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.
