Between Two Sets
Problem Statement :
There will be two arrays of integers. Determine all integers that satisfy the following two conditions: The elements of the first array are all factors of the integer being considered The integer being considered is a factor of all elements of the second array These numbers are referred to as being between the two arrays. Determine how many such numbers exist. Example a=[2,4] b=[24,36] There are two numbers between the arrays: 6 and 12. 6%2=0, 6%6=0, 24%6=0 and 36%6=0 for the first value. 12%2=0, 12%6=0 and 24%12=0, 36%12=0 for the second value. Return 2. Function Description Complete the getTotalX function in the editor below. It should return the number of integers that are betwen the sets. getTotalX has the following parameter(s): int a[n]: an array of integers int b[m]: an array of integers Returns int: the number of integers that are between the sets Input Format The first line contains two space-separated integers, n and m, the number of elements in arrays a and b. The second line contains n distinct space-separated integers a[i] where 0 <= i < n. The third line contains m distinct space-separated integers b[j] where 0 <= j < m. Constraints 1 <= n,m <= 10 1 <= a[i] <= 100 1 <= b[j] <= 100
Solution :
Solution in C :
C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int n,i,min,max,j,c=0,f;
int m;
scanf("%d %d",&n,&m);
int *a = malloc(sizeof(int) * n);
for(int a_i = 0; a_i < n; a_i++){
scanf("%d",&a[a_i]);
}
int *b = malloc(sizeof(int) * m);
for(int b_i = 0; b_i < m; b_i++){
scanf("%d",&b[b_i]);
}
max=a[0];
for(i=0;i<n;i++)
{
if(a[i]>max)
max=a[i];
}
min=b[0];
for(i=0;i<m;i++)
{
if(b[i]<min)
min=b[i];
}
for(i=max;i<=min;i++)
{
f=0;
for(j=0;j<n;j++)
{
if(i%a[j]!=0)
{
f=1;
break;
}
}
if(f==0)
{
for(j=0;j<m;j++)
{
if(b[j]%i!=0)
{
f=1;
break;
}
}
}
if(f==0)
c++;
}
printf("%d",c);
return 0;
}
C++ :
#include <iostream>
#include <vector>
#include <cmath>
#include <ctime>
#include <cassert>
#include <cstdio>
#include <queue>
#include <set>
#include <map>
#include <fstream>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#include <numeric>
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
using namespace std;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.precision(10);
cout << fixed;
#ifdef LOCAL_DEFINE
freopen("input.txt", "rt", stdin);
#endif
int n, m;
cin >> n >> m;
vi a(n); forn(i, n) cin >> a[i];
vi b(m); forn(i, m) cin >> b[i];
int ans = 0;
for1(k, 100) {
bool ok = true;
for (int x: a) ok &= k % x == 0;
for (int x: b) ok &= x % k == 0;
if (ok) ++ans;
}
cout << ans << '\n';
#ifdef LOCAL_DEFINE
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
return 0;
}
Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
int[] a = new int[n];
for(int a_i=0; a_i < n; a_i++){
a[a_i] = in.nextInt();
}
int[] b = new int[m];
for(int b_i=0; b_i < m; b_i++){
b[b_i] = in.nextInt();
}
Arrays.sort(a);
Arrays.sort(b);
int count=0;
for (int i=a[n-1]; i<=b[0]; i++){
boolean flag=true;
for (int j=0; j<n; j++){
if (i%a[j]!=0){
flag=false;
break;
}
}
if (flag==true){
for (int j=0; j<m; j++){
if (b[j]%i!=0){
flag=false;
break;
}
}
}
if (flag==true)
count++;
}
System.out.println (count);
}
}
python3 :
if __name__=='__main__':
n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
ans = 0
for i in range(1,101):
flag = True
for j in a:
if i%j!=0:
flag = False
break
if flag:
for k in b:
if k%i!=0:
flag = False
break
if flag:
ans+=1
print(ans)
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