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Author: satyam_343
Tester: apoorv_me
Editorialist: iceknight1093






Given an array A and an integer M, find the lexicographically largest array B whose elements are between 1 and M, and A_i = \gcd(B_1, B_2, B_3, \ldots, B_i) for every i.
It’s guaranteed that a solution exists for every given input.


First, recall that \gcd(B_1, B_2, B_3, \ldots, B_i) is always a multiple of \gcd(B_1, B_2, B_3, \ldots, B_i, B_{i+1}).
In other words, we know that input is such that A_i is a multiple of A_{i+1} for each i, since we’re told the answer exists. Keep this fact in mind.

Note that we have no choice for B_1: it must always be A_1 itself, since that’s the only way the condition on prefix gcd can be satisfied for the first index.
Now, let’s try to find the remaining elements.

We want the lexicographically maximum array, so it’s best to make a greedy choice at each step.
Suppose we’ve found the values of B_1, B_2, \ldots B_{i-1}, and we want to find B_i.

B_i should be chosen such that \gcd(B_1, B_2, \ldots, B_i) = A_i.
However, we already know that \gcd(B_1, B_2, \ldots, B_{i-1}) = A_{i-1}, since that’s how we constructed the first i-1 elements.
So, we just want to choose B_i such that \gcd(A_{i-1}, B_i) = A_i. In particular, B_i should be a multiple of A_i.
Recall that we already know that A_{i-1} is a multiple of A_i, so we just need to choose B_i to be a multiple of A_i that’s as large as possible; yet also shares no other factors with A_{i-1} than A_i.

Clearly, the absolute best we can do is to choose A_i\cdot \left\lfloor\frac{M}{A_i}\right\rfloor, i.e, the largest multiple of A_i that’s \leq M.
However, this might not always work - we might have its GCD with A_{i-1} be some larger multiple of A_i.
In such a case, we can just bruteforce to find the next best thing!
That is, while \gcd(B_i, A_{i-1}) \neq A_i, keep reducing B_i by A_i.

This is clearly a correct solution, the only concern is speed.
And indeed, this is fast enough - very fast, actually.


Let A_{i-1} = x\cdot A_i and B_i = y\cdot A_i.
Then, \gcd(A_{i-1}, B_i) = A_i\cdot \gcd(x, y), so our objective is to ensure that \gcd(x, y) = 1.

Suppose we take K steps to find an answer.
That would mean that \gcd(x, y-i) \gt 1 for all 0 \leq i \lt K; in particular, x must share a prime factor with every integer in that range.

Now, x \leq 10^9 means it has at most 9 distinct prime factors.
Further, till 10^9, the gap between consecutive primes is at most about 300.
So, after 300\cdot 10 = 3000 steps, we’ll surely find a prime number that’s not a factor of x and be done.
In practice, the number of steps required will be far far less than 3000 because primes just aren’t that spaced out (also, large prime gaps only happen for larger primes, and x can’t have many large primes in its factorization) - I would expect the actual number of steps to be of the order \mathcal{O}(\log{10^9}).

The 3000 upper bound is good enough though, because we don’t actually trigger it at each index - it’s only done when A_i \neq A_{i-1} which happens \log( A_1) times at most, since the element (at least) halves each time we move to a strictly smaller one.


\mathcal{O}(N + 3000) GCD operations per testcase (in practice, the 3000 is far smaller and more like 50).


Author's code (C++)
#pragma GCC optimod_intze("O3,unroll-loops")
#include <bits/stdc++.h>   
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;   
using namespace std;
#define ll long long
const ll INF_MUL=1e15;
const ll INF_ADD=1e18;
#define pb push_back               
#define mp make_pair          
#define nline "\n"                           
#define f first                                          
#define s second                                             
#define pll pair<ll,ll> 
#define all(x) x.begin(),x.end()     
#define vl vector<ll>           
#define vvl vector<vector<ll>>    
#define vvvl vector<vector<vector<ll>>>          
#ifndef ONLINE_JUDGE    
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#define debug(x);  
void _print(ll x){cerr<<x;}  
void _print(char x){cerr<<x;}   
void _print(string x){cerr<<x;}    
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());   
template<class T,class V> void _print(pair<T,V> p) {cerr<<"{"; _print(p.first);cerr<<","; _print(p.second);cerr<<"}";}
template<class T>void _print(vector<T> v) {cerr<<" [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T>void _print(set<T> v) {cerr<<" [ "; for (T i:v){_print(i); cerr<<" ";}cerr<<"]";}
template<class T>void _print(multiset<T> v) {cerr<< " [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T,class V>void _print(map<T, V> v) {cerr<<" [ "; for(auto i:v) {_print(i);cerr<<" ";} cerr<<"]";} 
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef tree<pair<ll,ll>, null_type, less<pair<ll,ll>>, rb_tree_tag, tree_order_statistics_node_update> ordered_pset;
const ll MOD=998244353;
const ll MAX=500500;
void solve(){    
    ll n,m; cin>>n>>m; 
    vector<ll> a(n+5,0);
    for(ll i=1;i<=n;i++){
    vector<ll> ans(n+5,0);
    for(ll i=2;i<=n;i++){
        ll ops=0;
    ll use=0;
    for(ll i=1;i<=n;i++){
        cout<<ans[i]<<" \n"[i==n];
int main()                                                                               
    #ifndef ONLINE_JUDGE                 
    freopen("input.txt", "r", stdin);                                           
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                        
    ll test_cases=1;               
Tester's code (C++)
#ifndef LOCAL
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")

#include <bits/stdc++.h>
using namespace std;

#ifdef LOCAL
#include "../debug.h"
#define dbg(...) "11-111"

struct input_checker {
	string buffer;
	int pos;

	const string all = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
	const string number = "0123456789";
	const string lower = "abcdefghijklmnopqrstuvwxyz";

	input_checker() {
		pos = 0;
		while (true) {
			int c = cin.get();
			if (c == -1) {
			buffer.push_back((char) c);

	int nextDelimiter() {
		int now = pos;
		while (now < (int) buffer.size() && buffer[now] != ' ' && buffer[now] != '\n') {
		return now;

	string readOne() {
		assert(pos < (int) buffer.size());
		int nxt = nextDelimiter();
		string res;
		while (pos < nxt) {
			res += buffer[pos];
		return res;

	string readString(int minl, int maxl, const string &pattern = "") {
		assert(minl <= maxl);
		string res = readOne();
		assert(minl <= (int) res.size());
		assert((int) res.size() <= maxl);
		for (int i = 0; i < (int) res.size(); i++) {
			assert(pattern.empty() || pattern.find(res[i]) != string::npos);
		return res;

	int readInt(int minv, int maxv) {
		assert(minv <= maxv);
		int res = stoi(readOne());
		assert(minv <= res);
		assert(res <= maxv);
		return res;

	long long readLong(long long minv, long long maxv) {
		assert(minv <= maxv);
		long long res = stoll(readOne());
		assert(minv <= res);
		assert(res <= maxv);
		return res;

	auto readInts(int n, int minv, int maxv) {
		assert(n >= 0);
		vector<int> v(n);
		for (int i = 0; i < n; ++i) {
			v[i] = readInt(minv, maxv);
			if (i+1 < n) readSpace();
		return v;

	auto readLongs(int n, long long minv, long long maxv) {
		assert(n >= 0);
		vector<long long> v(n);
		for (int i = 0; i < n; ++i) {
			v[i] = readLong(minv, maxv);
			if (i+1 < n) readSpace();
		return v;

	void readSpace() {
		assert((int) buffer.size() > pos);
		assert(buffer[pos] == ' ');

	void readEoln() {
		assert((int) buffer.size() > pos);
		assert(buffer[pos] == '\n');

	void readEof() {
		assert((int) buffer.size() == pos);

int32_t main() {
    ios_base::sync_with_stdio(0);   cin.tie(0);

    input_checker input;
    int T = input.readInt(1, (int)1e4); input.readEoln();
    int sum_N = 0;
    while(T-- > 0) {
        int n = input.readInt(1, (int)1e4); input.readSpace();
        int m = input.readInt(1, (int)1e9); input.readEoln();
        vector<int> a = input.readInts(n, 1, m); input.readEoln();
        auto b = a;
        for(int i = 1 ; i < n ; i++) {
            b[i] = (m / a[i]) * a[i];
            while(__gcd(a[i - 1], b[i]) > a[i])
                b[i] -= a[i];

        for(int i = 0 ; i < n ; i++)
            cout << b[i] << " \n"[i == n - 1];
    assert(sum_N <= (int)5e4);

    return 0;
Editorialist's code (Python)
from math import gcd
for _ in range(int(input())):
    n, m = map(int, input().split())
    a = list(map(int, input().split()))
    ans = [0]*n
    ans[0] = a[0]
    for i in range(1, n):
        ans[i] = a[i]*(m//a[i])
        while gcd(ans[i], a[i-1]) != a[i]: ans[i] -= a[i]

Clearly, the absolute best we can do is to choose Ai⋅floor(Ai/M)
can you explain this ?
and i think it will be floor(M/Ai) ?

Now, x≤10^0 means it has at most 9 distinct factors.
Here you are taking about prime factors?

Screenshot 2023-10-19 055745

why we are reducing by Ai only ?

You’re right, it is \left\lfloor\frac{M}{A_i}\right\rfloor
I thought that I’d fixed that typo but it looks like it wasn’t saved, should be fixed now.


Because we’re looking for multiples of A_i - a couple of lines above that, I’ve explained why B_i should be a multiple of A_i.
So, we start out at the largest possible multiple of A_i and check if it works.
If it doesn’t, you move to the next largest multiple, which is obtained by just subtracting A_i.
If you prefer algebra, you’re moving from A_i\cdot x to A_i\cdot (x-1) = A_i\cdot x - A_i.

How did you arrive at this conclusion of maximum limit we need to traverse to find the element not sharing any prime factors as 300*(9+1) ? I dont understand the intuition behind this

As mentioned there, the maximum number of times you decrement till you find a prime number is bounded by about 300 (see prime gap).
Also mentioned, x has at most 9 distinct prime factors, because the product of the smallest 10 primes exceeds 10^9.

So, if you take 3000 steps, you’ll surely reach at least 10 distinct primes yeah? One each 300 steps, at the very least.
Of them, at least one is guaranteed to not be a factor of x (since it has only 9 prime factors at most); and if a prime p is not a factor of x then \gcd(x, p) = 1 for sure (recall that our objective is to find a coprime number, after all).

Thankyou so much. Understood it well. Appreciate it :slight_smile:

Thank you sir

I had used a similar approach but instead of checking gcd(B_i, A_{i-1})!=A_i I was checking gcd(B_i, r)!=1 where r = A_{i-1}/A_{i}. I was facing TLE in this case. Can someone help me figure out the reason for that.

Could you tell me a case where this fails? Thanks :smile:

#define int long long
using namespace std;

void solve()
	int n,m; cin >> n >> m;
	int a[n];
	for(int i=0;i<n;i++) cin >> a[i];
	int b[n];
	b[0] = a[0];
	for(int i=0;i<n-1;i++)
		int rem = m % a[i + 1];
		int max_possible = m - rem;
		if(a[i] == a[i + 1])
			b[i + 1] = max_possible;
			if(max_possible % a[i] == 0) max_possible -= a[i + 1];
			b[i + 1] = max_possible;
	for(int &u : b) cout << u << " ";
	cout << '\n';

int32_t main()
	int t = 1; 
	cin >> t;