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();
}