingZeros(b);
}
return a<<Math.min(ashift, bshift);
}
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 C2().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))){
// when nextLine, (isSpaceChar(b) && 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)); }
}
In C :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int a[100000],b[100000];
long long dp1[100000],dp2[100000],dp[100000];
int main(){
int n,m,q,r1,r2,c1,c2,ans,i,j;
scanf("%d%d%d",&n,&m,&q);
for(i=0;i<n;i++)
scanf("%d",a+i);
for(i=0;i<m;i++)
scanf("%d",b+i);
while(q--){
memset(dp1,0,sizeof(dp1));
memset(dp2,0,sizeof(dp2));
memset(dp,0,sizeof(dp));
scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
for(i=r1;i<=r2;i++)
for(j=1;j*j<=a[i];j++)
if(a[i]%j==0){
dp1[j-1]++;
if(j*j!=a[i])
dp1[a[i]/j-1]++;
}
for(i=c1;i<=c2;i++)
for(j=1;j*j<=b[i];j++)
if(b[i]%j==0){
dp2[j-1]++;
if(j*j!=b[i])
dp2[b[i]/j-1]++;
}
for(i=99999,ans=0;i>=0;i--){
dp[i]+=dp1[i]*dp2[i];
if(dp[i]>0){
ans++;
dp[0]-=dp[i];
for(j=2;j*j<=i+1;j++)
if((i+1)%j==0){
dp[j-1]-=dp[i];
if(j*j!=i+1)
dp[(i+1)/j-1]-=dp[i];
}
}
}
printf("%d\n",ans);
}
return 0;
}
In Python3 :
#!/bin/python3
import math
import os
import random
import re
import sys
def f(k):
if gf[k]>=0:
return gf[k]
res=ga[k]*gb[k]
if res==0:
gf[k]=0
return 0
for i in range(k+k,m+1,k):
res-=f(i)
gf[k]=res
return res
if __name__ == '__main__':
nmq = input().split()
n = int(nmq[0])
m = int(nmq[1])
q = int(nmq[2])
a = list(map(int, input().rstrip().split()))
b = list(map(int, input().rstrip().split()))
for q_itr in range(q):
r1C1R2C2 = input().split()
r1 = int(r1C1R2C2[0])
c1 = int(r1C1R2C2[1])
r2 = int(r1C1R2C2[2])
c2 = int(r1C1R2C2[3])
# Write Your Code Here
sa=set(a[r1:r2+1])
sb=set(b[c1:c2+1])
na=len(a)
nb=len(b)
mxa=max(sa)
mxb=max(sb)
m=min(mxa,mxb)
mm=max(mxa,mxb)
ga=[0]*(m+1)
gb=[0]*(m+1)
ga[1]=na
gb[1]=nb
for i in range(2,m+1):
for j in range(i,mm+1,i):
if j in sa:
ga[i]+=1
if j in sb:
gb[i]+=1
gf=[-1]*(m+1)
f(1)
r=sum([(x>0) for x in gf])
print(r)
View More Similar Problems
Compare two linked lists
You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis
View Solution →Merge two sorted linked lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →Get Node Value
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t
View Solution →Delete duplicate-value nodes from a sorted linked list
This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -
View Solution →Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
View Solution →Find Merge Point of Two Lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share
View Solution →