DISTGCD-Editorial

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

Setter: Utkarsh Gupta
Tester: Lavish Gupta, Abhinav sharma, Takuki Kurokawa
Editorialist: Devendra Singh

DIFFICULTY:

1297

PREREQUISITES:

GCD, Euclidean algorithms, Calculation of divisors of a number

PROBLEM:

Chef has two distinct positive integers A and B.

Chef wonders how many distinct values are possible for the expression \mathrm{gcd}(A+X,B+X), where X can take any non-negative integer value.

Help Chef find this value.

Here, \mathrm{gcd} stands for Greatest Common Divisor.

QUICK EXPLANATION

The answer to the problem is number of divisors of abs(B-A).

EXPLANATION:

Let A \lt B ( If they are not just swap them as gcd(x,y)=gcd(y,x)).This means X+A \lt X+B.

Proof: If we subtract a smaller number from a larger (we reduce a larger number), GCD doesn’t change.

Let a and b be two numbers such that a\leq b. Let GCD of a and b be g.
Let a= x\cdot g and b=y\cdot g where y\geq x and x and y are coprime. gcd (a, b-a) = gcd(x\cdot g, (y-x)\cdot g) = g= gcd(a,b)

Thus, gcd(A+X,B+X) is same as gcd(A+X,B-A). Now, gcd(A+X,B-A) has to be a divisor of B-A. We can find all the divisors of B-A and each divisor is a possible value for the expression gcd(A+X,B+X)(=gcd(A+X,B-A)).

Let d be any divisor of B-A, Take X=(10^9+7)*d-A (10^9+7 is a prime number and can be replaced by other large prime numbers which are co prime with B-A) This gives us gcd(10^9+7*d-A+A ,B-A)= gcd(10^9+7*d ,B-A) = d.
Thus, each divisor is a possible value for the expression gcd(A+X,B+X)(=gcd(A+X,B-A)).

Calculation of all divisors of a number N can be done in O(\sqrt{N}) time.
The answer to the problem is number of divisors of abs(B-A).

TIME COMPLEXITY:

O(\sqrt{(abs(B-A))}) for each test case.

SOLUTION:

Setter's solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
using namespace std;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
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;
            }

            if(!(l <= x && x <= r))
            {
                cerr << l << ' ' << r << ' ' << x << '\n';
                assert(1 == 0);
            }

            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 solve()
{
    int a,b;
    a=readInt(1,1000000000,' ');
    b=readInt(1,1000000000,'\n');
    assert(a!=b);
    int d=abs(a-b);
    ll ans=0;
    for(ll i=1;i*i<=d;i++)
    {
        if((d%i)==0)
        {
            ans++;
            if((d/i)!=i)
                ans++;
        }
    }
    cout<<ans<<'\n';
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL),cout.tie(NULL);
    int T=readInt(1,1000,'\n');
    while(T--)
        solve();
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Editorialist's Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
ll INF = 1e18;
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void sol(void)
{
  int a, b, ans = 0;
  cin >> a >> b;
  if (a > b)
    swap(a, b);
  for (int i = 1; i * i <= (b - a); i++)
  {
    if ((b - a) % i == 0)
      ans++;
    if ((b - a) % i == 0 && (b - a) != (i * i))
      ans++;
  }
  cout << ans << '\n';
  return;
}
int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int test = 1;
  cin >> test;
  while (test--)
    sol();
}

3 Likes

Why my code not giving correct submission?
4/6 test case passed.

I have computed the total no. of divisors by using prime factorization method.

Here is my code:
https://www.codechef.com/viewsolution/66650884

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

int solve(int n)
{
    
    vector<int>v;
    for(int i=2 ; i*i<=n;i++)
    {
        if(n % i == 0)
        {
            int ct = 0;
            while(n % i == 0)
            {
                ct++;
                n = n/i;
            }
            v.push_back(ct+1);
        }
    }
    if(n > 2)
    v.push_back(2);
    
    int ans = 1;
    for(auto i : v)
    ans = ans * i;
    
    return ans;
}

int main() {
	// your code goes here
	int t;
	cin >> t;
	while(t--)
	{
	    int a , b , d;
	    cin >> a >> b;
	    d = abs(a - b);
	    cout << solve(d) << endl;
	}
	return 0;
}

@devendra7700