The product mystery - Codechef
Problem Statement :
Given two positive numbers B and C, what is the minimum positive value of A, such that A⋅B is divisible by C. Here, A⋅B denotes the value obtained when A is multiplied by B. Input Format The first line will contain an integer T - number of test cases. Then the test cases follow. The first and only line of each test case contains two integers B and C. Output Format For each test case, output the minimum value of A such that A⋅B is divisible by C. Constraints 1≤T≤105 1≤B,C≤109 Sample Input 1 2 2 4 8 12 Sample Output 1 2 3
Solution :
Solution in C++ :
#include<bits/stdc++.h>
#include <iterator>
#include <iostream>
#include <numeric>
#include <math.h>
#define ll long long
#define ull long
#define mpa make_pair
#define pb push_back
#define ff first
#define pii pair<ll,ll>
#define dd long double
#define trace(x) cerr << #x << " : " << x << endl
#define ss second
#define boost ios_base::sync_with_stdio(0)
#define forp(i,a,b) for(ll i=a;i<=b;i++)
#define rep(i,n) for(ll i=0;i<n;++i)
#define ren(i,n) for(ll i=n-1;i>=0;i--)
#define forn(i,a,b) for(ll i=a;i>=b;i--)
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(typeof((c).begin()) i = (c).begin(); i != (c).end();
#define sc(x) scanf("%lld",&x)
#define clr(x,val) memset(x,val,sizeof(x))
#define pr(x) printf("%lld\n",x)
#define gc getchar
#define pdd pair<dd,dd>
#define prec(x) cout<<fixed<<setprecision(x)
#define fre freopen("input.txt","r",stdin),freopen("output.txt","w",stdout)
#define arr array
using namespace std;
ll a[200005];
void solve(ll tc){
ll b,c;
cin>>b>>c;
ll gc=__gcd(b,c);
cout<<c/gc<<endl;
}
int main(){
boost;
//pre_cum();
//prec(20);
//fre;
ll t=1;
ll tc=1;
cin>>t;
while(t--){
solve(tc);
tc++;
}
return 0;
}
Solution in Java :
import java.util.*;
import java.io.*;
public class Main {
static long startTime = System.currentTimeMillis();
// for global initializations and methods starts here
// global initialisations and methods end here
static void run() {
boolean tc = true;
AdityaFastIO r = new AdityaFastIO();
//FastReader r = new FastReader();
try (OutputStream out = new BufferedOutputStream(System.out)) {
//long startTime = System.currentTimeMillis();
int testcases = tc ? r.ni() : 1;
int tcCounter = 1;
// Hold Here Sparky------------------->>>
// Solution Starts Here
start:
while (testcases-- > 0) {
// a*b = c
long b = r.nl();
long c = r.nl();
out.write((c / gcd(b, c) + " ").getBytes());
out.write(("\n").getBytes());
}
// Solution Ends Here
} catch (IOException e) {
e.printStackTrace();
}
}
static class AdityaFastIO {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public BufferedReader br;
public StringTokenizer st;
public AdityaFastIO() {
br = new BufferedReader(new InputStreamReader(System.in));
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public AdityaFastIO(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public String readLine() throws IOException {
byte[] buf = new byte[100000001]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int ni() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg) return -ret;
return ret;
}
public long nl() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg) return -ret;
return ret;
}
public double nd() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.') while ((c = read()) >= '0' && c <= '9') ret += (c - '0') / (div *= 10);
if (neg) return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null) return;
din.close();
}
}
public static void main(String[] args) throws Exception {
run();
}
static int[] readIntArr(int n, AdityaFastIO r) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = r.ni();
return arr;
}
static long[] readLongArr(int n, AdityaFastIO r) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++) arr[i] = r.nl();
return arr;
}
static List<Integer> readIntList(int n, AdityaFastIO r) throws IOException {
List<Integer> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.ni());
return al;
}
static List<Long> readLongList(int n, AdityaFastIO r) throws IOException {
List<Long> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.nl());
return al;
}
static long mod = 998244353;
static long modInv(long base, long e) {
long result = 1;
base %= mod;
while (e > 0) {
if ((e & 1) > 0) result = result * base % mod;
base = base * base % mod;
e >>= 1;
}
return result;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int ni() {
return Integer.parseInt(word());
}
long nl() {
return Long.parseLong(word());
}
double nd() {
return Double.parseDouble(word());
}
}
static int MOD = (int) (1e9 + 7);
static long powerLL(long x, long n) {
long result = 1;
while (n > 0) {
if (n % 2 == 1) result = result * x % MOD;
n = n / 2;
x = x * x % MOD;
}
return result;
}
static long powerStrings(int i1, int i2) {
String sa = String.valueOf(i1);
String sb = String.valueOf(i2);
long a = 0, b = 0;
for (int i = 0; i < sa.length(); i++) a = (a * 10 + (sa.charAt(i) - '0')) % MOD;
for (int i = 0; i < sb.length(); i++) b = (b * 10 + (sb.charAt(i) - '0')) % (MOD - 1);
return powerLL(a, b);
}
static long gcd(long a, long b) {
if (a == 0) return b;
else return gcd(b % a, a);
}
static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
static long lower_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] >= x) r = m;
else l = m;
}
return r;
}
static int upper_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] <= x) l = m;
else r = m;
}
return l + 1;
}
static void addEdge(ArrayList<ArrayList<Integer>> graph, int edge1, int edge2) {
graph.get(edge1).add(edge2);
graph.get(edge2).add(edge1);
}
public static class Pair implements Comparable<Pair> {
int first;
int second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(Pair o) {
// TODO Auto-generated method stub
if (this.first != o.first)
return (int) (this.first - o.first);
else return (int) (this.second - o.second);
}
}
public static class PairC<X, Y> implements Comparable<PairC> {
X first;
Y second;
public PairC(X first, Y second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(PairC o) {
// TODO Auto-generated method stub
return o.compareTo((PairC) first);
}
}
static boolean isCollectionsSorted(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) <= list.get(i - 1)) return false;
return true;
}
static boolean isCollectionsSortedReverseOrder(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) >= list.get(i - 1)) return false;
return true;
}
}
Solution in Python :
import math
for _ in range(int(input())):
b, c = map(int, input().split())
print(c // math.gcd(b, c))
View More Similar Problems
AND xor OR
Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value
View Solution →Waiter
You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the
View Solution →Queue using Two Stacks
A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que
View Solution →Castle on the Grid
You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):
View Solution →Down to Zero II
You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.
View Solution →Truck Tour
Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr
View Solution →