CHFGCD - Editorial

PROBLEM LINK:

Practice
Contest: Division 3
Contest: Division 2
Contest: Division 1

Author: Souradeep
Tester: Aryan Choudhary
Editorialist: Vichitr Gandas

DIFFICULTY:

CAKEWALK

PREREQUISITES:

GCD

PROBLEM:

Given two positive integers X and Y. Now perform some operations (possibly zero) on them. In each operation, choose X or Y and add 1 to it. Find the minimum number of operations such that the greatest common divisor of X and Y is greater than 1.

EXPLANATION

We can make gcd > 1 in at most 2 operations by making both numbers even, if they are odd and hence gcd \ge 2. But its possible that we are able to get it with less operations.
So check

  • If gcd(A,B)>1, then 0 operations needed.
  • Else if gcd(A+1, B) > 1 or gcd(A, B+1)>1 then 1 operation needed.
  • Otherwise 2 operations needed.

TIME COMPLEXITY:

O(\log{(min(A,B))}) per test case

SOLUTIONS:

Setter's Solution
#include <bits/stdc++.h>
using namespace std;
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long
 
const int N = 1005;

int32_t main()
{
	IOS;
	int t;
	cin >> t;
	while(t--)
	{
		int x, y;
		cin >> x >> y;
		if(__gcd(x, y) > 1)
			cout << 0 << endl;
		else
		{
			if(__gcd(x + 1, y) > 1 || __gcd(x, y + 1) > 1)
				cout << 1 << endl;
			else
				cout << 2 << endl;
		}
	}
	return 0;
}
Tester's Solution
/* in the name of Anton */

/*
  Compete against Yourself.
  Author - Aryan (@aryanc403)
  Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
    #include <header.h>
#else
    #pragma GCC optimize ("Ofast")
    #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
    //#pragma GCC optimize ("-ffloat-store")
    #include<bits/stdc++.h>
    #define dbg(args...) 42;
#endif

using namespace std;
#define fo(i,n)   for(i=0;i<(n);++i)
#define repA(i,j,n)   for(i=(j);i<=(n);++i)
#define repD(i,j,n)   for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
    cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}

long long readInt(long long l, long long r, char endd) {
    long long x=0;
    int cnt=0;
    int fi=-1;
    bool is_neg=false;
    while(true) {
        char g=getchar();
        if(g=='-') {
            assert(fi==-1);
            is_neg=true;
            continue;
        }
        if('0'<=g&&g<='9') {
            x*=10;
            x+=g-'0';
            if(cnt==0) {
                fi=g-'0';
            }
            cnt++;
            assert(fi!=0 || cnt==1);
            assert(fi!=0 || is_neg==false);

            assert(!(cnt>19 || ( cnt==19 && fi>1) ));
        } else if(g==endd) {
            if(is_neg) {
                x=-x;
            }
            assert(l<=x&&x<=r);
            return x;
        } else {
            assert(false);
        }
    }
}
string readString(int l, int r, char endd) {
    string ret="";
    int cnt=0;
    while(true) {
        char g=getchar();
        assert(g!=-1);
        if(g==endd) {
            break;
        }
        cnt++;
        ret+=g;
    }
    assert(l<=cnt&&cnt<=r);
    return ret;
}
long long readIntSp(long long l, long long r) {
    return readInt(l,r,' ');
}
long long readIntLn(long long l, long long r) {
    return readInt(l,r,'\n');
}
string readStringLn(int l, int r) {
    return readString(l,r,'\n');
}
string readStringSp(int l, int r) {
    return readString(l,r,' ');
}

void readEOF(){
    assert(getchar()==EOF);
}

vi readVectorInt(int n,lli l,lli r){
    vi a(n);
    for(int i=0;i<n-1;++i)
        a[i]=readIntSp(l,r);
    a[n-1]=readIntLn(l,r);
    return a;
}

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
mt19937 rng(seed=chrono::steady_clock::now().time_since_epoch().count());
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}

class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{    return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y ));   }};

void add( map<lli,lli> &m, lli x,lli cnt=1)
{
    auto jt=m.find(x);
    if(jt==m.end())         m.insert({x,cnt});
    else                    jt->Y+=cnt;
}

void del( map<lli,lli> &m, lli x,lli cnt=1)
{
    auto jt=m.find(x);
    if(jt->Y<=cnt)            m.erase(jt);
    else                      jt->Y-=cnt;
}

bool cmp(const ii &a,const ii &b)
{
    return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}

const lli mod = 1000000007L;
// const lli maxN = 1000000007L;

    lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
    lli m;
    string s;
    vi a;
    //priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .

int main(void) {
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    // freopen("txt.in", "r", stdin);
    // freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
T=readIntLn(1,1e5);
while(T--)
{

    const lli x=readIntSp(1,1e9),y=readIntLn(1,1e9);
    if(__gcd(x,y)>1){
        cout<<0<<endl;
        continue;
    }

    if(__gcd(x+1,y)>1||__gcd(x,y+1)>1){
        cout<<1<<endl;
        continue;
    }
    cout<<2<<endl;
}   aryanc403();
    readEOF();
    return 0;
}

Editorialist's Solution
/*
 * @author: vichitr
 * @date: 25th July 2021
 */

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define fast ios::sync_with_stdio(0); cin.tie(0);

void solve() {
	int A, B; cin >> A >> B;
	int ans;
	if (__gcd(A, B) > 1)
		ans = 0;
	else if (__gcd(A + 1, B) > 1 or __gcd(A, B + 1) > 1)
		ans = 1;
	else
		ans = 2;
	cout << ans << '\n';
}

signed main() {
	fast;

#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif

	int t = 1;
	cin >> t;
	for (int tt = 1; tt <= t; tt++) {
		// cout << "Case #" << tt << ": ";
		solve();
	}
	return 0;
}

If you have other approaches or solutions, let’s discuss in comments.If you have other approaches or solutions, let’s discuss in comments.

2 Likes

What if the input is 17 29?
It will need 12 operations on 17

I’ve tried the same approach and i was not able to get AC, can anyone tell me what am i doing wrong.
link to code here

the output will either be 0, 1, 2 only as we try to make these numbers even and the minimum gcd, even numbers can have is 2 .

The goal is to make both numbers, even so that their gcd>1, here 17+1=18 & 29+1=30, total 2 operations.

Oh, got it.
I interpreted the question wrong.

1 Like

why is this logic wrong ?
ll gcd(ll a,ll b)
{
if(a==0)
return b;
return gcd(b%a,a);
}

int main()
{
fio
wit{
ll x,y;
cin>>x>>y;
if(gcd(x,y)>1)
cout<<“0”<<endl;
else if(x%2==0||(y%2==0))
{
cout<<“1”<<endl;
}
else
{ cout<<“2”<<endl;}

}
return 0; 

}

Half of the code is missing, so I’ll use Solution: 49213027 | CodeChef, in which case: consider the test input:

1
3 9
1 Like

OMG this was a bad mistake , thanks for the help.

1 Like

Why 7 27 output is 2 instead of 1?
gcd(7, 28) is 7 > 1

It should be 1 - where are you seeing that the output should be 2?

sorry my bad :upside_down_face:

1 Like

though used the same approach but could not get ac. can any one explain what was my mistake?
my code - Fj10CB - Online C++0x Compiler & Debugging Tool - Ideone.com

It fails on the test input:

1
5 21

Try this
1
3 5

Output should be 1 but your code outputs 2

but my code is giving ans 2 for 5 and 21. isn’t correct?

No: 5+1=6, and \gcd(6,21)=3.

1 Like

Please Help. Why is this giving WA. Cant think of any case it will fail.
#include <bits/stdc++.h>
using namespace std;
#define ll long long

int main() {
int t;cin>>t;
while(t–)
{
ll x,y,c=0;cin>>x>>y;
while(__gcd(x,y)==1)
{
if(x<y)swap(x,y);
if(x%2==0)y++;
else x++;
c++;
}
cout<<c<<endl;
}
}

https://www.codechef.com/viewsolution/49217035
Pls help it’s giving me TLE